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Sergeu [11.5K]
3 years ago
10

A 2.0kg mass is attached to a horizontal spring having a spring constant of 0.05Nm.

Physics
1 answer:
Alex Ar [27]3 years ago
3 0

Good.  You can do some very interesting experiments with that equipment.

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How are energy, force, and the motion of objects related?
OLEGan [10]

Answer:

When two objects interact, each one exerts a force on the other that can cause energy to be transferred to or from the object. For example, when energy is transferred to an Earth-object system as an object is raised, the gravitational field energy of the system increases. This energy is released as the object falls; the mechanism of this release is the gravitational force. Likewise, two magnetic and electrically charged objects interacting at a distance exert forces on each other that can transfer energy between the interacting objects.

Explanation:

Even when an object is sitting still, it has energy stored inside that can be turned into kinetic energy (motion). ... A force is a push or pull that causes an object to move, change direction, change speed, or stop. Without a force, an object that is moving will continue to move and an object at rest will remain at rest.

5 0
3 years ago
Read 2 more answers
How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?
Anestetic [448]

Answer:

21 m

Explanation:

The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

d=vt

where

d is the distance covered

v is the speed

t is the time

The frog in this problem has a speed of

v = 2.1 m/s

and therefore, after t = 10 s, the distance it covered is

d=(2.1)(10)=21 m

3 0
3 years ago
1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diame
IgorC [24]

Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

4 0
2 years ago
large truck exerts 7000 n of force on a piston with an area of 0.4m squared and the is ton can only support 16,000 pa of pressur
sasho [114]
7000 N / 0,4m^2  = 17 500 Pa.
6 0
3 years ago
A boat sails along the shore. To an observer, the boat appears to move at a speed of 22 m/s, and a man on the boat walking forwa
seropon [69]
The boat is moving at 22 m/s while the man is moving at 23.1 m/s.

That means the man, relative to the boat, is moving at 23.1-22 = 1.1 m/s.

v =d/t, so t = d/v --> t = 3/1.1 = 2.7 s
7 0
2 years ago
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