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3 years ago

When you lift an object, you add the energy of lifting to the object. Tor F

Physics

1 answer:

PtichkaEL [24]3 years ago

6 0

It's true, when we lift an object we add energy to it.

because, when we lift an object by applying force , the object attains a height and hence the energy gets stored in it, in the form gravitational potential energy .

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A barbel weighing 215N is raised to a height of 2.0 M above the ground. It is then dropped from that height.

aleksandrvk [35]

Answer:

430 J
6.26 m/s

Explanation:

A. The kinetic energy is the same as the initial potential energy:

PE = mgh = (215 N)(2.0 M) = 430 J

B. The velocity achieved by falling from a height h is given by ...

v = √(2gh)

v = √(2·9.8 m/s^2·2 m) = √(39.2 m^2/s^2)

v ≈ 6.26 m/s

8 0

3 years ago

Suppose that the radius of the circular path is r when the speed of the rocket is v and the acceleration of the rocket has magni

SCORPION-xisa [38]

Answer:

Explanation:

4 0

3 years ago

Aspirin is a compound composed of carbon, hydrogen, and oxygen atoms

vekshin1

Answer:

A homogeneous Mixture

Explanation:

The acid that contains the acetylsalicylic acid is a <u>mixture,</u> but it isnt a compound. though aspirin is. (hopefully this helps? qwq)

8 0

2 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago

If the net force acting on an object is 0 N, you can be sure thr forces acting on the object are,

Bas_tet [7]

Answer:

A. Balanced

Explanation:

7 0

3 years ago