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densk [106]
3 years ago
15

A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe

n the cord is horizontal. At the bottom of its path, the ball strikes a 2.50 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.
Physics
1 answer:
Komok [63]3 years ago
6 0

Answer:

a) The speed of the ball is 2.47 m/s (in -x direction)

b) The speed of the block, both just after the collision is 1.236 m/s (in +x direction)

Explanation:

Please look at the solution in the attached Word file.

Download docx
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NemiM [27]

Answer:

The number of molecules displaced in a vibration makes the amplitude of a sound.

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A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to
katovenus [111]
The correct answer should be c.The kinetic energy of the water molecules decreases.

If the temperature drops that means that the molecules are coming together. If the temperature rises then it means that the molecules are spreading. If the kinetic energy falls down that means that they are slower which means that they are cooler.
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A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T
slamgirl [31]

Answer:

u=14.48m/s

Explanation:

From the question we are told that:

Height of window h=2m

Height of window off the ground h_g=7.5m

Time to fall and drop t=1.3s

 

Generally the Newton's equation motion  is mathematically given by

 s=ut+\frac{1}{2}at^2

Where

h=ut+\frac{1}{2}at^2

2=u1.3-\frac{1}{2}*9.8*1.3^2

2=u1.3-8.281

u=7.91m/s^2  

Generally the Newton's equation motion  is mathematically given by

2as=v^2-u^2

Where

-2gh_g=v^2-u^2

-2*9.8*7.5=(7.91)^2-u^2

-147=62.5681-u^2

u=\sqrt{209.5681}

u=14.48m/s

Therefore the  ball’s initial speed

u=14.48m/s

8 0
2 years ago
A student is swimming south applying a force of 256 N. The water exerts a westward force of 104 N. If the student has a mass of
grigory [225]

Answer:

a=2.9\ m/sec^2

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force F_r acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

\vec F_r=m\vec a

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force F_r is computed as the hypotenuse of a right triangle

|F_r|=\sqrt{256^2+104^2}

|F_r|=276.32\ Nw

The acceleration can be obtained from the formula

F_r=ma

Note we are using only magnitudes here

\displaystyle a=\frac{F_r}{m}

\displaystyle a=\frac{276.32Nw}{95.3Kg}

\boxed{a=2.9\ m/sec^2}

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Answer:

0 v

Explanation:

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