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Alexxx [7]
2 years ago
11

What is an object doing at the moment its instantaneous acceleration is zero?

Physics
1 answer:
Oliga [24]2 years ago
3 0

Answer: Option A

Explanation:

Acceleration is defined as the rate of change of the vellocity.

Then, if the acceleration is equal zero, the object has a constant velocity or a velocity equal to zero (also constant, of course)

Then, te correct option would be option A; resting, because an object resting has instantaneuos acceleration of zero.

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What is an enzyme and what is it from
k0ka [10]

Answer:

Enzymes help break down food that we consume. They are found in the stomach or small intestine.

4 0
1 year ago
Which of the following helps to explain how hydraulic lifts work?
Gnesinka [82]

Answer : C. Pascal's principle.

Explaination : Pascal's principle (well-known as Pascal's law) states that if a closed container contains a fluid at rest, then a small change in pressure at one side of the fluid is transmitted to each and every part of the fluid and also to the walls of the container without any loss. In a hydraulic lift, we need the same mechanism to work and so we take the help of Pascal's principle.


Hence, the correct option is C. Pascal's principle.

8 0
3 years ago
Read 2 more answers
Light always travels in a straight line
Licemer1 [7]

 Light travels in straight lines. Once a light has been produced, it will keep moving in a straight line until it hits something else. Shadows are evidence of light traveling in straight lines. An object blocks light so that it can’t reach the surface where we see the shadow.

8 0
2 years ago
Read 2 more answers
A long thin rod of mass M and length L is situated along the y axis with one end at the origin. A small spherical mass m1 is pla
Yuri [45]

Answer:

= - 5.65\times 10^{-2} J

Explanation:

Given data:

L =2.00 *10^4 m

d = 18*10^4 m

M = 18  *10^6 kg

m_1 = 8*10^6 kg

Gravitational energy is given as

U =- G \frac{m_1 m_2}{r}

mass per unit length is given as

\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m

calculating potential energy

dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}

=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}

=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}

=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}

=-G*m_1*\sigma ln(1.125)

=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)

= - 5.65\times 10^{-2} J

5 0
3 years ago
What is the initial position
labwork [276]

Answer:

Initial position of a body is the position of the body before accelerating or increasing its velocity the position changes and then that position is the final position.

hope it is helpful...

6 0
3 years ago
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