Answer:
31.1 N
Explanation:
m = mass attached to string = 0.50 kg
r = radius of the vertical circle = 2.0 m
v = speed of the mass at the highest point = 12 m/s
T = force of the string on the mass attached.
At the highest point, force equation is given as
[tex]T + mg =\frac{mv^{2}}{r}[/tex]
Inserting the values
[tex]T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}[/tex]
T = 31.1 N
Answer: 41N
Explanation :
T= mv^2/R + mgcos θ
At the highest point on the circle θ=0
Cos 0 = 1
T= mv^2/R + mg
m = 0.5kg
Velocity at the highest point (amplitude)= 12m/s
T = 0.5× 12^2/2 + 0.5×10
0.5×144/2 +5
T = 0.5×72 + 5
T = 36+5
T = 41N
A proton and an alpha particle (q = +2e, m = 4 u) are fired directly toward each other from far away, each with an initial speed of 0.010c. What is their distance of closest approach, as measured between their centers?
Answer:
Distance of closest approach, [tex]r=1.91\times 10^{-14}\ m[/tex]
Explanation:
It is given that,
Charge on proton, [tex]q_p=e[/tex]
Charge on alpha particle, [tex]q_a=2e[/tex]
Mass of proton, [tex]m_p=1.67\times 10^{-27}\ kg[/tex]
Mass of alpha particle, [tex]m_a=4m_p=6.68\times 10^{-27}\ kg[/tex]
The distance of closest approach for two charged particle is given by :
[tex]r=\dfrac{k2e^2(m_p+m_a)}{2m_am_pv_p^2}[/tex]
[tex]r=\dfrac{9\times 10^9\times 2(1.6\times 10^{-19})^2(1.67\times 10^{-27}+6.68\times 10^{-27})}{2\times 6.68\times 10^{-27}\times 1.67\times 10^{-27}(0.01\times 3\times 10^8)^2}[/tex]
[tex]r=1.91\times 10^{-14}\ m[/tex]
So, their distance of closest approach, as measured between their centers [tex]1.91\times 10^{-14}\ m[/tex]. Hence, this is the required solution.
Mass m1 is has an initial velocity v1i. Mass m1 collides with a stationary mass m2 . The net external force on the two particle system is zero. If the two masses stick together after the collision, then the final speed vf of the two masses is:
Answer:
vf = (m1*v1i)/(m1 + m2)
Explanation:
take * as multiplication
from the conservation of linear momentum, we know that:
sum of pi = sum of pf, where sum of pi = m1*v1i + m2*v2i
pf = vf(m1 + m2) since the masses stick
together.
v2i = o, since mass m2 is initially stationary.
m1*v1i + m2*(0) = vf(m1 + m2)
m1*v1i = vf(m1 + m2)
therefore
vf = m1*v1i/(m1 + m2)
At one instant, an electron (charge = –1.6 x 10–19 C) is moving in the xy plane, the components of its velocity being vx = 5.0 x 105 m/s and vy = 3.0 x 105 m/s. A magnetic field of 0.80 T is in the positive y direction. At that instant, what is the magnitude of the magnetic force on the electron?
A magnetic field of 0.80 T is there, then the magnitude of the magnetic force on the electron will be equal to - 6.4 x 10⁻¹⁴ k.
What is Charge?Charged material exerts a force when it is introduced to an electromagnetic field because of the chemical property of electric charge. You can have a negative or a positive electric charge. Unlike charge is created one another while like charges repel one another. Neutral refers to an item that carries no net charge.
q = - 1.6 x 10⁻¹⁹ c
v(x) = 5 x 10⁵ m/s,
v(y) = 3 x 10⁵ m/s,
B = 0.8 T along Y-axis
The velocity vector is given by
v = 5 x 10⁵ i + 3 x 10⁵ j
B = 0.8 j
Now calculate the force,
F = q (v x B)
F = -1.6 x 10⁻¹⁹ {(5 x 10⁵ i + 3 x 10⁵ j) x (0.8 j)}
F = - 1.6 x 10⁻¹⁹ (5 x 0.8 x 10^5 k)
F = - 6.4 x 10⁻¹⁴ k
The force's amplitude and direction are along the negative Z axis Is 6.4 x 10⁻¹⁴ N.
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The magnitude of the magnetic force on the electron, given its charge, velocity components, and a magnetic field of 0.80 T in the positive y direction, is 8.0 x 10^(-14) N.
What is the magnitude of the magnetic force on the electron?The magnetic force on a charged particle moving in a magnetic field can be calculated using the following formula:
F = q * v * B * sin(θ)
Where:
- F is the magnetic force.
- q is the charge of the particle.
- v is the velocity of the particle.
- B is the magnetic field strength.
- θ is the angle between the velocity vector and the magnetic field vector.
In this case, the charge of the electron, q, is -1.6 x 10^(-19) C, the velocity components are given as vx = 5.0 x 10^5 m/s and vy = 3.0 x 10^5 m/s, and the magnetic field strength is 0.80 T in the positive y direction.
The angle θ between the velocity and the magnetic field is 90 degrees (π/2 radians) because the electron is moving in the xy plane, and the magnetic field is in the positive y direction.
Now, let's calculate the magnitude of the magnetic force:
F = (-1.6 x 10^(-19) C) * (5.0 x 10^5 m/s) * (0.80 T) * sin(π/2)
F = (-8.0 x 10^(-14) N) * 1
F = -8.0 x 10^(-14) N
So, the magnitude of the magnetic force on the electron is 8.0 x 10^(-14) N.
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13. A proton moves at 7.50×107 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.800 m. What is the field strength
The magnetic force acting on a charged particle moving perpendicular to the field is:
[tex]F_{b}[/tex] = qvB
[tex]F_{b}[/tex] is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.
The centripetal force acting on a particle moving in a circular path is:
[tex]F_{c}[/tex] = mv²/r
[tex]F_{c}[/tex] is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.
If the magnetic force is acting as the centripetal force, set [tex]F_{b}[/tex] equal to [tex]F_{c}[/tex] and solve for B:
qvB = mv²/r
B = mv/(qr)
Given values:
m = 1.67×10⁻²⁷kg (proton mass)
v = 7.50×10⁷m/s
q = 1.60×10⁻¹⁹C (proton charge)
r = 0.800m
Plug these values in and solve for B:
B = (1.67×10⁻²⁷)(7.50×10⁷)/(1.60×10⁻¹⁹×0.800)
B = 0.979T
The magnetic field strength of the proton is 0.979Tesla
The magnetic force acting on a charged particle moving perpendicular to the field is expressed using the equation.
Fm = qvB
The centripetal force traveled by the proton in a circular path is expressed as:
Fp = mv²/r
To get the field strength, we will equate both the magnetic force and the centripetal force as shown:
Fm = Fp
qvB = mv²/r
qB = mv/r
m is the mass of a proton
v is the velocity = 7.50×10⁷ m/s
m is the mass on the proton = 1.67 × 10⁻²⁷kg
q is the charge on the proton = 1.60×10⁻¹⁹C
r is the radius = 0.800m
Substitute the given parameters into the formula as shown:
[tex]B=\frac{mv}{qr}\\B = \frac{1.67 \times 10^{-27} \times 7.5 \times 10^7}{1.60 \times 10^{-19} \times 0.8}[/tex]
[tex]B=0.979Tesla[/tex]
On solving, the magnetic field strength of the proton is 0.979Tesla
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A cosmic ray proton moving toward the Earth at 5.00×107 m/s experiences a magnetic force of 1.70×10−16 N . What is the strength of the magnetic field if there is a 45º angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface?
Explanation:
It is given that,
Speed of proton, [tex]v=5\times 10^7\ m/s[/tex]
Magnetic force, [tex]F=1.7\times 10^{-16}\ N[/tex]
(1) The strength of the magnetic field if there is a 45º angle between it and the proton’s velocity. The magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
[tex]B=\dfrac{F}{qv\ sin\theta}[/tex]
[tex]B=\dfrac{1.7\times 10^{-16}\ N}{1.6\times 10^{-19}\ C\times 5\times 10^7\ m/s\ sin(45)}[/tex]
B = 0.00003 T
or
B = 0.03 mT
The magnitude of Earth's magnitude of 25 to 65 Tesla. The value obtained in part (a) is not consistent with the known strength of the Earth’s magnetic field on its surface.
A ballet dancer has a maximum net torque of 45 Nm, and the moment of inertia of his body is 30 kg mA2. if he starts twirling from rest and keeps accelerating for 95 seconds, how fast will he be rotating? a) b) if he holds rocks in his hands, which variables in this problem would change, in which direction, and why?
Answer:
(a) 142.5 rad/s
Explanation:
Torque = 45 Nm
Moment of inertia = 30 kgm^2
w0 = 0, t = 95 second
(a) Let it is rotating with angular speed w.
Torque = moment of inertia x angular acceleration
45 = 30 x α
α = 1.5 rad/s^2
Use first equation of motion for rotational motion
w = w0 + α t
w = 0 + 1.5 x 95
w = 142.5 rad/s
(b) if he hold rocks, then the moment of inertia change and then angular acceleration change and then final angular velocity change.
Consider a grain of table salt which is made of positive
and negative ions (Na+and Cl−). Suppose each of these ions carries a charge of 1.60x10^−19 C and are 5.29x10^−11mapart. What is the magnitude of the electrostatic force
between them?
Answer:
Force, [tex]F=8.23\times 10^{-8}\ N[/tex]
Explanation:
It is given that,
Each ion in Na⁺ and Cl⁻ has a charge of, [tex]q=1.6\times 10^{-19}\ C[/tex]
Distance between two ions, [tex]d=5.29\times 10^{-11}\ m[/tex]
We need to find the electrostatic force. It is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{(1.6\times 10^{-19})^2}{(5.29\times 10^{-11})^2}[/tex]
[tex]F=8.23\times 10^{-8}\ N[/tex]
So, the magnitude of electrostatic force between them is [tex]F=8.23\times 10^{-8}\ N[/tex]. Hence, this is the required solution.
Ball A (7kg) and ball B (3kg) hit each other in a perfectly elastic collision. If ball A has an initial velocity of 12 m/s and ball B has an initial velocity of -1 m/s. What are the final velocities of each ball?
Explanation:
Mass of ball A, [tex]m_A=7\ kg[/tex]
Mass of ball B, [tex]m_B=3\ kg[/tex]
Initial velocity of ball A, [tex]u_A=12\ m/s[/tex]
Initial velocity of ball B, [tex]u_B=-1\ m/s[/tex]
We need to find the final velocity of each ball. For a perfectly elastic collision, the coefficient of restitution is equal to 1. It is given by :
[tex]e=\dfrac{v_B-v_A}{u_A-u_B}[/tex]
[tex]v_A\ and\ v_B[/tex] are final velocities of ball A and B
[tex]1=\dfrac{v_B-v_A}{13}[/tex]
[tex]v_B-v_A=13[/tex]...........(1)
Using the conservation of linear momentum as :
[tex]m_Au_A+m_Bu_B=m_Av_A+m_Bu_B[/tex]
[tex]7(12)+3(-1)=7v_A+3u_B[/tex]
[tex]7v_A+3u_B=81[/tex]..............(2)
On solving equation (1) and (2) using calculator we get :
[tex]v_A=4.2\ m/s[/tex]
[tex]v_B=17.2\ m/s[/tex]
So, the final velocities of ball A and B are 4.2 m/s and 17.2 m/s. Hence, this is the required solution.
A cosmic ray proton moving toward the Earth at 3.5x 10^7 ms experiences a magnetic force of 1.65x 10^-16 N. What is the strength of the magnetic field if there is a 45° angle between it and the proton's velocity?
Answer:
Magnetic field, [tex]B=4.16\times 10^{-5}\ T[/tex]
Explanation:
It is given that,
Velocity of proton, [tex]v=3.5\times 10^7\ m/s[/tex]
Magnetic force, [tex]F=1.65\times 10^{-16}\ N[/tex]
Charge of proton, [tex]q=1.6\times 10^{-19}\ C[/tex]
We need to find the strength of the magnetic field if there is a 45° angle between it and the proton's velocity. The formula for magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
[tex]B=\dfrac{F}{qv\ sin\theta}[/tex]
[tex]B=\dfrac{1.65\times 10^{-16}}{1.6\times 10^{-19}\times 3.5\times 10^7\times sin(45)}[/tex]
B = 0.0000416 T
[tex]B=4.16\times 10^{-5}\ T[/tex]
Hence, this is the required solution.
The strength of the magnetic field, given the force on the proton, its charge, and velocity, and the angle between the velocity and the magnetic field, is approximately 4.21 x 10^-5 T.
Explanation:The strength of a magnetic field can be calculated using the formula for the force exerted on a moving charge in a magnetic field, which is F = q * v * B * sin(θ). Here, F is the magnetic force, q is the charge of the particle, v is the speed of the particle, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field direction.
In this problem, the magnetic force (F) is given as 1.65 x 10^-16 N, the charge of a proton (q) is +1.6 x 10^-19 C, the speed of the proton (v) is 3.5 x 10^7 m/s, and the angle between the velocity and the magnetic field direction (θ) is 45 degrees. Hence we can rearrange the formula to find the magnetic field (B), getting B = F / (q * v * sin(θ)).
Replacing the values into the equation gives: B = 1.65 x 10^-16 N / (+1.6 x 10^-19 C * 3.5 x 10^7 m/s * sin(45°)) which gives the strength of the magnetic field as approximately 4.21 x 10^-5 T.
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The container is fitted with a piston so that the volume can change. When the gas is heated at constant pressure, it expands to a volume of 25 L. What is the temperature of the gas in kelvins?
Answer:
[tex]T_{2}[/tex] =[tex]\frac{25T_{1}}{V_{1}}[/tex]
where
[tex]V_{1}[/tex] is initial volume in liters
[tex]T_{1}[/tex] is initial temperature in kelvins
Explanation:
Let the initial volume be [tex]V_{1}[/tex] and the initial temperature be [tex]P_{1}[/tex]
Now by ideal gas law
[tex]\frac{P_{1}V_{1}}{T_{1}}=nR..............(i)[/tex]
Similarly let
[tex]V_{2}[/tex] be final volume
[tex]T_{2}[/tex] be the final volume
thus by ideal gas law we again have
[tex]\frac{P_{2}V_{2}}{T_{2}}=nR..............(ii)[/tex]
Equating i and ii we get
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex]
For system at constant pressure the above expression reduces to
[tex]\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}[/tex]
Solving for [tex]T_{2}[/tex] we get
[tex]T_{2}[/tex] =[tex]\frac{25T_{1}}{V_{1}}[/tex]
where
[tex]V_{1}[/tex] is initial volume in liters
[tex]T_{1}[/tex] is initial temperature in kelvins
A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height. Find (a) the ball’s initial velocity and (b) the height it reaches.
Answer:
Initial velocity = [tex]29.43m/s[/tex] b) Height it reaches = 44.145 m
Explanation:
Using the first equation of motion we have
[tex]v=u+at[/tex]
here
v is the final velocity
u is the initial velocity
a is the acceleration of the object
t is time
When the ball reaches it's highest point it's velocity will become 0 as it will travel no further
Also the acceleration in our case is acceleration due to gravity ([tex]-9.8m/s^{2}[/tex]) as the ball moves in it's influence alone with '-' indicating downward direction
Thus applying the values we get
[tex]0=u-(9.81)m/s^{2}\times 3\\\\u=19.43m/s[/tex]
b)
By 3rd equation of motion we have
[tex]v^{2}-u^{2}=2gs[/tex]
here s is the distance covered
Applying the value of u that we calculated we get
[tex]s=\frac{-u^{2}}{2g}\\\\s=\frac{-29.43^{2}}{-2\times 9.81}\\\\s=44.145m[/tex]
Two point charges q1 and q2 are held in place 4.50 cm apart. Another point charge Q = -1.85 mC, of mass 5.50 g, is initially located 3.00 cm from both of these charges and released from rest. You observe that the initial acceleration of Q is 314 m>s2 upward, parallel to the line connecting the two point charges. Find q1 and q2.
Answer:
[tex]q_1 = 6.22 \times 10^{-11} C[/tex]
[tex]q_2 = -6.22 \times 10^{-11} [/tex]
Explanation:
Force on the charge Q = -1.85 mC is along the line joining the two charges
so here we can say that net force on it is given by
[tex]F = ma[/tex]
[tex]F = (0.00550)(314)[/tex]
[tex]F = 1.727 N[/tex]
Now this is the force due to two charges which are in same magnitude but opposite sign
so this force is given as
[tex]F = 2\frac{kq_1q}{r^2} cos\theta[/tex]
here we know that
[tex]F = 2\frac{(9\times 10^9)(1.85 \times 10^{-3})(q)}{0.03^2}cos\theta[/tex]
here we know that
[tex]cos\theta = \frac{2.25}{3} = 0.75[/tex]
now we have
[tex]1.727 = 2\frac{(9\times 10^9)(1.85 \times 10^{-3})(q)}{0.03^2}(0.75)[/tex]
[tex]1.727 = 2.775 \times 10^{10} q[/tex]
[tex]q = 6.22 \times 10^{-11} C[/tex]
A spacecraft of mass 1500 kg orbits the earth at an altitude of approximately 450 km above the surface of the earth. Assuming a circular orbit, what is the attractive force that the earth exerts to keep the spacecraft+ in orbit? Answer: (a) 1.28x 10^7 (N) (b) 2.99 x 10^7 (N) (c) 3.56 x 10^7 (N). (d) 4.11 x 10^7 (N) (e) 5.06x 10^7 (N)
Answer:
1.28 x 10^4 N
Explanation:
m = 1500 kg, h = 450 km, radius of earth, R = 6400 km
Let the acceleration due to gravity at this height is g'
g' / g = {R / (R + h)}^2
g' / g = {6400 / (6850)}^2
g' = 8.55 m/s^2
The force between the spacecraft and teh earth is teh weight of teh spacecraft
W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N
Determine the length of a copper wire that has a resistance of 0.172 ? and cross-sectional area of 7.85 × 10-5 m2. The resistivity of copper is 1.72 × 10-8 ?·m.
Final answer:
To determine the length of a copper wire given its resistance, cross-sectional area, and the resistivity of copper, use the formula L = (R·A)/ρ. The length of the copper wire, based on the provided resistance of 0.172 Ω, cross-sectional area of 7.85 × [tex]10^-^5[/tex] m², and resistivity of 1.72 × [tex]10^-^8[/tex]Ω·m, is calculated to be 7.85 meters.
Explanation:
The question involves determining the length of a copper wire given its resistance, cross-sectional area, and the resistivity of copper. The formula to calculate the length of the wire is derived from Ohm's law, which in terms of resistivity (ρ) is expressed as R = ρL/A, where R is the resistance, L is the length of the wire, A is the cross-sectional area, and ρ is the resistivity of the material.
We are provided with the resistance (R) as 0.172 Ω, the cross-sectional area (A) as 7.85 × [tex]10^-^5[/tex] m², and the resistivity (ρ) of copper as 1.72 × [tex]10^-^8[/tex] Ω·m. By rearranging the formula to solve for L, we get L = (R·A)/ρ. Substituting the given values into the formula, we find the length of the copper wire.
Let's calculate: L = (0.172 Ω × 7.85 × [tex]10^-^5[/tex] m^2) / (1.72 × [tex]10^-^8[/tex] Ω·m) = 7.85 m. Therefore, the length of the copper wire is 7.85 meters.
Consider two cubes, one of aluminum and one of copper. If each cube measures 8 cm along an edge, calculate the mass of each cube.
Answer:
≈1,39 and 4.56 kg.
Explanation:
for more details see the attachment. Note, the ρ(Cu) means 'Density of copper', ρ(Al) means 'Density of aluminium'.
An automobile tire has a radius of 0.344 m, and its center moves forward with a linear speed of v = 20.1 m/s. (a) Determine the angular speed of the wheel. (b) Relative to the axle, what is the tangential speed of a point located 0.135 m from the axle?
Answer:
The angular speed and tangential speed are 58.69 rad/s and 7.92 m/s.
Explanation:
Given that,
Radius = 0.344 m
Speed v= 20.1 m/s
(I). We need to calculate the angular speed
Firstly we will calculate the time
Using formula of time
[tex]t = \dfrac{d}{v}[/tex]
[tex]t=\dfrac{2\pi\times r}{v}[/tex]
[tex]t =\dfrac{2\times3.14\times0.344}{20.1}[/tex]
[tex]t=0.107[/tex]
The angular velocity of the tire
[tex]\omega=\dfrac{2\pi}{t}[/tex]
[tex]\omega=\dfrac{2\times3.14}{0.107}[/tex]
[tex]\omega=58.69\ rad/s[/tex]
Now, using formula of angular velocity
(II). We need to calculate the tangential speed of a point located 0.135 m from the axle
The tangential speed
[tex]v = r\omega[/tex]
Where,
r = distance
[tex]\omega[/tex]= angular velocity
Put the value into the formula
[tex]v= 0.135\times58.69[/tex]
[tex]v=7.92\ m/s[/tex]
Hence, The angular speed and tangential speed are 58.69 rad/s and 7.92 m/s.
The windpipe of a typical whooping crane is about 4.6 ft long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? Assume a temperature of 33°C. ans in hz
Answer:
Frequency, f = 481.8 Hz
Explanation:
Given that,
Length of windpipe, l = 4.6 ft = 0.182 m
We need to find the lowest resonant frequency of this pipe at 33 degrees Celcius. Firstly, we will find the speed of sound at 33 degrees Celcius as :
[tex]v=331+0.6T[/tex]
[tex]v=331+0.6\times 33[/tex]
v = 350.8 m/s
At resonance, wavelength is equal to 4 times length of pipe i.e.
λ = 4 l
We need that, [tex]f=\dfrac{v}{\lambda}[/tex]
[tex]f=\dfrac{350.8\ m/s}{4\times 0.182\ m}[/tex]
f = 481.8 Hz
So, the resonant frequency of the windpipe is 481.8 Hz. Hence, this is the required solution.
The lowest resonant frequency of the whooping crane's windpipe, assuming it is closed at one end and a temperature of 33°C, is approximately 62.82 Hz.
Explanation:The lowest resonant frequency of the whooping crane's windpipe, which can be thought of as a pipe that is closed at one end, can be found using the formula for the fundamental frequency of such a pipe: f = v/λ, where f is the frequency, v is the speed of sound, and λ is the wavelength. In this case, the speed of sound is dependent on the temperature and can be approximated as v = 331.4 + 0.6*T m/s, where T is the temperature in degrees Celsius.
At a temperature of 33°C, the speed of sound, v, is approximately 351.8 m/s. Note that we need to convert the length of the windpipe from feet to meters, with 4.6 ft being approximately 1.4 m. The fundamental wavelength for a pipe closed at one end is four times the length of the pipe (λ=4L), so in this case, λ=4*1.4 m = 5.6 m.
Substituting these values into the frequency formula gives us f = v/λ = 351.8 m/s / 5.6 m = 62.82 Hz. So, the lowest resonant frequency of the whooping crane's windpipe is approximately 62.82 Hz.
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A cyclist is coasting at 12 m/s when she starts down a 450-m-long slope that is 30 m high. The cyclist and her bicycle have a combined mass of 70 kg. A steady 12 N drag force due to air resistance acts on her as she coasts all the way to the bottom. What is her speed at the bottom of the slope?
Answer:
The speed of her at the bottom is 24.035 m/s.
Explanation:
Given that,
Speed of cyclist = 12 m/s
Height = 30 m
Distance d = 450 m
Mass of cyclist and bicycle =70 kg
Drag force = 12 N
We need to calculate the speed at the bottom
Using conservation of energy
K.E+P.E=drag force+K.E+P.E
Potential energy is zero at the bottom.
K.E+P.E=drag force+K.E
[tex]\dfrac{1}{2}mv^2+mgh=Fx+\dfrac{1}{2}mv^{2}[/tex]
[tex]\dfrac{1}{2}\times70\times12^2+70\times9.8\times30=12\times450+\dfrac{1}{2}\times70\timesv^2[/tex]
[tex]25620=5400+35v^2[/tex]
[tex]35v^2=25620-5400[/tex]
[tex]35v^2=20220[/tex]
[tex]v^2=\dfrac{20220}{35}[/tex]
[tex]v=\sqrt{\dfrac{20220}{35}}[/tex]
[tex]v=24.035\ m/s[/tex]
Hence, The speed of her at the bottom is 24.035 m/s.
Answer:
Speed at the bottom of the slope is 24.03 m /sec
Explanation:
We have given speed of the cyclist v = 12 m/sec
She starts down a 450 m long that is 30 m high
Si distance = 450 m and height h = 30 m
Combined mass of cyclist and bicycle m = 70 kg
Drag force = 12 N
We have to find the speed at the bottom
According to conservation theory
KE +PE = work done by drag force + KE + PE
As we know that at the bottom there will be no potential energy
So [tex]\frac{1}{2}\times 70\times 12^2+70\times 9.8\times 30=12\times 450+0+\frac{1}{2}\times 70\times v^2[/tex]
[tex]\frac{1}{2}\times 70\times v^2=20220[/tex]
[tex]v=24.03m/sec[/tex]
A basket of negligible weight hangs from a vertical spring scale of force constant 1500 N/m. Part A If you suddenly put a 3.0-kg brick in the basket, find the maximum distance that the spring will stretch. (Express your answer in units of cm.)
Answer:
3.92 cm
Explanation:
k = spring constant of the spring scale = 1500 N/m
m = mass of the brick = 3 kg
x = stretch caused in the spring
h = height dropped by the brick = x
Using conservation of energy
Spring potential energy gained by the spring scale = Potential energy lost
(0.5) k x² = mgh
(0.5) k x² = mgx
(0.5) k x = mg
(0.5) (1500) x = (3) (9.8)
x = 0.0392 m
x = 3.92 cm
A 100-lb load is suspended by two chains in a room. The angle between each and the horizontal ceiling is 45°. What is the magnitude of the force each chain must be support?
Magnitude of force on each chain suspended with 100 ib load in a room, must be support is 314.54 N.
What is static equilibrium state of hanging load?If the load, is hanging on chain or rope, and is in the equilibrium state, then the sum of all the horizontal component of it must is equal to zero.
Given information-
The load suspended by two chains in a room is 100-ib.
The angle between each chain and the horizontal ceiling is 45°.
The image attached below shows the given situation.
First convert the 100-ib into unit of N as,
[tex]F=100\rm ib=100\times4.4482\rm N\\F=444.82N[/tex]
As the load suspended by two chains in a room is 100-ib and the angle between each chain and the horizontal ceiling is 45°. Thus, the vertical component of the,
[tex]\sin (45)=\dfrac{F_y}{444.82}\\F_y=314.54\rm N[/tex]
Hence, the magnitude of the force each chain must be support is 314.54 N.
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Thus, the magnitude of the force each chain must support is approximately 70.7 lb.
To find the magnitude of the force each chain must support, you can use the principles of equilibrium.
Given:
A 100-lb load is suspended by two chains.The angle between each chain and the horizontal ceiling is 45°.Since the system is in equilibrium, the vertical forces must balance the weight of the load, and the horizontal components of the forces must cancel each other out.
1. Determine the vertical component of the force in each chain:
Each chain supports part of the total load. Let [tex]\( F \)[/tex] be the force in each chain.
Since the chains are symmetric, each chain supports half the load.
The vertical component of the force in each chain is given by:
[tex]\[ F_{\text{vertical}} = F \sin(45^\circ) \][/tex]
where [tex]\( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)[/tex].
2. Calculate the vertical component for each chain:
For equilibrium, the sum of the vertical components of the forces in the two chains must equal the total load:
[tex]\[ 2 \times F \sin(45^\circ) = 100 \text{ lb} \][/tex]
Substitute [tex]\( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)[/tex] :
[tex]\[ 2 \times F \times \frac{\sqrt{2}}{2} = 100 \][/tex]
Simplify:
[tex]\[ F \sqrt{2} = 100 \][/tex]
Solve for [tex]\( F \)[/tex]:
[tex]\[ F = \frac{100}{\sqrt{2}} \][/tex]
Rationalize the denominator:
[tex]\[ F = \frac{100 \sqrt{2}}{2} = 50 \sqrt{2} \][/tex]
3. Approximate the force:
Using [tex]\( \sqrt{2} \approx 1.414 \)[/tex]:
[tex]\[ F \approx 50 \times 1.414 = 70.7 \text{ lb} \][/tex]
A simple pendulum is made from a 0.75-m-long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elapses before it attains its greatest speed?
Answer:
It takes 0.43 seconds before the pendulum attains the maxium speed.
Explanation:
L= 0.75m
g= 9.8 m/s²
T= 2π * √(L/g)
T=1.73 sec
T(vmax) = T/4
T(vmax) = 0.43 sec
The time elapsed before a simple pendulum attains its greatest speed is given by t = T/4 = π√(L/g)/2, where T is the period and L is the length of the pendulum.
Explanation:When a simple pendulum is released from rest, it oscillates back and forth. The time it takes for the pendulum to reach its greatest speed depends on the length of the pendulum. The formula for the period of a simple pendulum is given by:
T = 2π√(L/g)
Where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. To find the time it takes for the pendulum to reach its greatest speed, we can divide the period by 4. This is because the pendulum reaches its greatest speed when it is at the bottom of its swing, which is halfway through one period.
So, the time elapsed before the pendulum attains its greatest speed is:
t = T/4 = (2π√(L/g))/4 = π√(L/g)/2
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A force of 1.200×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 615 N. If he starts from rest and is on a level road, what speed ???? will he be going after 25.0 m?
This question requires Newton's Second Law and the formula for kinetic energy to solve. Unfortunately, we lack the man's mass in the problem as posed, making it impossible to determine the velocity. If it is a class question, the student should consult with the professor considering possible typographical errors.
Explanation:This question is about Newton's Second Law of Motion, which states that the acceleration of an object is equal to the net force acting on it divided by the object's mass. In this case, the forces acting on the man are a forward force of 1.200×10³ N and a backwards air resistance force of 615 N. The net force the man feels is therefore (1.200×10³ - 615) N = 585 N.
We also know that the formula for work done (Work = Force x Distance) and that this work is transferred into kinetic energy (Work = 1/2 x Mass x Velocity²). He moved 25.0 m, but we are not given the man's mass to find the velocity directly, making it impossible to solve unless we have the mass. However, if you received this question in class and believe that there is a typo, please consult with your professor for clarification.
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The man will be going approximately 17.1 m/s after 25.0 m.
Sure, here is the solution to the problem:
Given:
Force applied (F_a) = 1.200 × 10³ N
Force resisting (F_r) = 615 N
Distance (d) = 25.0 m
To find:
Final velocity (v)
Solution:
Step 1: Calculate the net force (F_net)
The net force is the difference between the applied force and the resisting force.
F_net = F_a - F_r
F_net = 1.200 × 10³ N - 615 N
F_net = 585 N
Step 2: Calculate the work done (W)
The work done is equal to the net force times the distance.
W = F_net × d
W = 585 N × 25.0 m
W = 14,625 J
Step 3: Calculate the final velocity (v)
The work done is also equal to the change in kinetic energy (ΔKE).
ΔKE = W
(1/2)mv² = 14,625 J
where:
m is the mass of the man and the bicycle
Since the man and the bicycle are initially at rest, their initial velocity (u) is 0. Therefore, the final velocity (v) can be calculated as follows:
v² = 2W / m
v = √(2W / m)
Assuming a mass of 100 kg for the man and the bicycle, we can calculate the final velocity as follows:
v = √((2 × 14,625 J) / (100 kg))
v = √(292.5 J/kg)
v ≈ 17.1 m/s
Given that a pipe having a diameter of 1.5 feet and a height of 10 feet, what is the psi at 5 feet ?
Answer:
The pressure is 2.167 psi.
Explanation:
Given that,
Diameter = 1.5 feet
Height = 10 feet
We need to calculate the psi at 5 feet
Using formula of pressure at a depth in a fluid
Suppose the fluid is water.
Then, the pressure is
[tex]P=\rho g h[/tex]
Where, P = pressure
[tex]\rho[/tex] = density
h = height
Put the value into the formula
[tex]P=1000\times9.8\times1.524[/tex]
[tex]P=14935.2\ N/m^2[/tex]
Pressure in psi is
[tex]P=2.166167621\ psi[/tex]
[tex]P=2.167\ psi[/tex]
Hence, The pressure is 2.167 psi.
A ball is on the end of a rope that is 1.72 m in length. The ball and rope are attached to a pole and the entire apparatus, including the pole, rotates about the pole's symmetry axis. The rope makes an angle of 64.0° with respect to the vertical. What is the tangential speed of the ball
Answer:
Tangential Speed equals 5.57m/s
Explanation:
In the figure shown for equilibrium along y- axis we have
[tex]\sum F_{y}=0[/tex]
Resolving Forces along y axis we have
[tex]Tcos(\theta )=mg............(i)[/tex]
Similarly along x axis
[tex]\sum F_{x}=ma_{x}[/tex]
[tex]Tsin(\theta )=m[tex]\frac{v^{2} }{r}[/tex]............(ii)[/tex]
Dividing ii by i we have
[tex]tan(\theta )=\frac{v^{2}}{rg}[/tex]
In the figure below we have [tex]r=lsin(\theta )[/tex]
Thus solving for v we have
[tex]v=\sqrt{lgsin(\theta) tan(\theta )}[/tex]
Applying values we get
v=5.576m/s
Batman (mass = 88.8 kg) jumps straight down from a bridge into a boat (mass = 530 kg) in which a criminal is fleeing. The velocity of the boat is initially +14.5 m/s. What is the velocity of the boat after Batman lands in it?
Answer:
The velocity of the boat after Batman lands in it is 2.08 m/s.
Explanation:
Given that,
Mass of batman [tex]m_{1}= 88.8\ kg[/tex]
Mass of boat [tex]m_{2}=530\ kg[/tex]
Initial velocity = 14.5 m/s
We need to calculate the velocity of boat
Using conservation of momentum
[tex]m_{1}u=(m_{1}+m_{2})v[/tex]...(I)
Where, [tex]m_{1}[/tex]=mass of batman
[tex]m_{2}[/tex] =mass of boat
u=initial velocity
v = velocity of boat
Put the value in the equation
[tex]88.8\times14.5=530+88.8\times v[/tex]
[tex]v=\dfrac{88.8\times14.5}{530+88.8}[/tex]
[tex]v=2.08\ m/s[/tex]
Hence, The velocity of the boat after Batman lands in it is 2.08 m/s.
A 360.0 $g$ block is dropped onto a vertical spring with a spring constant k = 254.0 $N/m$. The block becomes attached to the spring, and the spring compresses 0.26 $m$ before momentarily stopping. While the spring is being compressed, what work is done by the block's weight?
Answer:
8.6 J
Explanation:
Work done by the block = change in energy in the spring
W = ½ kx²
W = ½ (254.0 N/m) (0.26 m)²
W = 8.6 J
In a particular case of Compton scattering, a photon collides with a free electron and scatters backwards. The wavelength after the collision is exactly double the wavelength before the collision. What is the wavelength of the incident photon?
Answer:
hence initial wavelength is [tex]\lambda =4.86\times10^{-12}m[/tex]
Explanation:
shift in wavelength due to compton effect is given by
[tex]\lambda ^{'}-\lambda =\frac{h}{m_{e}c}\times(1-cos\theta )[/tex]
λ' = the wavelength after scattering
λ= initial wave length
h= planks constant
m_{e}= electron rest mass
c= speed of light
θ= scattering angle = 180°
compton wavelength is
[tex]\frac{h}{m_{e}c}= 2.43\times10^{-12}m[/tex]
[tex]\lambda '-\lambda =2.43\times10^{-12}\times(1-cos\theta )[/tex]
[tex]\lambda '-\lambda =2.43\times10^{-12}\times(1+1 )[/tex] ( put cos 180°=-1)
also given λ'=2λ
putting values and solving we get
[tex]\lambda =4.86\times10^{-12}m[/tex]
hence initial wavelength is [tex]\lambda =4.86\times10^{-12}m[/tex]
Suppose you take a trip that covers 1650 km and takes 35 hours to make it. What is your average speed in km/h? Please round your answer to the nearest whole number (integer).
Answer:
47.14 Km/h
Explanation:
distance = 1650 km
time = 35 hours
The average speed is defined as the ratio of total distance to the total time taken.
Average speed = total distance / total time
Average speed = 1650 / 35 = 47.14 Km/h
A speed skater goes around a turn with a 31 m radius. The skater has a speed of 14 m/s and experiences a centripetal force of 460 N. What is the mass of the skater?
Answer:
Mass of the skater, m = 72.75 kg
Explanation:
It is given that,
Radius of the circular path, r = 31 m
Speed of the skater, v = 14 m/s
Centripetal force, F = 460 N
We need to find the mass of the skater. It can be determined using the formula of centripetal force. It is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]m=\dfrac{Fr}{v^2}[/tex]
[tex]m=\dfrac{460\ N\times 31\ m}{(14\ m/s)^2}[/tex]
m = 72.75 kg
So, the mass of the skater is 72.75 kg. Hence, this is the required solution.
The centripetal acceleration is found to be approximately 6.32 m/s², which helps determine the mass of the skater to be around 72.78 kg.
To find the mass of the speed skater, we will use the formula for centripetal force:
F = m * ac, where:
F is the centripetal force (460 N)
m is the mass of the skater (unknown)
ac is the centripetal acceleration
Centripetal acceleration can be calculated using:
ac = v² / r, where:
v is the speed of the skater (14 m/s)
r is the radius of the turn (31 m)
First, calculate the centripetal acceleration:
ac = (14 m/s)² / 31 m = 196 m²/s² / 31 m ≈ 6.32 m/s²
Now, substitute into the centripetal force formula to solve for the mass (m):
460 N = m * 6.32 m/s²
Solving for m:
m = 460 N / 6.32 m/s² ≈ 72.78 kg
Therefore, the mass of the skater is approximately 72.78 kg.
For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency (6.8×109 Hz ) and wavelength (0.044 m ). Recall that the Avogadro constant is 6.022×1023 mol−1. Express the energy in joules to two significant figures.
Answer:
2.7 J
Explanation:
The energy of one photon is given by
[tex]E=hf[/tex]
where
h is the Planck constant
f is the frequency
For the photons in this problem,
[tex]f=6.8\cdot 10^9 Hz[/tex]
So the energy of one photon is
[tex]E_1=(6.63\cdot 10^{-34})(6.8\cdot 10^9 )=4.5\cdot 10^{-24} J[/tex]
The number of photons contained in 1.0 mol is
[tex]N_A = 6.022\cdot 10^{23} mol^{-1}[/tex] (Avogadro number)
So the total energy of [tex]N_A[/tex] photons contained in 1.0 mol is
[tex]E=N_A E_1 =(6.022\cdot 10^{23})(4.5\cdot 10^{-24})=2.7 J[/tex]