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ycow [4]
3 years ago
14

How much work is required to lift a 2 kg mass to a height of 10 meters?

Physics
1 answer:
scoray [572]3 years ago
6 0
Use the equation potential energy =m*g*h
m-mass,
g-gravitational acceleration,
h-height
Potential energy = 2*10*10
=200
kg {m}^{2}   {s}^{ - 2}
This is the unit of the energy
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A dog has a mass of 20 kg. If the dog is pushed across the ice with a force of 40 N, what is its acceleration?
olasank [31]

Answer:

The acceleration is 2 m/s2.

Explanation:

We calculate the acceleration (a), with the data of mass (m) and force (F), through the formula:

F = m x a  ---> a= F/m

a = 40 N/20 kg                   <em>  1N= 1 kg x m/s2</em>

a= 40 kgx m/s2/ 20 kg

<em>a= 2 m/s2</em>

7 0
3 years ago
How would you calculate an object’s mechanical energy? a. Add its kinetic and potential energies. b. Multiply its kinetic and po
dalvyx [7]
A. Add it's Kinetic and Potential energies
6 0
3 years ago
Read 2 more answers
A physics student swings a tennis ball connected to a rope in a vertical circle with a constant speed of 6.29 m/s. The ball has
Alex777 [14]

Answer:

r = 0.5 m

Explanation:

First we find the angular speed of the ball by using its period:

ω = θ/t

For the time period:

ω = angular speed = ?

θ = angular displacement = 2π rad

t = time period = 0.5 s

Therefore,

ω = 2π rad/0.5 s

ω = 12.56 rad/s

Now, for the radius:

v = rω

r = v/ω

where,

v = linear speed = 6.29 m/s

r = radius = ?

r = (6.29 m/s)/(12.56 rad/s)

<u>r = 0.5 m</u>

8 0
3 years ago
Starting from your campsite you walk 3.0 km east, 6.0 km north, 1.0 km east, and then 4.0 km west. How far are you from your cam
Hatshy [7]
Think of it like a graph. You start at the origin which is (0,0).  go three to the east which now you are (3,0). Then, six to the north. Now, you are at (3,6).  1 to the east, ((4,6).  Then you go 4 to the west which is back tracking. So, you end at (0,6) which is saying you are now 6 km north from your campsite. 

Hope this helps!
6 0
3 years ago
How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?
Illusion [34]
Power is defined as the rate at which the body is doing work:
P=\frac{W}{t}
Work is defined as displacement done by the force times that displacement:
W=F\cdot h
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
W=62N\cdot10m=620J
of work.
Now we just divide that by 5s to get how much power is required:
P=\frac{620J}{5s}=124W
5 0
3 years ago
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