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likoan [24]
3 years ago
12

Water at the top of a slope has potential energy. true or false

Physics
2 answers:
ASHA 777 [7]3 years ago
8 0
<span>The statement is TRUE. Water does have potential energy at the top of a slope. The reason why is that potential energy is energy possessed by a body based on its position relative to a zero point. In this case, water at the top of the slope is at an elevation above ground (zero point). The energy is not kinetic (moving) energy since the water is not moving.</span>
Serggg [28]3 years ago
8 0

Answer:

true :)

Explanation:

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Naily [24]

Answer:

Saved Energy ?

Explanation:

5 0
2 years ago
A large truck is moving at 22.0 m/s. if its momentum is 125,000 kg • meters per second, what is the truck's mass? 176 kg 2750 kg
solong [7]

The  mass of the large truck is determined as 5680 kg.

<h3>Mass of the truck</h3>

The mass of the truck is calculated as follows;

P = mv

where;

  • P is momentum
  • m is mass
  • v is velocity

m = P/v

m = 125000/22

m = 5680 kg

Thus, the  mass of the large truck is determined as 5680 kg.

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

7 0
1 year ago
I need the answer to number #23 please will give BRAINLIEST
Murrr4er [49]
Yeah lowkey i think it’s B
3 0
3 years ago
Explain how depression arises from a combination of both genetic and environmental factors
anastassius [24]
Genetics may cause you to become depressed easier, but your environment affects your mood and can affect your long term attitude.  Please mark Brainliest!!!
6 0
3 years ago
Read 2 more answers
Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?
shtirl [24]

Answer:

L=6.21m

Explanation:

For the simple pendulum problem we need to remember that:

\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0,

where \theta is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

\omega^{2}=\frac{g}{L},

where \omega is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

\omega=\frac{2\pi}{T},

where T is the oscillation period. Now, we can easily solve for L:

(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m

3 0
3 years ago
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