Answer:
Specific Gravity = m/[m(s)-m(os)]
Explanation:
Specific gravity, also called relative density, is the ratio of the density of a substance to the density of a reference substance. By this definition we need to find out the ratio of density of the object of mass m to the density of the surrounding liquid.
m = mass of the object
Weight in air
W (air) = mg, where g is the gravitational acceleration
Weight with submerged with only one mass
m(s)g + Fb = mg + m(b)g, consider this to be equation 1
where Fb is the buoyancy force
Weight with submerged with both masses
m(os)g + Fb’ = mg + m(b)g, consider this to be equation 2
equation 1 – equation 2 would give us
m(s)g – m(os)g = Fb’ – Fb
where Fb = D x V x g, where D is the density of the liquid the object is submerged in, g is the force of gravity and V is the submerged volume of the object
m(s)g – m(os)g = D(l) x V x g
m(s) – m(os) = D(l) x V
we know that Mass = Density x V, which in our case would be, D(b) x V, which also means
V = Mass/D(b), where D(b) is the density of the mass
Substituting V into the above equation we get
m(s) – m(os) = [D(l) x m)/ D(b)]
Rearranging to get the ratio of density of object to the density of liquid
D(b)/D(l) = m/[m(s)-m(os)], where D(b)/D(l) denotes the specific gravity
The wheel of a car has a radius of 20.0 cm. It initially rotates at 120 rpm. In the next minute it makes 90.0 revolutions. (a) What is the angular acceleration? (b) How much further does the car travel before coming to rest? There is no slipping.
Answer:
Explanation:
Given that,
Radius of the wheel, r = 20 cm = 0.2 m
Initial speed of the wheel, [tex]\omega_i=120\ rpm=753.98\ rad/s[/tex]
Displacement, [tex]\theta=90\ rev=565.48\ rad[/tex]
To find,
The angular acceleration and the distance covered by the car.
Solution,
Let [tex]\alpha[/tex] is the angular acceleration of the car. Using equation of rotational kinematics as :
[tex]\theta=\omega_i t+\dfrac{1}{2}\alpha t^2[/tex]
[tex]565.48=753.98\times 60+\dfrac{1}{2}\alpha (60)^2[/tex]
[tex]\alpha =-24.81\ rad/s^2[/tex]
Let t is the time taken by the car before coming to rest.
[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }[/tex]
[tex]t=\dfrac{0-753.98}{-24.81}[/tex]
t = 30.39 seconds
Let v is the linear velocity of the car. So,
[tex]v=r\times \omega_i[/tex]
[tex]v=0.2\times 753.98[/tex]
v = 150.79 m/s
Let d is the distance covered by the car. It can be calculated as :
[tex]d=v\times t[/tex]
[tex]d=150.79\ m/s\times 30.39\ s[/tex]
d = 4582.5 meters
or
d = 4.58 km
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 cheerleader lifts his 43.8 kg partner straight up off the ground a distance of 0.737 m before releasing her. Assume the partner’s velocity is zero at the beginning and the end of the lift. The acceleration of gravity is 9.8 m/s 2 . If he does this 27 times, how much work has he done? Answer in units of J.
Answer:
The cheerleader does a total work of 8,532 J (8,541 J without any intermediate rounding)
Explanation:
Hi there!
The equation of work is the following:
W = F · d
Where:
W = work.
F = applied force.
d = distance.
The cheerleader´s partner is on the ground due to the force of gravity acting on her, also known as weight. To lift the partner at a constant velocity, the cheerleader has to suppress this force applying on the partner a force that is equal to the partner´s weight but in opposite direction.
The weight of the partner is calculated as follows:
Weight = m · g
Where m is the mass of the partner and g is the acceleration due to gravity.
Let´s calculate the weight of the partner:
Weight = 43.8 kg · 9.8 m/s²
Weight = 429 N
]Then, the work done in one lift by the cheerleader is the following:
W = F · d
W = 429 N · 0.737 = 316 J
If the cheerleader does 27 lifts, the work done will be (316 J · 27) 8,532 J
The work done by the cheerleader to lift his partner 27 times is calculated as the product of force (which itself is the product of the partner's mass and gravity) and distance, hence resulting in a total work done of 8570.61 Joules.
Explanation:In Physics, energy exerted to move an object is often calculated as work done. Work is defined as the product of the force applied to an object and the distance over which it is applied, and its unit of measurement is Joules (J). Regarding the given problem, the work done by the cheerleader to lift his partner is calculated using the formula: Work = force x distance, where force is derived from the product of the partner's mass and gravity (Force = mass x gravity). So, the work done for one lift would be calculated as follows:
Work(std.) = Force * distance = mass * gravity * distance = 43.8 kg * 9.8 m/s ² * 0.737 m = 317.43 J.
To find out the total work done after 27 lifts, simply multiply this result by 27: Total Work = 27 * Work(std.) = 27 * 317.43 J = 8570.61 J. Thus, the cheerleader does 8570.61 Joules of work by lifting his partner 27 times.
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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 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
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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.
Two long, straight wires are parallel and 16 cm apart. One carries a current of 2.9 A, the other a current of 5.7 A. (a) If the two currents flow in opposite directions, what is the magnitude (in N/m) and direction of the force per unit length of one wire on the other?
Explanation:
Given that,
Distance between two long wires, d = 16 cm = 0.16 m
Current in one wire, [tex]I_1=2.9\ A[/tex]
Current in wire 2, [tex]I_2=5.7\ A[/tex]
The magnetic force per unit length of one wire on the other is given by the following expression as :
[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi d}[/tex]
[tex]\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.9\times 5.7}{2\pi \times 0.16}[/tex]
[tex]\dfrac{F}{l}=2.06\times 10^{-5}\ N/m[/tex]
The current is flowing in opposite direction, the magnetic force acting on it is repulsive. Hence, this is the required solution.
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.
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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.
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.
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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
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.
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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.
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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
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.
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(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.
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.
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A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.65, and the coefficient of kinetic friction is 0.35. What horizontal force must be exerted on the box for it to accelerate at 1.2 m/s2?
A) 60 N
B) 230 N
C) 120 N
D) 170 N
E) 490 N
To solve the exercise it is necessary to apply the concepts related to Newton's Second Law, as well as the definition of Weight and Friction Force.
According to the problem there is a movement in the body and it is necessary to make a sum of forces on it, so that
[tex]\sum F = ma[/tex]
There are two forces acting on the body, the Force that is pushing and the opposing force that is that of friction, that is
[tex]F - F_f = ma[/tex]
To find the required force then,
[tex]F=F_f+ma[/tex]
By definition we know that the friction force is equal to the multiplication between the friction coefficient and the weight, that is to say
[tex]F = \mu mg +ma[/tex]
[tex]F = 0.35*50*0.8+50*1.2[/tex]
[tex]F=(171.5N)+(50Kg)(1.2m/s^2)[/tex]
[tex]F=231.5N[/tex]
[tex]F\approx 230N[/tex]
Therefore the horizontal force applied on the block is B) 230N
To find the horizontal force necessary to accelerate the box at 1.2 m/s^2, we consider the frictional forces. The maximum static friction can be calculated using the coefficient of static friction and the weight of the box. The horizontal force required is equal to the maximum static friction.
Explanation:To find the horizontal force required to accelerate the 50-kg box at 1.2 m/s^2, we need to consider the frictional forces acting on the box. First, we calculate the maximum static friction using the formula:
Maximum static friction = coefficient of static friction × normal force
Normal force = mass × gravity
Maximum static friction = 0.65 × (50 kg × 9.8 m/s^2)
Horizontal force = maximum static friction = 0.65 × (50 kg × 9.8 m/s^2)
So, the horizontal force needed is approximately 318.5 N.
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A single-turn circular loop of wire of radius 5.0 cm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.05480.0548-\text{s}s time interval, the magnitude of the field increases uniformly from 200 to 300 mT. Determine the magnitude of the emf induced in the loop.
Answer:
Induced EMF,[tex]\epsilon=0.0143\ volts[/tex]
Explanation:
Given that,
Radius of the circular loop, r = 5 cm = 0.05 m
Time, t = 0.0548 s
Initial magnetic field, [tex]B_i=200\ mT=0.2\ T[/tex]
Final magnetic field, [tex]B_f=300\ mT=0.3\ T[/tex]
The expression for the induced emf is given by :
[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]
[tex]\phi[/tex] = magnetic flux
[tex]\epsilon=\dfrac{d(BA)}{dt}[/tex]
[tex]\epsilon=A\dfrac{d(B)}{dt}[/tex]
[tex]\epsilon=A\dfrac{B_f-B_i}{t}[/tex]
[tex]\epsilon=\pi (0.05)^2\times \dfrac{0.3-0.2}{0.0548}[/tex]
[tex]\epsilon=0.0143\ volts[/tex]
So, the induced emf in the loop is 0.0143 volts. Hence, this is the required solution.
Final answer:
The magnitude of the induced emf in the loop is -0.0184 V.
Explanation:
The magnitude of the emf induced in a single-turn circular loop of wire can be calculated using Faraday's law of electromagnetic induction. According to the law, the magnitude of the induced emf is equal to the rate of change of magnetic flux through the loop. The formula for calculating the induced emf is:
emf = -N * ΔB/Δt
Where emf is the induced electromotive force, N is the number of turns in the loop, ΔB is the change in magnetic field strength, and Δt is the change in time. Plugging in the given values, we have:
emf = -1 * (300 - 200) * 10^-3 / 0.0548
emf = -0.0184 V
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
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]
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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.
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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]
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
What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze on an effective area of 51 cm2? Assume the pressure in the balloon is one atmosphere before you squeeze.
To create a gauge pressure of 62.5 cm H₂O on an area of 51 cm², you must exert a force of approximately 31.27 newtons on the balloon. This is obtained by converting the pressure to pascals, the area to square meters and using the formula Pressure = Force/Area.
To find the force you must exert on the balloon to create a gauge pressure of 62.5 cm H₂O when squeezing an area of 51 cm², you can use the relationship between pressure, force, and area given by the formula Pressure (P) = Force (F) / Area (A). First, convert the gauge pressure from cm H₂O to pascals (Pa) since 1 cm H₂O is approximately 98.0665 Pa. Then multiply the converted pressure by the area in m² to find the force in newtons (N).
Step 1: Convert cm H₂O to pascals (Pa):
62.5 cm H₂O * 98.0665 Pa/cm H₂O = 6131.65 Pa
Step 2: Convert the area from cm²to m² (since there are 10,000 cm2 in 1 m²):
51 cm² / 10,000 = 0.0051 m²
Step 3: Calculate the force:
Force (F) = Pressure (P) * Area (A)
F = 6131.65 Pa * 0.0051 m² = 31.27 N
Therefore, you must exert a force of approximately 31.27 newtons on the balloon.
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.
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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 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
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]
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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
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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.
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