To solve the problem, it is necessary to apply the related concepts to Newton's second law as well as the Normal and Centripetal Force experienced by passengers.
By Newton's second law we understand that
[tex]F = mg[/tex]
Where,
m= mass
g = Gravitational Acceleration
Also we have that Frictional Force is given by
[tex]F_r = \mu N[/tex]
In this particular case the Normal Force N is equivalent to the centripetal Force then,
[tex]N = \frac{mv^2}{r}[/tex]
Applying this to the information given, and understanding that the Weight Force is statically equivalent to the Friction Force we have to
[tex]F = F_r[/tex]
[tex]mg = \mu N[/tex]
[tex]mg = \mu \frac{mv^2}{r}[/tex]
Re-arrange to find v,
[tex]v= \sqrt{\frac{gr}{\mu}}[/tex]
[tex]v = \sqrt{\frac{(9.8)(2)}{0.25}}[/tex]
[tex]v = 8.9m/s[/tex]
From the last expression we can realize that it does not depend on the mass of the passengers.
The minimum speed required for passengers to remain stuck to the wall of a cylindrical carnival ride with a radius of 2.0 meters and a coefficient of static friction of 0.25 is d) 8.9 m/s.
To determine the minimum speed required to prevent passengers from sliding down the wall of a cylindrical carnival ride with a 2.0-m radius, we can use the principles of centripetal force and friction.
The centripetal force needed to keep a rider in circular motion is provided by the normal force (N) which acts horizontally. This force is balanced by the frictional force (f) acting vertically upwards to counteract the gravitational force (mg) pulling the rider down.
The frictional force is given by:
f = μN
Where μ is the coefficient of static friction (0.25). The normal force is equivalent to the centripetal force needed for circular motion:
N = mv² / r
Thus, the frictional force equation becomes:
μ(mv² / r) = mg
Solving for v:
μv² / r = g
v² = rg / μ
[tex]v = \sqrt{\frac{rg}{\mu}}[/tex]
Substituting the given values (r = 2.0 m, g = 9.8 m/s², μ = 0.25):
[tex]v = \sqrt{\frac{2.0 \cdot 9.8}{0.25}}[/tex]
v = √(78.4)
v = 8.9 m/s
Therefore, the correct option is d) as the minimum speed required is 8.9 m/s.
The position vector of a particle of mass 2 kg is given as a function of time by ~r = (5 m) ˆı + (5 m/s)t ˆ . Determine the magnitude of the angular momentum of the particle with respect to the origin at time 7 s
The angular momentum of the particle with respect to the origin is 50 kgm²/s.
What is angular momentum?
The angular momentum of an object is the product of moment of inertia and angular velocity.
L = mvr
where;
m is the massv is the velocityr is the radiusr = 5i + 5tj
v = dr/dt
v = 5 m/s
L = m(v x r)
v x r = 5j x (5i + 5(7)j)
v x r = 5j x (5i + 35j)
v x r = -25k
|v x r| = 25
L = m(v x r)
L = 2 x 25
L = 50 kgm²/s
Thus, the angular momentum of the particle with respect to the origin is 50 kgm²/s.
Learn more about angular momentum here: https://brainly.com/question/7538238
Final answer:
The magnitude of the angular momentum of the particle with respect to the origin at time 7 s is 50 kg m²/s.
Explanation:
To find the magnitude of the angular momentum of a particle with a certain position vector as a function of time, we use the formula ℒ = r × p, where r is the position vector and p is the linear momentum vector (p = mv, with m as mass and v as velocity).
Given the position vector r(t) = (5 m)î + (5 m/s)tì for a mass of 2 kg, we first find the velocity vector by differentiating the position vector with respect to time, v(t) = dr/dt = 0î + (5 m/s)ì. The linear momentum at time t is p(t) = (2 kg)(v(t)) = (2 kg)(0î + (5 m/s)ì) = 0î + (10 kg m/s)ì.
To find the angular momentum at 7 s, we evaluate r at t = 7 s: r(7 s) = (5 m)î + (5 m/s)(7 s)ì = (5 m)î + (35 m)ì. Then, the angular momentum ℒ(7 s) = r(7 s) × p(7 s) = ((5 m)î + (35 m)ì) × (0î + (10 kg m/s)ì), which must be computed using the cross product. ℒ(7 s) ends up being ((5 m)(10 kg m/s))ê = (50 kg m²/s)ê. Therefore, the magnitude of the angular momentum is 50 kg m²/s.
The noble gases neon (atomic mass 20.1797 u) and krypton (atomic mass 83.798 u) are accidentally mixed in a vessel that has a temperature of 79.2°C. What are the average kinetic energies and rms speeds of neon and krypton molecules in the vessel?
(a) average kinetic energies Kav, Ne = J Kav, Kr = J
(b) rms speeds vrms, Ne = m/s vrms, Kr = m/s
Answer:
(a) Kav Ne = Kav Kr = 7.29x10⁻²¹J
(b) v(rms) Ne= 659.6m/s and v(rms) Kr= 323.7m/s
Explanation:
(a) According to the kinetic theory of gases the average kinetic energy of the gases can be calculated by:
[tex] K_{av} = \frac{3}{2}kT [/tex] (1)
where [tex] K_{av} [/tex]: is the kinetic energy, k: Boltzmann constant = 1.38x10⁻²³J/K, and T: is the temperature
From equation (1), we can calculate the average kinetic energies for the krypton and the neon:
[tex] K_{av} = \frac{3}{2} (1.38\cdot 10^{-23} \frac{J}{K})(352.2K) = 7.29\cdot 10^{-21}J [/tex]
(b) The rms speeds of the gases can be calculated by:
[tex] K_{av} = \frac{1}{2}mv_{rms}^{2} \rightarrow v_{rms} = \sqrt \frac{2K_{av}}{m} [/tex]
where m: is the mass of the gases and [tex]v_{rms}[/tex]: is the root mean square speed of the gases
For the neon:
[tex] v_{rms} = \sqrt \frac{2(7.29 \cdot 10^{-21}J)}{20.1797 \frac{g}{mol} \cdot \frac {1mol}{6.022\cdot 10^{23}molecules} \cdot \frac{1kg}{1000g}} = 659.6 \frac{m}{s} [/tex]
For the krypton:
[tex] v_{rms} = \sqrt \frac{2(7.29 \cdot 10^{-21}J)}{83.798 \frac{g}{mol} \cdot \frac {1mol}{6.022\cdot 10^{23}molecules} \cdot \frac{1kg}{1000g}} = 323.7 \frac{m}{s} [/tex]
Have a nice day!
The average kinetic energy and root mean square speed of both neon and krypton gases can be calculated using constants such as the Boltzmann constant, the molecular mass of the gases, and the temperature of the system in Kelvins.
Explanation:The average kinetic energy (Kav) of a gas molecule, irrespective of the type of gas, is given by the equation Kav = 3/2 kT, where k is Boltzmann's constant and T is the absolute temperature. The root mean square speed (vrms) of the molecules in a gas is given by vrms = sqrt(3kT/m), where m is the molecular mass of the gas.
The temperature should be converted to Kelvins by adding 273.15 to the Celsius temperature, hence T = 79.2°C + 273.15 = 352.35K.
For neon, the molecular mass, m = 20.1797 u. Plugging in these values, we can calculate the average kinetic energy and rms speed for neon molecules in the vessel. Similar calculations are done for krypton with m = 83.798 u.
Learn more about Kinetic Energy and Speed of Gases here:https://brainly.com/question/33824895
#SPJ11
A planet has two moons with identical mass. Moon 1 is in a circular orbit of radius r. Moon 2 is in a circular orbit of radius 2r. The magnitude of the gravitational force exerted by the planet on Moon 2 is which of the following compared with the gravitational force exerted by the planet on Moon 1?
half as largetwice as largeone-fourth as largefour times as largethe same
Answer:
Half as large.
Explanation:
Using Newton's law of universal gravitation, if the mass of the planet is M and of the Moons 1 and 2 is m, them the force exerted by the planet on them will be:
[tex]F_1=\frac{GMm}{r}[/tex]
[tex]F_2=\frac{GMm}{2r}[/tex]
Which clearly shows that the force that the planet exerts on the Moon 2 is half the force it exerts on the Moon 1.
The gravitational force exerted by the planet on Moon 2, which orbits twice as far from the planet as Moon 1, is one-fourth as large as the force exerted on Moon 1 due to the inverse-square law of gravity.
Explanation:This question is about the gravitational force that a planet exerts on its moons. Each moon has an identical mass but they differ in their distances from the planet - the radius of their orbits. One moon orbits at a radius of r, while the second moon orbits at a radius of 2r.
The gravitational force that a planet exerts on an object is inversely proportional to the square of the distance between the center of the planet and the object. So, if you double the distance, the gravitational force becomes one-fourth as large. Therefore, the magnitude of the gravitational force exerted by the planet on Moon 2, which is twice as far from the planet as Moon 1, is one-fourth as large as the gravitational force exerted by the planet on Moon 1.
Learn more about Gravitational Force here:https://brainly.com/question/32609171
#SPJ3
A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard ground, her stopping time would have been much shorter. Using the impulse-momentum theorem as your guide, determine which one of the following statements is correct.
a. the air mattress exerts the same impulse, but a greater net average force, on the high-jumper than does the hard groundb. the air mattress exerts a greater impulse, and a greater net average force, on the high-jumper than does the hard groundc. the air mattress exerts a smaller impulse, and a smaller net average force, on the high-jumper than does the hard groundd. the air mattress exerts a greater impulse, but a smaller net average force, on the high-jumper than does the hard grounde. the air mattress exerts the same impulse, but a smaller net avg force, on the hj than hg
Answer:
e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.
Explanation:
This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.
The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.
Mathematical expression for the Newton's second law of motion is given as:
[tex]F=\frac{dp}{dt}[/tex] ............................................(1)
where:
dp = change in momentum
dt = time taken to change the momentum
We know, momentum:
[tex]p=m.v[/tex]
Now, equation (1) becomes:
[tex]F=\frac{d(m.v)}{dt}[/tex]
∵mass is constant at speeds v << c (speed of light)
[tex]\therefore F=m.\frac{dv}{dt}[/tex]
and, [tex]\frac{dv}{dt} =a[/tex]
where: a = acceleration
[tex]\Rightarrow F=m.a[/tex]
also
[tex]F\propto \frac{1}{dt}[/tex]
so, more the time, lesser the force.
& Impulse:
[tex]I=F.dt[/tex]
[tex]I=m.a.dt[/tex]
[tex]I=m.\frac{dv}{dt}.dt[/tex]
[tex]I=m.dv=dp[/tex]
∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.
So, the impulse in both the cases will be same.
Final answer:
According to the impulse-momentum theorem, the air mattress exerts the same impulse as the hard ground on the high-jumper, but with a smaller net average force because the stopping time is extended on the air mattress.
Explanation:
The key to understanding this scenario lies in the impulse-momentum theorem, which states that the change in momentum (impulse) of an object is equal to the average net external force applied to it, times the duration of time this force acts upon the object.
In both cases, landing on the hard ground and landing on an air mattress, the change in momentum of the high-jumper is the same, since she comes to rest from the same initial velocity in both scenarios. Thus, the overall impulse received by the high-jumper must be the same in both situations. However, the difference lies in the time over which the stopping force is applied.
When landing on the air mattress, the stopping time is longer than on the hard ground. According to the impulse-momentum theorem (Ap = Fnet At), since the impulse (Ap) is the same but the time (t) is greater when landing on the air mattress, the average net force (Fnet) must be smaller in comparison with landing on the hard ground. Therefore, the correct statement is that the air mattress exerts the same impulse but a smaller net average force than the hard ground.
A landing craft with mass 1.21×104 kg is in a circular orbit a distance 5.90×105 m above the surface of a planet. The period of the orbit is 5900 s . The astronauts in the lander measure the diameter of the planet to be 9.80×106 m . The lander sets down at the north pole of the planet.
What is the weight of an astronaut of mass 84.6 kg as he steps out onto the planet's surface?
Answer:
W = 661.6 N
Explanation:
The weight of a body is the force of attraction of the plant on the body, so we must use the law of gravitational attraction
F = G m M / r²
Where G is the gravitational attraction constant that values 6.67 10-11 N m² / kg², M is the mass of the planet and r is the distance from the center of the planet.
Let's look for the mass of the planet, for this we write Newton's second law for the landing craft
F = m a
Acceleration is centripetal a = v² / r
G m M / r² = m (v² / r)
The ship rotates rapidly (constant velocity module), let's use uniform kinematic relationships
v = d / t
The distance of a circle is
d = 2π r
v = 2π r / t
We replace
G m M / r² = m (4π² r² / t² r)
G M = 4 π² r³ / t²
M = 4π² r³ / G t²
The measured distance r from the center of the plant is
r = R orbit + R planet
r = 5.90 10⁵ + ½ 9.80 10⁶
r = 5.49 10⁶ m
M = 4 π² (5.49 10⁶)³ / (6.67 10⁻¹¹ (5.900 10³)²)
M = 6,532 10²¹ / 2,321 10⁺³
M = 2.814 10²⁴ kg
With this data we calculate the astronaut's weight
W = (G M / R²) m
W = (6.67 10⁻¹¹ 2,816 10²⁴ /(4.90 10⁶)2) 84.6
W = 7.82 84.6
W = 661.57 N
To calculate the weight of the astronaut on the surface of the planet, we first need to find out the gravitational acceleration at the planet's surface using the parameters of the landing craft's orbit. After we have the value of gravitational acceleration, we multiply it by the mass of the astronaut.
Explanation:To answer your question, we first need to calculate the gravitational acceleration at the surface of the planet, denoted by g. We can use the orbital parameters of the landing craft to find g using the following formula where G represents the universal gravitation constant, M is the mass of the planet, and r is the radius of the planet:
g = GM/r^2
After we solve this, we can then calculate the weight of the astronaut on the surface using W = mg, where m is the mass of the astronaut.
Learn more about gravitational acceleration here:https://brainly.com/question/30429868
#SPJ11
At an amusement park, the wheelie carries passengers in a circular path of radius r = 11.2 m. If the angular speed of the wheelie is 0.550 revolutions/s, (a) What is the tangential velocity of the passengers due to the circular motion? (b) What is the acceleration of the passengers?
Answer:
(a) Tangential velocity will be 38.648 m/sec
(b) Acceleration will be [tex]133.617m/sec^2[/tex]
Explanation:
We have given radius r = 11.2 m
Angular speed [tex]\omega =0.550rev/sec=0.550\times 2\pi =3.454rad/sec[/tex]
(a) We have to find the tangential velocity
We know that tangential velocity is given by
[tex]v_t=\omega r=3.454\times 11.2=38.684m/sec[/tex]
(b) We know that acceleration is given by
[tex]a=\frac{v^2}{r}=\frac{38.684^2}{11.2}=133.617m/sec^2[/tex]
A runner whose mass is 54 kg accelerates from a stop to a speed of 7 m/s in 3 seconds. (A good sprinter can run 100 meters in about 10 seconds, with an average speed of 10 m/s.) (a) What is the average horizontal component of the force that the ground exerts on the runner's shoes? (b) How much work is done on the point-particle system by this force?
Answer:
a. [tex]F=126N[/tex]
b. [tex]E_K=1323J[/tex]
Explanation:
Given:
[tex]m=54kg[/tex]
[tex]v=7 m/s[/tex]
[tex]t= 3s[/tex]
The runner force average to find given the equations
a.
[tex]F=m*a[/tex]
[tex]a=\frac{v}{t}[/tex]
[tex]F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}[/tex]
[tex]F=126N[/tex]
b.
Work done by the system by this force so
[tex]W=F*d[/tex]
[tex]W=E_K[/tex]
[tex]E_K=\frac{1}{2}*m*v^2[/tex]
[tex]E_K=\frac{1}{2}*54kg*(7m/s)^2[/tex]
[tex]E_K=1323J[/tex]
(a) The average horizontal component of the force that the ground exerts on the runner's shoes is 126 N.
(b) The work done on the point-particle system by this force is 1,323 J.
Net force on the runner
The net force on the runner is determined from Newton's second law of motion as shown below.
Ff = ma
The average horizontal component of the force that the ground exerts on the runner's shoes is calculated as follows;
Ff = m(v/t)
Ff = 54 x (7/3)
Ff = 126 N
Work done by the force of frictionThe work done by frictional force is calculated as follows;
W = ¹/₂mv²
W = 0.5 x 54 x 7²
W = 1,323 J
Learn more about kinetic energy here: https://brainly.com/question/25959744
A reaction will be spontaneous only at low temperatures if both ΔH and ΔS are negative. For a reaction in which ΔH = −320.1 kJ/mol and ΔS = −98.00 J/K · mol, determine the temperature (in °C) below which the reaction is spontaneous.
Answer:
2993 °C
Explanation:
A reaction is spontaneous if the free Gibbs energy (ΔG) is negative. ΔG is related to the enthalpy of the reaction (ΔH) and the entropy of the reaction (ΔS) through the following expression.
ΔG = ΔH - T . ΔS
For 1 mol, if ΔG < 0, then
ΔH - T . ΔS < 0
ΔH < T . ΔS
-320.1 × 10³ J < T . (-98.00 J/K)
T < 3266 K
To convert Kelvin to Celsius we use the following expression.
K = °C + 273.15
°C = K - 273.15 = 3266 - 273.15 = 2993 °C
Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the angular acceleration when it has an angle of 10.0 degrees from the vertical? A typical pencil has an average length of 15.0 cm and an average mass of 10.0g. Calculate the time for the pencil to hit the ground, assuming that it falls from standing perfectly vertical and maintains this angular acceleration.
Answer:
α = 17 rad / s² , t = 0.4299 s
Explanation:
Let's use Newton's second angular law or torque to find angular acceleration
τ = I α
W r = I α
The weight is applied in the middle of the pencil,
sin 10 = r / (L/2)
r = L/2 sin 10
The pencil can be approximated by a bar that rotates on one end, in this case its moment of inertia is
I = 1/3 M L²
Let's calculate
mg L / 2 sin 10 = (1/3 m L²) α
α f = 3/2 g / L sin 10
α = 3/2 9.8 / 0.150 sin 10
α = 17 rad / s²
If the pencil has a constant acceleration we can use the angular kinematics relationships, as part of the rest wo = 0
θ = w₀ t + ½ α t²
t = RA (2θ / α )
The angle from the vertical to the ground is
θ = π / 2
t = √ (2 π / (2 α ))
t = √ (π / α )
t = √ (π / 17)
t = 0.4299 s
The time for the pencil to hit the ground if it falls from standing perfectly vertical and maintains the angular acceleration is 0.4299s.
How to calculate the time taken?Newton's second angular law will be used to find the angular acceleration. This will be:
τ = I α
Wr = I α
Here, the weight is applied in the middle of the pencil, therefore,
sin 10 = r / (L/2)
Make r the subject of the formula
r = L/2 sin 10
Then, the moment of inertia will be:
I = 1/3 M L²
mg L / 2 sin 10 = (1/3 m L²) α
αf = 3/2 g / L sin 10
α = [3/2 × 9.8] / [0.150 sin10]
α = 17 rad/s²
The angle from the vertical to the ground will be denoted as:
θ = π / 2
t = √ (2 π / (2 α ))
t = √(π / α )
t = √(π / 17)
t = √3.142 / 17
t = [tex]\sqrt{0.1848}[/tex]
t = 0.4299s
Therefore, the time for the pencil to hit the ground is 0.4299s.
Learn more about time on:
https://brainly.com/question/13893070
The water flowing through a 1.8 cm (inside diameter) pipe flows out through three 1.2 cm pipes. (a) If the flow rates in the three smaller pipes are 27, 19, and 12 L/min, what is the flow rate in the 1.8 cm pipe? (b) What is the ratio of the speed of water in the 1.8 cm pipe to that in the pipe carrying 27 L/min?
To solve this problem it is necessary to apply the Discharge of flow equations.
From the theory the flow rate is defined as
Q = AV
Where,
A =Area
V = Velocity
PART A) The question is telling us about the total fluid flow rate then
[tex]Q_T = Q_1+Q_2+Q_3[/tex]
[tex]Q_T = 27+19+12[/tex]
[tex]Q_T = 58L/min[/tex]
PART B) The radius would be given between another pipe with a flow rate of 27L / min.
For proportionality ratio we have to
[tex]\frac{Q_T}{Q'} = \frac{A_TV_T}{A'V'}[/tex]
[tex]\frac{V_T}{V'} = \frac{A_'Q_T}{A_TQ'}[/tex]
[tex]\frac{V_T}{V'} = \frac{(\pi r_T^2)Q_T}{(\pi r'^2)Q'}[/tex]
[tex]\frac{V_T}{V'} = \frac{1.2*^2 58}{1.8^2 27}[/tex]
[tex]\frac{V_T}{V'} = 0.9547[/tex]
An AC generator provides emf to a resistive load in a remote factory over a two-cable transmission line. At the factory a step-down transformer reduces the voltage from its transmission-line value Vt to a much lower value that is safe and convenient for use in the factory. The transmission line resistance is 0.3 Ω per cable, and the power of the generator is Pt = 250,000 watts (rms).1 If 80,000 volts Vt = (rms), what are:
(a) the voltage drop ∆V across the transmission line?
(b) the rate at which energy is dissipated in the line as thermal energy?
(c) If instead Vt were 8000 V (rms), what would be the voltage drop and power dissipated in the transmission line?
(d) Repeat part c for 800 volts Vt = (rms)
Answer:
a) 1.875 V b) 5.86 W c) 18.75 V 586 W d) 187.5 V 5.86 kW.
Explanation:
a) As the load is purely resistive, we can get the Irms, applying directly the definition of electrical power, as follows:
I = P / V = 250,000 W / 80,000 V = 3.125 A
Applying Ohm’s Law to the resistance of the conductors in the transmission line, we have:
∆V = I. R = 3.125 A . 0.6 Ω = 1.875 V
b) The rate at which energy is dissipated in the line as thermal energy, can be obtained applying Joule’s law, using the RMS value of the current that we have already got:
Pd = I2. R = (3.125)2 A . 0.6Ω = 5.86 W
c) If Vt were 8,000 V instead of 80,000 V, we would have a different value for I, as follows:
I= 250,000 W / 8,000 V = 31.25 A
So, we would have a different ∆V:
∆V = 31.25 A . 0.6 Ω = 18.75 V
As the RMS current is different, we will have a different value por the dissipated power in the line:
P = (31.25)2 . 0.6 Ω = 586 W
d) As above, we will have a new value for I, as follows:
I = 250,000 W / 800 V = 312. 5 A
The new voltage loss in the transmission line will be much larger:
∆V = 312.5 A . 0.6 Ω = 187.5 V
Consequently, we will have a higher dissipated power:
P = (312.5)2 . 0.6 Ω = 58.6 kW
To find the voltage drop across the transmission line and the power dissipated, we use Ohm's Law and the power equation. The voltage drop across the line is ∆V = I * R, and the power dissipated is Ploss = I² * R. Substituting the values, we can calculate the voltage drop and power dissipated for different values of Vt.
Explanation:To find the voltage drop across the transmission line, we can use Ohm's Law. The total resistance of the transmission line is 0.3 Ω per cable, so the resistance for the entire line is 0.6 Ω. The voltage drop is given by: ∆V = I * R, where I is the current flowing through the transmission line. Since power P = V * I, we can rewrite the equation as: Vt * I = Pt, which gives us I = Pt / Vt. Substituting the values, we get I = 250,000 watts / 80,000 volts = 3.125 A.
(a) The voltage drop ∆V across the transmission line is therefore: ∆V = I * R = 3.125 A * 0.6 Ω = 1.875 V.
(b) The rate at which energy is dissipated in the line as thermal energy is equal to the power loss due to resistance, given by: Ploss = I² * R = (3.125 A)² * 0.6 Ω = 5.859375 W.
(c) If Vt is 8000 V (rms), the current flowing through the transmission line is: I = Pt / Vt = 250,000 watts / 8000 volts = 31.25 A. The voltage drop across the line is then ∆V = I * R = 31.25 A * 0.6 Ω = 18.75 V. The power dissipated in the line is: Ploss = I² * R = (31.25 A)² * 0.6 Ω = 585.9375 W.
(d) If Vt is 800 V (rms), the current flowing through the transmission line is: I = Pt / Vt = 250,000 watts / 800 volts = 312.5 A. The voltage drop across the line is ∆V = I * R = 312.5 A * 0.6 Ω = 187.5 V. The power dissipated in the line is: Ploss = I² * R = (312.5 A)² * 0.6 Ω = 58593.75 W.
Learn more about Transmission line voltage drop and power dissipation here:https://brainly.com/question/34016783
#SPJ11
A box with a frontal area of 2.5 ft2 and with a drag coefficient of CD =0.9 is fastened on the roof of your car. What is the aerodynamic drag force in pounds due to the box alone at a velocity of 88 ft/sec?
To solve this problem it is necessary to apply the concepts related to the drag force.
By definition we know that drag force can be expressed as
[tex]F_D = \frac{1}{2} \rho C_D AV^2[/tex]
Where,
\rho = Density
[tex]C_D =[/tex]Drag Coefficient
A = Area
V = Velocity
Our values are given as
[tex]A = 2.5ft^2[/tex]
[tex]V = 88ft/s[/tex]
[tex]C_D = 0.9[/tex]
[tex]\rho = 0.0765lb/ft^3 \rightarrow[/tex] Air at normal temperature
Replacing at the equation we have,
[tex]F_D = \frac{1}{2} \rho C_D AV^2[/tex]
[tex]F_D = \frac{1}{2} (0.0765lb/ft^3) (0.9) (2.5ft^2) (88ft/s)^2[/tex]
[tex]F_D = 666.468lbf[/tex]
The aerodynamic drag force is 666.468Lbf
A 0.40-kg mass is attached to a spring with a force constant of k = 277 N/m, and the mass–spring system is set into oscillation with an amplitude of A = 3.0 cm. Determine the following. (a) mechanical energy of the system J (b) maximum speed of the oscillating mass m/s (c) magnitude of the maximum acceleration of the oscillating mass
To solve this problem it is necessary to apply the concepts related to the kinetic energy expressed in terms of simple harmonic movement, as well as the concepts related to angular velocity and acceleration and linear acceleration and velocity.
By definition we know that the angular velocity of a body can be described as a function of mass and spring constant as
[tex]\omega = \sqrt{\frac{k}{m}}[/tex]
Where,
k = Spring constant
m = mass
From the given values the angular velocity would be
[tex]\omega = \sqrt{\frac{277}{0.4}}[/tex]
[tex]\omega = 26.31rad/s[/tex]
The kinetic energy on its part is expressed as
[tex]E = \frac{1}{2} m\omega^2A^2[/tex]
Where,
A = Amplitude
[tex]\omega[/tex] = Angular Velocity
[tex]m = Mass[/tex]
PART A) Replacing previously given values the energy in the system would be
[tex]E = \frac{1}{2} m\omega^2A^2[/tex]
[tex]E = \frac{1}{2} (0.4)(26.31)^2(3*10^{-2})^2[/tex]
[tex]E= 0.1245J[/tex]
PART B) Through the amplitude and angular velocity it is possible to know the linear velocity by means of the relation
[tex]v = A\omega[/tex]
[tex]v = (3*10^{-2})(26.31)[/tex]
[tex]v = 0.7893m/s[/tex]
PART C) Finally, the relationship between linear acceleration and angular velocity is subject to
[tex]a = A\omega^2[/tex]
[tex]a = (3*10^{-2})(26.31)^2[/tex]
[tex]a = 20.76m/s^2[/tex]
(a) The mechanical energy of the system is 1.245 J (b) The maximum speed of the oscillating mass is 0.6225 m/s (c) The magnitude of the maximum acceleration of the oscillating mass is 20.775m/s².
(a) The mechanical energy of a mass†“spring system in simple harmonic motion (SHM) is given by the equation:
[tex]\[ E = \frac{1}{2}kA^2 \][/tex]
where [tex]$k$[/tex] is the spring constant and [tex]$A$[/tex]is the amplitude of oscillation. Plugging in the given values:
[tex]\[ E = \frac{1}{2} \times 277 \, \text{N/m} \times (0.03 \, \text{m})^2 \][/tex]
[tex]\[ E = \frac{1}{2} \times 277 \times 0.0009 \][/tex]
[tex]\[ E = 1.245 \, \text{J} \][/tex]
(b) The maximum speed of the mass occurs at the equilibrium position (when the displacement is zero) and is given by:
[tex]\[ v_{max} = A\sqrt{\frac{k}{m}} \][/tex]
where[tex]$m$[/tex] is the mass of the object. Using the given values:
[tex]\[ v_{max} = 0.03 \, \text{m} \times \sqrt{\frac{277 \, \text{N/m}}{0.40 \, \text{kg}}} \][/tex]
[tex]\[ v_{max} = 0.03 \times \sqrt{692.5} \][/tex]
[tex]\[ v_{max} = 0.03 \times 27.0676 \][/tex]
[tex]\[ v_{max} = 0.6225 \, \text{m/s} \][/tex]
(c) The maximum acceleration of the mass occurs at the maximum displacement (amplitude) and is given by:
[tex]\[ a_{max} = \frac{F_{max}}{m} \][/tex]
where [tex]$F_{max}$[/tex]is the maximum force exerted by the spring, which is [tex]$kA$[/tex]. Thus:
[tex]\[ a_{max} = \frac{kA}{m} \][/tex]
Using the given values:
[tex]\[ a_{max} = \frac{277 \, \text{N/m} \ times 0.03 \, \text{m}}{0.40 \, \text{kg}} \][/tex]
[tex]\[ a_{max} = \frac{8.31}{0.40} \][/tex]
[tex]\[ a_{max} = 20.775 \, \text{m/s}^2 \][/tex]
A woman walks into a carpet store wearing high-heeled shoes with a circular heel of diameter 0.987 cm. To the dismay of the store manager, she balances on one heel on an expensive carpet sample. If she has a mass of 53.0 kg, determine the pressure she exerts on the carpet sample.
To solve this problem it is necessary to apply the concepts related to the Pressure which describes the amount of Force made on an area unit.
Its mathematical expression can be defined as
[tex]P = \frac{F}{A}[/tex]
Where,
F= Force
A = Area
Our values are given as
[tex]d = 0.987*10^{-2}m[/tex]
With the diameter we can find the Area of the circular heel, that is
[tex]A = \pi (\frac{0.987*10^{-2}}{2})^2[/tex]
[tex]A = 7.6511*10^{-5}m^2[/tex]
To find the force by weight we need to apply the Newton's second Law
[tex]F = mg[/tex]
[tex]F = 53*9.8[/tex]
[tex]F = 519.4N[/tex]
Finally the pressure would be
[tex]P = \frac{F}{A}[/tex]
[tex]P = \frac{519.4}{7.6511*10^{-5}}[/tex]
[tex]P = 6788566.35Pa[/tex]
[tex]P = 6.788MPa[/tex]
Therefore the pressure that she exerts on the carpet sample is 6.788Mpa
Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes have a large drag coefficient, by design. One model expands to a rectangle 1.8 m square, with a drag coefficient of 1.4. A runner completes a 200 m run at 5.0 m/s with this chute trailing behind. How much thermal energy is added to the air by the drag force?
To develop this problem it is necessary to apply the concepts related to the Aerodynamic Drag Force.
By definition the Drag Force is defined as
[tex]F_D = \frac{1}{2} \rho A C_D v^2[/tex]
Where,
A = Area
[tex]\rho[/tex]= Density
[tex]C_d[/tex] = Drag coefficient
v = Velocity
According to our values we have,
[tex]A = 1.8*1.8=3.24m^2[/tex]
[tex]C_D = 1.4[/tex]
[tex]V = 5m/s[/tex]
Replacing we have
[tex]F_D = \frac{1}{2} 1.23*3.24*1.4*5^2[/tex]
[tex]F_D = 69.74[/tex]
By definition we know that the thermal energy is given by the force applied in a given displacement then
[tex]W = F*d[/tex]
[tex]W = 69.74*200[/tex]
[tex]W = 13948J[/tex]
Therefore the thermal energy is added to the air by the drag force is 13.9kJ
The thermal energy added to the air by the drag force from a runner's parachute can be calculated by determining the work done against this force during a run. This involves first calculating the drag force with known parameters, then using the force to calculate the work done over the run's distance. This quantity represents the thermal energy imparted to the air.
Explanation:The calculation for thermal energy added to the air by the drag force requires understanding the basic principles of work and energy. The work done by the drag force is given by the formula W = Fd, where W is work, F is force, and d is distance. In this case, the force is the drag force internalized by the parachute. The drag force is calculated using the equation F = 0.5 * p * v² * C * A, where:
By substituting these values into the formula, we can compute the drag force. Thereafter, we can use this force in our initial work equation to calculate the work done over a 200 m run. It's important to note that the work done against the drag force is transformed into heat (thermal energy). Thus, the work done equals the thermal energy added to the air by the drag force.
Learn more about Thermal Energy from Drag Force here:https://brainly.com/question/2505689
#SPJ3
(b) Write an expression for the net external torque acting on the system of the two blocks + wheel + string about the wheel's axle. What are its rotation sense and vector direction? [NOTE: The string is part of the system, so the string tension forces provide internal torques, not external torques.]
Answer:
τext =(M*R*g)*(m₂ - m₁) / (m₁ + m₂)
The wheel is rotating clockwise. Its vector direction is - k which is perpendicular to the plane of the wheel.
Explanation:
Given info
m₁ = mass of block 1
m₂ = mass of block 2
m₁ < m₂
M = Mass of the wheel
R = Radius of the wheel
I = Moment of inertia of the wheel = M*R²
We assume that the wheel is rotating clockwise since m₁ < m₂
In order to get an expression for the net external torque acting on the system we can apply the following equation
τext = I*α
where α is the angular acceleration of the wheel which can be found as follows
at = R*α ⇒ α = at / R
then we have to compute the acceleration of the system (at), using the Newton's 2nd Law
Block 1:
∑Fy = m₁*at (↑+) ⇒ T - m₁*g = m₁*at (I)
Block 2:
∑Fy = m₂*at (↓+) ⇒ - T + m₂*g = m₂*at (II)
If we apply
(I) + (II) ⇒ at = g*(m₂ - m₁) / (m₁ + m₂)
Now, we get α:
α = at / R = (g*(m₂ - m₁) / (m₁ + m₂)) / R
⇒ α = (g / R)*(m₂ - m₁) / (m₁ + m₂)
Finally
τext = I*α = (M*R²)*((g / R)*(m₂ - m₁) / (m₁ + m₂))
⇒ τext =(M*R*g)*(m₂ - m₁) / (m₁ + m₂)
The wheel is rotating clockwise. Its vector direction is - k which is perpendicular to the plane of the wheel.
guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the guitar, there is a distance of 70 cm between fixed end points. The guitar string is tightened to a tension of 255 N.
What is the frequency of the fundamental wave on the guitar string?
Answer:
Fundamental frequency= 174.5 hz
Explanation:
We know
fundamental frequency=[tex]\frac{velocity}{2 *length}[/tex]
velocity =[tex]\sqrt{\frac{tension}{mass per unit length} }[/tex]
mass per unit length=[tex]\frac{3.5}{1000*1.22}[/tex]=0.00427[tex]\frac{kg}{m}[/tex]
Now calculating velocity v=[tex]\sqrt{\frac{255}{0.00427} }[/tex]
=244.3[tex]\frac{m}{sec}[/tex]
Distance between two nodes is 0.7 m.
Plugging these values into to calculate frequency
f = [tex]\frac{244.3}{2 *0.7}[/tex] =174.5 hz
A typical meteor that hits the earth's upper atmosphere has a mass of only 2.5 g, about the same as a penny, but it is moving at an impressive 40 km/s. As the meteor slows, the resulting thermal energy makes a glowing streak across the sky, a shooting star. The small mass packs a surprising punch.
At what speed would a 900 kg compact car need to move to have the same kinetic energy?
Answer:
Answer:u=66.67 m/s
Explanation:
Given
mass of meteor m=2.5 gm\approx 2.5\times 10^{-3} kg
velocity of meteor v=40km/s \approx 40000 m/s
Kinetic Energy of Meteor
K.E.=\frac{mv^2}{2}
K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}
K.E.=2\times 10^6 J
Kinetic Energy of Car
=\frac{1}{2}\times Mu^2
=\frac{1}{2}\times 900\times u^2
\frac{1}{2}\times 900\times u^2=2\times 10^6
900\times u^2=4\times 10^6
u^2=\frac{4}{9}\times 10^4
u=\frac{2}{3}\times 10^2
u=66.67 m/s
A hollow ball with a diameter of 3.72 cm has an average density of 0.0841 g/cm3. What force must be applied to hold it stationary while it is completely submerged under water? (Enter the magnitude in N, and select the direction from the options given.)
To calculate the force required to hold a hollow ball stationary while it is completely submerged underwater, we can use Archimedes' principle. The buoyant force on the ball is equal to the weight of the water it displaces.
Explanation:To calculate the force required to hold a hollow ball stationary while it is completely submerged underwater, we can use Archimedes' principle. The buoyant force on the ball is equal to the weight of the water it displaces. Since the ball is completely submerged, it displaces a volume of water equal to its own volume. The formula to calculate the force is:
Force = Density of water × Volume of ball × Acceleration due to gravity
Using the given diameter of the ball, we can calculate its volume by using the formula for the volume of a sphere: Volume = (4/3) × π × (radius)3
Once we have the volume, we can substitute it into the formula to find the force required.
Learn more about Calculating force on a submerged object here:https://brainly.com/question/15076913
#SPJ3
A barge floating in fresh water (rho = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and height H = 2.0 m. When empty the bottom of the barge is located H0 = 0.45 m below the surface of the water. When fully loaded with coal the bottom of the barge is located H1 = 1.8 m below the surface. a) Write an equation for the buoyant force on the empty barge in terms of the known data.
b) Determine the mass of the barge in kilograms.
c) Find the mass of the coal in terms of the given data.
d) Find the mass of the coal in kilograms.
e) How far would the barge be submerged (in meters) if mc,2 = 250000 kg of coal had been placed in the empty barge?
This answer provides the equations and steps to calculate the buoyant force, the mass of the barge, the mass of the coal, and how far the barge would be submerged.
Explanation:a) The equation for the buoyant force on the empty barge can be written as:
FB = ρgV
where FB is the buoyant force, ρ is the density of water, g is the acceleration due to gravity, and V is the volume of the barge submerged in water.
b) To determine the mass of the barge, we can use the equation:
FB = mg
where FB is the buoyant force, m is the mass of the barge, and g is the acceleration due to gravity.
c) The mass of the coal can be found by subtracting the mass of the empty barge from the mass of the fully loaded barge.
d) To find the mass of the coal in kilograms, you can use the equation:
mc = ρcVc
where mc is the mass of the coal, ρc is the density of the coal, and Vc is the volume of the coal.
e) To calculate how far the barge would be submerged with 250,000 kg of coal, you can use the equation:
FB = mg
and solve for H, where H is the height the barge is submerged.
Learn more about Calculating buoyant force here:https://brainly.com/question/30637119
#SPJ3
A platinum sphere with radius 0.0135 0.0135 m is totally immersed in mercury. Find the weight of the sphere, the buoyant force acting on the sphere, and the sphere's apparent weight. The densities of platinum and mercury are 2.14 × 10 4 2.14×104 kg/m3 and 1.36 × 10 4 1.36×104 kg/m3, respectively.
Answer:
W=2.2 N
F=1.4 N
W'=0.8 N
Explanation:
Given that
Radius ,r = 0.0135 m
Density of the platinum ,ρ₁ = 2.14 x 10⁴ kg/m³
Density of the mercury ,ρ₂ = 1.36 x 10⁴ kg/m³
The weight of the sphere
W= m g
mass = m = volume x density
[tex]m=\dfrac{4}{3}\pi r^3\times \rho_1\ kg[/tex]
[tex]m=\dfrac{4}{3}\times \pi\times 0.0135^3\times 2.14\times 10^4\ kg[/tex]
m = 0.22 kg
W= 0.22 x 10 = 2.2 N (↓) ( take g =10 m/s²)
The buoyant force
[tex]F= \dfrac{4}{3}\pi r^3\times \rho_2\times g[/tex]
[tex]F=\dfrac{4}{3}\times \pi\times 0.0135^3\times 1.36\times 10^4\times 10[/tex]
F= 1.4 N (↑)
The apparent weight
W' = 2.2 - 1.4 N
W'= 0.8 N
Atoms in a solid are not stationary, but vibrate about their equilibrium positions. Typically, the frequency of vibration is about f = 4.11 x 1012 Hz, and the amplitude is about 1.23 x 10^-11 m. For a typical atom, what is its (a) maximum speed and (b) maximum acceleration
To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.
Our values are given as
[tex]f = 4.11 *10^{12} Hz[/tex]
[tex]A = 1.23 * 10^{-11}m[/tex]
The angular velocity of a body can be described as a function of frequency as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi 4.11 *10^{12}[/tex]
[tex]\omega=2.582*10^{13} rad/s[/tex]
PART A) The expression for the maximum angular velocity is given by the amplitude so that
[tex]V = A\omega[/tex]
[tex]V =( 1.23 * 10^{-11})(2.582*10^{13})[/tex]
[tex]V = = 317.586m/s[/tex]
PART B) The maximum acceleration on your part would be given by the expression
[tex]a = A \omega^2[/tex]
[tex]a =( 1.23 * 10^{-11})(2.582*10^{13})^2[/tex]
[tex]a= 8.2*10^{15}m/s^2[/tex]
Particle 1 has mass 4.6 kg and is on the x-axis at x = 5.7 m. Particle 2 has mass 7.2 kg and is on the y-axis at y = 4.2 m. Particle 3 has mass 2.6 kg and is at the origin. What is the direction of the net gravitational force on particle 3, measured as a counterclockwise angle from the +x axis?
To solve this problem it is necessary to apply the concepts related to the Gravitational Force, for this purpose it is understood that the gravitational force is described as
[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]
Where,
G = Gravitational Universal Force
[tex]m_i =[/tex] Mass of each object
To solve this problem it is necessary to divide the gravitational force (x, y) into the required components and then use the tangent to find the angle generated between both components.
Our values are given as,
[tex]m_1 =4.6 kg\\m_2 = 7.2 kg\\m_3 = 2.6 kg\\r_1 = 5.7 m\\r_2 = 4.2 m[/tex]
Applying the previous equation at X-Axis,
[tex]F_x = \frac{Gm_1m_3}{R_{1}^2}\\F_x = \frac{6.67*10^{-11}*4.6*2.6}{5.7^2}\\F_x = 2.46*10^{-11}N[/tex]
Applying the previous equation at Y-Axis,
[tex]F_y = \frac{Gm_2m_3}{R_2^2}\\F_y = \frac{6.67*10^{-11}*7.2*2.6}{4.2^2}\\F_y = 7.08*10^{-11} N[/tex]
Therefore the angle can be calculated as,
[tex]tan\theta = \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{7.08*10^{-11}}{2.46*10^{-11}}\\\theta = 71\°[/tex]
Then in the measure contrary to the hands of the clock the Force in the particle 3 is in between the positive direction of the X and the negative direction of the Y at 71 ° from the positive x-axis.
A 2.16 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 110 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 11.2 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?
Answer:
0.15694 m
Explanation:
m = Mass of block = [tex]2.16\times 10^{-2}\ kg[/tex]
v = Velocity of block = 11.2 m/s
k = Spring constant = 110 N/m
Here the kinetic energy of the fall and spring are conserved
[tex]\frac{1}{2}mv^2=\frac{1}{2}kA^2\\\Rightarrow mv^2=kA^2\\\Rightarrow A=\sqrt{\frac{mv^2}{k}}\\\Rightarrow A=\sqrt{\frac{2.16\times 10^{-2}\times 11.2^2}{110}}\\\Rightarrow A=0.15694\ m[/tex]
The amplitude of the resulting simple harmonic motion is 0.15694 m
Distance Using Hubble's Law II.Find the percent difference (% = |A-O|/A × 100) between the actual (A) distance values and calculated (O) distance values using the recessional velocities for the Virgo and Corona Borealis clusters (vVirgo = 1,200 km/s, dVirgo = 17 Mpc; vCorona Borealis = 22,000 km/s, dCorona Borealis = 310 Mpc). Use H = 70 km/s/Mpc.the percent difference for the Virgo = _________%the percent difference for the Corona Borealis =_________ %At which distance, the closer or further one, is Hubble's law more accurate for the objects? Closer or further
Answer:
Explanation:
Let us first calculate for Virgo
[tex]A= d_{virgo} = 17;M_{pc}, V_{virgo} = 1200 km/s, [/tex]
Using Hubble's law
[tex]v = H_{0}D[/tex] For Virgo
[tex]V_{virgo} = H_{0}D_{virgo}[/tex]
[tex]O = D_{virgo} = \frac{v_{virgo}}{H_{0}} = \frac{1200 km/s}{70 km/s/Mpc}= 17.143 Mpc [/tex]
Percentage difference for the Virgo
[tex]\% = \frac{|A-O|}{A}\times 100 = \frac{|17 Mpc-17.143 Mpc|}{17 Mpc}\times 100 = 0.84 \% [/tex]
Now for calculate for Corona Borealis
[tex]A= d_{Corona Borealis} = 310 Mpc, v_{Corona Borealis} = 22000 km/s, [/tex]
Using Hubble's law
[tex]v = H_{0}D[/tex] For Corona Borealis
[tex]v_{Corona Borealis } = H_{0}D_{Corona Borealis } \\O = D_{Corona Borealis } = \frac{v_{Corona Borealis }}{H_{0}} = \frac{22000 km/s}{70 km/s/Mpc}= 314.286 Mpc [/tex]
Percentage difference for the Virgo
[tex]\% = \frac{|A-O|}{A}\times 100 = \frac{|310 Mpc-314.286 Mpc|}{310 Mpc}\times 100 = 1.3825 \% [/tex]
So clearly Hubble's law is more accurate for the closer objects
Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After she has descended a vertical distance of 5.00 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 15.0 m?
Answer:
Final velocity at the bottom of hill is 15.56 m/s.
Explanation:
The given problem can be divided into four parts:
1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)
From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:
[tex]p_i = p_f[/tex]
[tex]m_1u_1 + m_2v_2 = (m_1 + m_2)v[/tex]
[tex]v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)} [/tex]
[tex]v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s[/tex]
2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.
The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:
[tex]E(i) = E(f)[/tex]
[tex]KE(i) + PE(i) = KE(f) + PE(f)[/tex]
[tex]0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)[/tex]
[tex]v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}[/tex]
Here, initial velocity is the final velocity from the first stage. Therefore:
[tex]v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s[/tex]
3. Use conservation of momentum to find the combined speed of Gayle and her brother.
Given:
Initial velocity of Gayle and sled is, [tex]u_1(i)=10.5[/tex] m/s
Initial velocity of her brother is, [tex]u_2(i)=0[/tex] m/s
Mass of Gayle and sled is, [tex]m_1=55.0[/tex] kg
Mass of her brother is, [tex]m_2=30.0[/tex] kg
Final combined velocity is given as:
[tex]v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}[/tex]
[tex]v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79 [/tex] m/s
4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.
Using conservation of energy, the final velocity at the bottom of the hill is:
[tex]E(i) = E(f)[/tex]
[tex]KE(i) + PE(i) = KE(f) + PE(f)[/tex]
[tex]0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)[/tex]
[tex]v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s[/tex]
Using principles of kinetic and potential energy, and assuming ideal conditions devoid of friction and air resistance, Gayle and her brother's final speed would be equal to the square root of twice the product of acceleration due to gravity and the total height of the hill.
Explanation:This question relates to the principles of kinetic energy and potential energy, as well as conservation of energy. When Gayle initially sleds down the hill, she converts potential energy (mgh) into kinetic energy (1/2 mv^2). Taking 'g' as acceleration due to gravity (9.8 m/s^2) and 'h' as the vertical height she descends (5.0 m), the speed she attains at the time her brother joins (v') can be calculated using √(2gh).
The potential energy at that point is their combined mass times gravity and their height, which is transformed to kinetic energy at the bottom. If 'H' is the total height of the hill (15.0 m), their final combined speed is given by √(2gH). Ignoring resistive forces, they maintain this speed for the entire ride down the hill.
Learn more about Conservation of Energy here:https://brainly.com/question/35373077
#SPJ11
Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the current in the other is 3.45 A.
Find the magnitude of the force per unit length that one wire exerts on the other.
The force per unit length between the two wires is [tex]6.0\cdot 10^{-5} N/m[/tex]
Explanation:
The magnitude of the force per unit length exerted between two current-carrying wires is given by
[tex]\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}[/tex]
where
[tex]\mu_0 = 4\pi \cdot 10^{-7} Tm/A[/tex] is the vacuum permeability
[tex]I_1, I_2[/tex] are the currents in the two wires
r is the separation between the two wires
For the wires in this problem, we have
[tex]I_1 = 1.75 A[/tex]
[tex]I_2 = 3.45 A[/tex]
r = 2.00 cm = 0.02 m
Substituting into the equation, we find
[tex]\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m[/tex]
Learn more about current and magnetic fields:
brainly.com/question/4438943
brainly.com/question/10597501
brainly.com/question/12246020
brainly.com/question/3874443
brainly.com/question/4240735
#LearnwithBrainly
Light striking a metal surface causes electrons to be emitted from the metal via the photoelectric effect.In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the metal are held constant. Assuming that the light incident on the metal surface causes electrons to be ejected from the metal, what happens if the intensity of the incident light is increased?Check all that apply.A. The work function of the metal decreases.B. The number of electrons emitted from the metal per second increases.C. The maximum speed of the emitted electrons increases.D. The stopping potential increases.
The number of electrons emitted from the metal per second increases if the intensity of the incident light is increased.
Answer: Option B
Explanation:
As a result of photoelectric effect, electrons are emitted by the light incident on a metal surface. The emitted electrons count and its kinetic energy can measure as the function of light intensity and frequency. Like physicists, at the 20th century beginning, it should be expected that the light wave's energy (its intensity) will be transformed into the kinetic energy of emitted electrons.
In addition, the electrons count emitting from metal must vary with light wave frequency. This frequency relationship was expected because the electric field oscillates due to the light wave and the metal electrons react to different frequencies. In other words, the number of electrons emitted was expected to be frequency dependent and their kinetic energy should be dependent on the intensity (constant wavelength) of light.
Thus, the maximum in kinetic energy of electrons emitted increases with increase in light's frequency and is experimentally independent of light intensity. So, the number of emitted electrons is proportionate to the intensity of the incident light.
Final answer:
If the intensity of the incident light on a metal surface in a photoelectric effect experiment is increased, while keeping frequency constant, the number of electrons emitted per second increases. This happens because a higher intensity means more photons are available to eject electrons. Neither the work function of the metal, the maximum speed of the emitted electrons, nor the stopping potential are affected by changes in light intensity.
Explanation:
The question relates to the effect of increasing the intensity of light in a photoelectric effect experiment while keeping the frequency of the incident light and the temperature of the metal constant. In this setup:
B. The number of electrons emitted from the metal per second increases. This is because the intensity of light corresponds to the number of photons striking the surface per unit time, and a higher intensity means more photons are available to eject electrons.It is important to note that:
The work function of the metal does not change with light intensity, as it is an inherent property of the metal.The maximum speed of the emitted electrons and the stopping potential are determined by the energy of the individual photons (which is related to frequency), not the overall intensity of the light.A 75 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it is descending vertically with a speed of 24 m/s. If the player was moving upward with a speed of 4.0 m/s just before impact, what will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic?
Answer:
31.66 m/s
Explanation:
mass of player, M = 75 kg
mass of ball, m = 0.45 kg
initial velocity of player, U = + 4 m/s
initial velocity of ball, u = - 24 m/s
Let the final speed of player is V and the ball is v.
use conservation of momentum
Momentum before collision = momentum after collision
75 x 4 - 0.45 x 24 = 75 x V + 0.45 x v
289.2 = 75 V + 0.45 v .... (1)
As the collision is perfectly elastic, coefficient of restitution,e = 1
So, [tex]e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}[/tex]
V - v = u - U
V - v = -24 - 4 = - 28
V = v - 28, put this value in equation (1), we get
289.2 = 75 (v - 28) + 045 v
289.2 = 75 v - 2100 + 0.45 v
2389.2 = 75.45 v
v = 31.66 m/s
Thus, the velocity of ball after collision is 31.66 m/.
A hunter is aiming horizontally at a monkey who is sitting in a tree. The monkey is so terrified when it sees the gun that it falls off the tree. At that very instant, the hunter pulls the trigger. What will happen?
a. The bullet will miss the monkey because although both the monkey and the bullet are falling downward due to gravity, the monkey is falling faster.
b. The bullet will miss the monkey because the monkey falls down while the bullet speeds straight forward.
c. The bullet will hit the monkey because both the monkey and the bullet are falling downward at the same rate due to gravity.
d. It depends on how far the hunter is from the monkey.
Answer:
C. The bullet will hit the monkey because both the monkey and the bullet are falling downward at the same rate due to gravity.
Explanation:
As we know that acceleration due to gravity is equal g m/s².The acceleration due to gravity is in downward direction always and the numerical value is equal to g =10 m/s².
Both the bullet and money moving downward with constant acceleration that is why bullet will hit the money.
Therefore the answer is C.