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fenix001 [56]
3 years ago
14

A water skier is pulled by a boat traveling with a constant velocity. Which one of the following statements is false concerning

this situation? A. The net vertical force on the skier is zero newtons. B. The net acceleration of the skier is zero m/s2. C. There is a net horizontal force on the skier in the direction the ​boat’s velocity. D. The water skier is in equilibrium. ​ E. The net force on the skier is zero newtons.
Physics
1 answer:
ycow [4]3 years ago
6 0

Answer:

C. There is a net horizontal force on the skier in the direction the ​boat’s velocity.

Explanation:

This is because, according to Newton's First law of motion, when an object which in this case is the skier tends to be in equilibrium because it travels at a constant velocity, there is no net acceleration of the skier, and this causes the net force on the skier to be equal to zero.

Therefore in this question, option C which states that "There is a net horizontal force on the skier in the direction the ​boat’s velocity" is a false statement.

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Rutherford discovered the nucleus of the atom by firing a particles at gold foil. An a particle has a charge of q = +2e and a ma
Readme [11.4K]

Answer:r_0=3.037\times 10^{-14}m

Explanation:

Given

charge on alpha particle=+2e

mass of alpha particle=6.64\times 10^{-27} kg

Charge on gold nucleus=+79e

Velocity at r=1m is 1.9\times 10^{7}

Using Energy conservation

Kinetic energy of particle will be converting to Potential energy as it approaches to nucleus

therefore

\frac{1}{2}mv^2+U_{r=1m}=U_{closest\ to\ nucleus}

\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )+\frac{K\left ( 2e\right )\left ( 79e\right )}{1}=\frac{K\left ( 2e\right )\left ( 79e\right )}{r_0}

\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )=\frac{9\times 10^9\times 158\times \left ( 1.6\times 10^{-19}\right )}{y}\left [\frac{1}{r_0}-\frac{1}{1}\right ]

on solving we get

\frac{1}{r_0}=3.292\times 10^{13}

r_0=3.037\times 10^{-14}m

8 0
3 years ago
A 54 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.3 m/s. The acceleration of
Marizza181 [45]

Answer:

h_{B} = 5.012\, m

Explanation:

It is assumed that pole vaulter began running at a height of zero. The physical model is formed after the Principle of Energy Conservation:

K_{A} = K_{B} + U_{B}

\frac{1}{2} \cdot m \cdot v_{A}^{2} = \frac{1}{2} \cdot m \cdot v_{B}^{2} + m \cdot g \cdot h_{B}

The previous expression is simplified and required height is found:

h_{B} = \frac{1}{2\cdot g} \cdot (v_{A}^{2}-v_{B}^{2})

h_{B} = \frac{1}{2 \cdot (9.807\, \frac{m}{s^{2}} )} \cdot [(10\, \frac{m}{s} )^{2}-(1.3\, \frac{m}{s} )^{2}]

h_{B} = 5.012\, m

5 0
4 years ago
Metabolic power is the rate at which your body "burns" fuel to power your activities. For many activities, your body is roughly
DerKrebs [107]

Answer:

Metabolic power 420.138 W

Explanation:

Given data:

considering drag coefficient Cd =  0.9

Assuming cross section of cyclist A = 0.50 m^2

Take density of air ρ = 1.2 kg/m3

We know that drag force is given as

Drag\ Force = Cd\times 1/2 \times \rho AV^2


D = 0.9\times 1/2\times 1.2\times 0.5\times 7.3^2


D = 14.388 N

Power = D\times V = 14.388\times 7.3 = 105.03 W

hence metabolic power is given as

Metabolic power = \frac{D}{\eta}

                   = \frac{105.03}{0.25}

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7 0
3 years ago
A coin dropped in the lift it takes time 0.5 s to reach the floor when lift is staionary it takes time t when lift is moving up
BARSIC [14]

Answer:

t₁ > t₂

Explanation:

A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then t₁ > t₂,  t₁ = t₂,  t₁ >> t₂ ,  t₂ > t₁

Solution:

Newton's law of motion is given by:

s = ut + (1/2)gt²;

where s is the the distance covered, u is initial velocity, g is the acceleration due to gravity and t is the time taken.

u = 0 m/s, t₁ is the time to reach ground when the light is stationary and t₂ is the time to reach ground when the lift is moving with a constant acceleration a.

hence:

When stationary:

s=\frac{1}{2}gt_1^2\\\\t_1^2=\frac{2s}{g}  \\\\When\ moving\ with\ acceleration(a):\\\\s=\frac{1}{2}(g+a)t_2^2\\\\t_2^2=\frac{2s}{g+a}

Hence t₂ < t₁, this means that t₁ > t₂.

4 0
3 years ago
1. What is the equation for gravitational potential energy?<br> E =
Pavel [41]
The gravitational potential energy of an object of mass m at height h on earth is given by PEg=mgh
4 0
3 years ago
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