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vitfil [10]
3 years ago
5

A vertical spring with k=490N/m is standing on the ground. You are holding a 5.0kg block just above the spring, not quite touchi

ng it. Part A How far does the spring compress if you let go of the block suddenly? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
WARRIOR [948]3 years ago
5 0

Answer:

Compression in the spring, x = 0.20 m

Explanation:

Given that,

Spring constant of the spring, k = 490 N/m

Mass of the block, m = 5 kg

To find,

Compression in the spring.

Solution,

Since the block is suddenly dropped on the spring gravitational potential energy of block converts into elastic potential energy of spring. Its expression is given by :

mgx=\dfrac{1}{2}kx^2

Where

x is the compression in the spring

x=\dfrac{2mg}{k}

x=\dfrac{2\times 5\times 9.8}{490}

x = 0.20 m

So, the compression in the spring due to block is 0.20 meters.

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Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio
Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

Explanation:

Given that,

First charge q_{1}= 20\mu C

second charge q_{2}= 8\mu C

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

E_{1}= \dfrac{kq_{1}}{r^2}

Put the valueinto the formula

E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}

E_{1}=18\times10^{5}\ N/C

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

E_{2}= \dfrac{kq_{2}}{r^2}

Put the valueinto the formula

E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}

E_{2}=7.2\times10^{5}\ N/C

Direction of electric field along AO

We need to calculate the net electric field at midpoint

E_{net}=E_{1}-E_{2}

E_{net}=(18-7.2)\times10^{5}\ N/C

E_{net}=10.8\times10^{5}\ N/C

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

8 0
3 years ago
A bus traveling at 24 m/s slows down to 12 m/s in 5.0 seconds. What is the acceleration?
Contact [7]

Answer:

a = -2.4 m/s²

Explanation:

Given,

The initial speed of the bus, u = 24 m/s

The final speed of bus, v = 12 m/s

Time taken to reach final speed is, t = 5.0 s

The acceleration of the body is given by the change in velocity by time

                                      a = (v - u) / t

                                         = (12 - 24) / 5

                                         = -2.4 m/s²

The negative sign in the acceleration indicates that the bus is decelerating.

Therefore, the acceleration of the bus is, a = -2.4 m/s²

8 0
3 years ago
What spheres are part of the earths system​
Mama L [17]

Answer:

northern and southern sphere

Explanation:

4 0
3 years ago
Read 2 more answers
A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l
Lelechka [254]

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>
  • Consider a body of mass m kept on a weighing machine in a lift.
  • The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
  • The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
  • Here we have given with the actual weight of the body as 100lbs.
  • This 100lb child is standing on the scale or the weighing machine, when it is riding .
  • During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
  • There is also<em> mg </em>downwards and a normal reaction in the upward direction.
  • when we equate both the upward force and downward force, we get,

                             ma=mg-N\\N=mg-ma    i.e. during riding the scale reads a weight less than that of actual weight.

  • When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

#SPJ4

7 0
2 years ago
What's at the bottom of the black hole? Explain .​
Brilliant_brown [7]

Answer:

Unknown

Explanation:

By definition, we can't observe what's inside there, because no light – no information of any kind – can escape a black hole. But astrophysical theories suggest that, at the core of a black hole, all the black hole's mass is concentrated into a tiny point of infinite density. This point is known as a singularity.

7 0
3 years ago
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