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2 years ago

Object's velocity change

Physics

1 answer:

marshall27 [118]2 years ago

3 0

the definition of acceleration in kinematics, allows to find that the correct answer is:

2. speed up ( acceleration)

The kinematics study the movement of the body, the acceleration is defined as the change of the speed by the time in the interval

a = Δv / t

Where the bold letters indicate vectors, a is the acceleration, v the velocity and t the time

Analyzing this expression we see that for there to be a change in velocity, there must be an acceleration of the body.

Let's analyze the different claims

1. True. If it is stopped and you start to move there is an acceleration, therefore there is a change in speed, but after you are moving the acceleration becomes zero and there is no change in speed

2. True. Whenever there is acceleration there is a change in speed

3. False. Moving slowly does not change the acceleration, therefore there is no change in speed

4. True. If you are moving and you stop at this moment there is an acceleration, therefore there is a change in speed, but after being stopped the acceleration is zero and there is no change in speed

5. False. If you change the direction at the instant of change there is an acceleration but after you go in the opposite direction there is not acceleration therefore there is no change in speed.

In conclusion using the definition of acceleration in kinematics, we can find the answer the condition for a change in velocity, the correct statement is:

2. speed up ( acceleration)

Learn more here: brainly.com/question/5063616

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What will the magnitude of the field be if the 10 nc charge is replaced by a 20 nc charge? Assume the system is big enough to co

MArishka [77]

Answer:

Same magnitude of the 10 nc charge cause the electric field is external.

Explanation:

To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.

As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.

F = qE

If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.

4 0

3 years ago

A 0.20 kg mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured

xxMikexx [17]

Answer:

b. 0.20 m/s.

Explanation:

Given;

initial mass, m = 0.2 kg

maximum speed, v = 0.3 m/s

The total energy of the spring at the given maximum speed is calculated as;

K.E = ¹/₂mv²

K.E = 0.5 x 0.2 x 0.3²

K.E = 0.009 J

If the mass is changed to 0.4 kg

¹/₂mv² = K.E

mv² = 2K.E

$v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 0.009}{0.4} } \\\\v = 0.21 \ m/s\\\\v \approx 0.20 \ m/s$

Therefore, the maximum speed is 0.20 m/s

4 0

3 years ago

A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio

Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

$f=2.85 Hz$

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$