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3 years ago

Science please helppp

Physics

1 answer:

Daniel [21]3 years ago

3 0

The red box must way more. Gravitational potential energy is the product of a an objects mass times the acceleration due to gravity (which is constant on earth) times its height. Since the objects are on the same shelf they are at the same height, and since gravitational acceleration is constant as long as we stay on planet earth, then the mass is the only possible thing that could have changed. This means that the red box must weigh more than the blue box.

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Help please <br><br><br> What is elastic potential energy

spin [16.1K]

Awnser:

Elastic Potential Energy. Elastic potential energy is Potential energy stored as a result of deformation of an elastic object,

Explanation:

4 0

3 years ago

A mass m = 550 g is hung from a spring with spring constant k = 2.8 N/m and set into oscillation at time t = 0. A second, identi

riadik2000 [5.3K]

Answer:

The second system must be set in motion $t=0.70s$ seconds later

Explanation:

The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:

$T=2\pi\sqrt(\frac{m}{k} )$

One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.

If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion $t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s)$ seconds later

6 0

3 years ago

If you are reading this please help ASAP !!!!!

NISA [10]

Answer:

Explanation:

5 0

3 years ago

Move 13 m west and then 8 m east?HELP!!!

Hoochie [10]

Answer:

Displacement will be 5m west

Distance would be 21m No direction

5 0

2 years ago

A car moving along a straight road at 30m/s slows uniformly to a speed of 10m/s in a time of 5s. determine the

nignag [31]

Answer:

A=ACCELERATION = -4 m/s^2

B=DISTANCE= 72 m

Explanation:

Solving for the acceleration of the car

A= (10 m/s-30 m/s) / 5s

A= -20 m/s / 5s

A= -4 m/s^2

Solving for the distance traveled after the third second

D= v1 * t + 1/2at^2

D= 30 m/s * 3 s + -2m/s^2 * (3s)^2

D= 90 m + - 18 m

D = 72 m

7 0

3 years ago