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EleoNora [17]
3 years ago
9

(01.02 MC) Nicole pushes her bike up a hill. Overhead, the sun exerts a gravitational force on Earth. Which statement is true ab

out the bike and Earth?
They both experience contact forces. They both experience non-contact forces.

The bike experiences a non-contact force and Earth experiences a contact force.

The bike experiences a contact force and Earth experiences a non-contact force.

They both experience non-contact forces.
Physics
2 answers:
jarptica [38.1K]3 years ago
7 0

When Nicole pushes the bike up a hill we know there is a friction force between hill and bike then due to this friction force It will have to push harder to move it upwards along the hill.

So this type of friction force during uphill is a type of contact force which is present of the bike.

Apart from this there are also few non contact force between bike and its surroundings

One of the force is gravitational force of Sun and Earth on the bike while will act towards their centers.

This force is a field force and it do not require any contact with the bike

So here correct answer would be

<em>They both experience contact forces. They both experience non-contact forces. </em>

seraphim [82]3 years ago
7 0

Answer:

non contact :C

Explanation:

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3 years ago
Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th
Kipish [7]

Answer:

Probability of tunneling is 10^{- 1.17\times 10^{32}}

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J

Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

Thus

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = 1.52\times 10^{-35}\ m

Now,

We can calculate the tunneling probability as:

P(t) = e^{\frac{- 2t}{\eta}}

P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}

P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

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3 years ago
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Lapatulllka [165]
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3 years ago
For a home sound system, two small speakers are located so that one is 52 cm closer to the listener than the other. What is the
galina1969 [7]

Answer:

The first frequency of audible sound in the speed sound is

f = 662 Hz

Explanation:

vs = 344 m/s

x =  52 cm * 1 / 100m = 0.52m

The wave length is the distance between the peak and peak so

d = 2x

d = 2*0.52 m

d = 1.04 m

So the frequency in the speed velocity is

f = 1 / T

f = vs / x = 344 m/s / 0.52m

f ≅ 662 Hz

7 0
3 years ago
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di
Pachacha [2.7K]

Answer:

m=146.277kg which is rounded to 146kg

Explanation:

Remember that F=ma

But F represents not 250N, but 250cos(35)N since the force is being pulled above the horizontal.

So 250cos(35)=204.7880111 approximately, and since a=1.4m/s^2, we have 204.7880111=m(1.4m/s^2). Then we divide both sides by the acceleration to get the mass. So m=146.2771508kg which the nearest number is 146kg

Mass is always in kg, unless stated otherwise.

4 0
2 years ago
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