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creativ13 [48]
2 years ago
14

Help me please Am I correct???​

Physics
1 answer:
trasher [3.6K]2 years ago
5 0

Answer:

They both experienced the same force as they weigh the same amount

Newtonian physics

Explanation:

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When an object is lifted 12 meters off the ground, it gains a certain amount of potential energy. If the same object is lifted 2
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I think the answer is twice as much
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What is the formula for volume using density and mass
lawyer [7]

The definition of density is

                                                       Density                  = (mass) / (volume)

Multiply each side by 'volume' :    (density) x (volume) = (mass)

Divide each side by 'density' :                         Volume = (mass) / (density)
 

5 0
3 years ago
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How could rescue workers use squeezing or compressing to get energy to their flashlights during rescue missions?
Neko [114]

Answer:

Explanation:

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A squeezing or compressing to get energy for flashlight can be regarded as "DYNAMO PROCESS" it involves spinning of "fly wheels" into the flashlight through consistent squeezing ,which is connected to a dynamo(Dynamo supply electrical current). Hence the needed light is seen on the bulb of the flashlight.

3 0
3 years ago
The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point
Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth,  R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

                                  w = 2π / T

                                  w = 2π / 24*3600

                                  w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                 v1 = R*w

                                 v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                 v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

                                 π/2  ........... s

                                 x     ............ 1/5 s

                                 x = π/2*5 = 18°    

- The radius of the earth R' at point where θ = 18° from the equator is:

                                R' = R*cos(18)

                                R' = (6.37 * 10 ^6)*cos(18)

                                R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                              v2 = R'*w

                              v2 = (6058230.0088)*(7.27 * 10^-5)

                              v2 = 440.433 m/s

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3 years ago
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Imma guess A! Idk if it’s 100% correct tho so I’d check that!
6 0
2 years ago
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