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nevsk [136]
3 years ago
11

• How much work isrequired to lift a 2kgobject 2m high?​

Physics
1 answer:
pychu [463]3 years ago
7 0

Answer You need to consider that the gravity on earth is 9.8 m/s/s. This means any object you let go on the earths surface will gain 9.8 m/s of speed every second. You need to apply a force on the object in the opposite direction to avoid this acceleration. If you are pushing something up at a constant speed, you are just resisting earths acceleration. The more massive and object is, the greater force is needed to accelerate it. The equation is Force = mass*acceleration. So for a 2kg object in a 9.8 m/s/s gravity you need 2kg*9.8m/s/s = 19.6 Newtons to counteract gravity. Work or energy = force * distance. So to push with 19.6 N over a distance of 2 meters = 19.6 N*2 m = 39.2 Joules of energy. There is an equation that puts together those two equations I just used and it is E = mgh

The amount of Energy to lift an object is (mass) * (acceleration due to gravity) * (height)

:Hence, the Work done to life the mass of 2 kg to a height of 10 m is 196 J. Hope it helps❤️❤️❤️

Explanation:

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A small first-aid kit is dropped by a rock climber who is descending steadily at 1.9 m/s. After 2.4 s, what is the velocity of t
Genrish500 [490]

Answer:

v=-21.65 m/s

Explanation:

From the exercise we have:

v_{o}=1.9m/s\\ g=9.81 m/s^{2}\\ t=2.4s

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v=v_{o}+gt

v=1.9m/s-(9.81m/s^{2})(2.4s)=-21.65m/s

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0.180-kg stone rets on a frictionless, horizontal surface. A bullet of mass 7.50 g, traveling horizontally at 390 m/s, strikes t
Musya8 [376]

Answer:

The magnitud of the velocity is

8.46m/s

and the direccion:

-28.3 degrees from the horizontal.

Explanation:

Fist we define our variables:

m_{s}=0.18kg\\m_{b}=7.5g = 0.0075kg\\v_{1b}= 390m/s -i\\v_{1s}=0m/s\\v_{2b}=210m/s -j\\v_{2s}=?

The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.

velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.

Using conservation of momentum:

m_{s}v_{1s}+m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}\\v_{1s}=0, so\\m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}

Clearing for the velocity of the stone after the crash:

v_{2s}=\frac{m_{b}v_{1b}-m_{b}v_{2b}}{m_{s}}

Substituting known values:

v_{2s}=\frac{0.0075kg(((390m/s-i)-210m/s-j)}{0.18kg}\\v_{2s}=(16.25m/s-i) - (8.75m/s-j)

The magnitud of the velocity is :

|v_{2s}|=\sqrt{(16.25m/s-i)^2 + (8.75m/s-j)^2}\\|v_{2s}|=18.46m/s

and the direction:

tan^{-1}(-8.75/16.25)=-28.3

this is -28.3 degrees from the +i direction or the horizontal direcction.

Note: i and j can also be seen as x and y axis.

3 0
3 years ago
Canopus, which is in the constellation of Carina and Argo Navis, is 310 ly away. You plan a sight-seeing vacation for Canopus an
vazorg [7]

Answer:

D'=2.933*10^{18}m

Explanation:

From the question we are told that:

Distance D=310ly

Where

1ly=9.46*10^{15}

Generally the equation for  meters will you traverse to reach your vacation destination is mathematically given by

D'=D*1ly

D'=310*9.46*10^{15}

D'=2.933*10^{18}m

6 0
2 years ago
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