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makvit [3.9K]
3 years ago
14

A train travels 98 kilometers in 1 hours, and then 78 kilometers in 3 hours. what is its average speed?

Physics
1 answer:
ozzi3 years ago
7 0

The average speed over the total four hour interval is

\dfrac{98\,\mathrm{km}+78\,\mathrm{km}}{4\,\mathrm h}=31\,\dfrac{\mathrm{km}}{\mathrm h}

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Answer:

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Ivan exerts 20 N of force to lift a basket 2 m. He then carries the basket 5 m across a room to place it on a shelf. Ivan does w
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As the distance between 2 objects increases, what happens to gravitational force?
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3 years ago
Which of the following is an example of a difference between planets in the solar system?
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Hope it helped brainiest plz and thank you.

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3 0
2 years ago
Could you help me with the question in the photo attached please.
s344n2d4d5 [400]

a) a=\frac{m_2 g-m_1 g sin \theta}{m_1+m_2}

b) Yes

c) a=\frac{m_2 g-m_1 g sin \theta - \mu_k m_1 g cos \theta}{m_1+m_2}

Explanation:

a)

To find the expression for the acceleration of the blocks, we have to write the equations of the forces acting on the two blocks.

In the following, we assume that the system accelerates downward (block 2) and up along the ramp (block 1).

For block 1, the equation of motion is:

T-m_1 g sin \theta = m_1 a

where:

T is the tension in the string (up along the ramp)

m_1 g sin \theta is the component of the weight parallel to the ramp (down along the ramp), with

m_1=mass of the block

g = acceleration due to gravity

\theta = angle of the ramp

a is the acceleration of the system

For block 2, we have:

m_2 g-T = m_2 a

where

m_2 g is the weight of block 2, where

m_2 is the mass of block 2

From eq. (2) we get

T=m_2 g- m_2 a

And substituting into eq.(1), we find an expression for the acceleration of the system:

m_2 g- m_2 a -m_1 g sin \theta = m_1 a\\(m_1+m_2)a=m_2g-m_1 g sin \theta\\a=\frac{m_2 g-m_1 g sin \theta}{m_1+m_2}

b)

We could have guessed the expression for the acceleration using Newton's second law:

F=ma

where, in this case:

F is the net force on the system, which is the difference between the weight of the block 2 and the component of the weight of block 1 acting along the ramp, so

F=m_2 g- m_1 g sin \theta

m is the total mass of the system, which is the sum of the masses of the two blocks:

m=m_1 + m_2

a is the acceleration

And solving for a:

a=\frac{F}{m}=\frac{m_2 g-m_1 g sin \theta}{m_1 + m_2}

c)

In this case, there is also kinetic friction acting along the ramp, on block 1.

The magnitude of the kinetic friction is:

F_f=\mu_k N

where

\mu_k is the coefficient of kinetic friction

N is the normal force acting on block 1

The normal force on block 1 can be found by writing the equation of the forces on the direction perpendicular to the ramp, we have:

m_1g cos \theta - N = 0

where m_1g cos \theta is the component of the weight of the block perpendicular to the ramp. Therefore,

N=m_1 g cos \theta

So the force of friction is

F_f=\mu_k m_1 g cos \theta

The direction of this force is in the direction opposite to the motion (up along the ramp), so the equation of motion for block 1 becomes

T-m_1 g sin \theta - \mu_k m_1 g cos \theta = m_1 a

Substituting again the expression for  the tension obtained in part a), we get:

m_2 g-m_2 a-m_1 g sin \theta - \mu_k m_1 g cos \theta = m_1 a\\a=\frac{m_2 g-m_1 g sin \theta - \mu_k m_1 g cos \theta}{m_1+m_2}

3 0
3 years ago
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