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Nikolay [14]
2 years ago
11

Jamie pushes a book off a table. The push is an example of a contact force because A. Jamie used energy. B. Jamie had to touch t

he book to move it. C. the book fell to the floor. D. the book was on a flat surface.
Physics
2 answers:
gizmo_the_mogwai [7]2 years ago
6 0
From the word "contact" clearly defines it as touch. three types of forces: contact -  field -  electro-static. thus the answer is C : Jamie hd to touch the book to move it'
 
olasank [31]2 years ago
3 0

Explanation :

Contact forces are the forces that act between the objects that are in contact.

When James pushes a book off a table, the push is an example of a contact force. This is because Jamie had to touch the book to move it. For experiencing a contact force, there should be contact between two objects.

The examples of contact forces are the frictional force, tension force, normal force etc.

While gravitational force, electric force and the magnetic force are non contact force.

So, the correct option is (B) " Jamie had to touch the book to move it ".

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B. red................
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Gravitational potential energy is often released by burning substances. true or false
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Gravitational potential energy is caused when an object is resting above the ground. It is released when the object is falling, not by burning substances.
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Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.4
Gre4nikov [31]

Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t  (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= (m+ \frac{1}{2} \lambda)

replacing ;

Δx = 2t   ; we have:

2t = (m+ \frac{1}{2} \lambda)

Given that thickness t = 700 nm

Then

2× 700 = (m+ \frac{1}{2} \lambda)     --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = (m+ \frac{1}{2} \lambda)     ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus;  the wavelength λ of the light when it is traveling in air = 560 nm

b)  

For the smallest thickness t_{min} ; \ \ \ m =0

∴ 2t_{min} =\frac{\lambda}{2}

t_{min} =\frac{\lambda}{4}

t_{min} =\frac{560}{4}

t_{min} =140 \ \  nm

Thus, the smallest thickness t of the air film = 140 nm

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2 years ago
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A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic frict
Harlamova29_29 [7]

Answer:

The mass of the crate is 5kg.

We know that the force of friction can be obtained by:

F = N*k

where k is the coefficient of friction, where we use the static one if the object is at rest, and the kinetic one if the object os moving. N is the normal force

If we tilt the base making an angle of 30° with the horizontal, now the normal force against the plank will be equal to the fraction of the weight in the direction normal to the surface of the plank.

Knowing that the angle is 30°, then the fraction of the weight that pushes against the normal is Cos(30°)*W = cos(30°)*5kg*9.8m/s^2 = 42.4N

The fraction of the force in the parallel direction to the plank (the force that would accelerate the crate downwards) is:

F = sin(30°)*5k*9,8m/s = 24.5N

now, the statical friction force is:

Fs = 42.4N*0.5 = 21.2N

The statical force is less than the 24.5N, so the crate will move downwards, then the force that acts on the crate is the kinetic force of friction:

Fk = 42.4N*0.4 = 16.96N

Then, the total force that acts on the crate is:

total force = F - Fk = 24.5N - 16.69N = 7.54N and the direction of this force points downside along the parallel direction of the plank.

3 0
2 years ago
A 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west at 8.25 m/s. what i
never [62]

Answer:

The recoil velocity is 0.2687 m/s.

Explanation:

∵ The person is sitting on an icy surface , we can assume that the surface is frictionless.

∴ There is no force acting acting on the person and book as a system in horizontal direction.

Hence , momentum is conserved for this system in horizontal direction of motion.

If 'i' and 'f' be the initial and final states of this system , then by principle of conservation of momentum(p)  -

p_{i}=p_{f}

System initially is at rest

∴p_{i}=0

∴ From the above 2 equations

p_{f}=0

We know that ,

Momentum(p)=Mass of the body(m)×velocity of the body(v)

Let m_{1} and m_{2} be the mass of the person and the book respectively and v_{1} and v_{2} be the final velocities of the person and book respectively.

∴p_{f}=m_{1}v_{1}+m_{2}v_{2}=0

From the question ,

m_{1} = 74.9 kg

m_{2} = 2.44 kg

v_{2} = 8.25 m/s

Substituting these values in the above equation we get ,

(74.9 × v_{1} )+ (2.44×8.25) = 0

∴v_{1}  = - 0.2687 m/s (Negative sign suggests that the motion of  the person is opposite to that of the book)

∴ The recoil velocity is 0.2687 m/s.

4 0
2 years ago
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