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shepuryov [24]
3 years ago
5

Which statement correctly describes the gravitational potential energy of the pendulum based on this diagram?

Physics
1 answer:
MrRa [10]3 years ago
8 0
I think it's 'C' but I won't know for sure until you let me see the diagram.
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Plz help it really easy
Mrac [35]

The answer is high to low.

3 0
3 years ago
Read 2 more answers
At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re
densk [106]

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a)  As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ -( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

⇒ ve = 1.08*10⁶ m/s

4 0
4 years ago
I need number two only
Bond [772]

Answer:

d: ice is less dense than liquid water.

Explanation:

hope this helps! ( brainliest would be appreciated :)

~mina

3 0
3 years ago
Read 2 more answers
A 398-kg boat is sailing 14.0° north of east at a speed of 1.50 m/s. 23.0 s later, it is sailing 33.0° north of east at a speed
kicyunya [14]

Answer:

Explanation:

We shall represent speed in vector form

First speed

v₁ = 1.5 cos 14 i + 1.5 sin 14 j

= 1.455 i + 0.363 j

v₂ = 4.4 cos 33 i + 4.4 sin 33 j

= 3.69 i + 2.39 j

v₂ - v₁

3.69 i + 2.39 j - 1.455 i - 0.363 j

= 2.235 i + 2.027 j

acceleration

=  v₂ - v₁ / time

= ( 2.235 i + 2.027 j  ) / 23

= .097 i + .088 j

force = mass x acceleration

= 398 x ( .097 i + .088 j )

= 38.6 i + 35.02 j

Magnitude of force F

F² = 38.6² + 35.02²

F = 52.11 N

Tan θ = 35.02 / 38.6

θ = 42° north of east.

4 0
3 years ago
How can a driver best be prepare to enter sharp curves
Eduardwww [97]
Slow down before entering the curve. Stay close to the outside. Use the gas pedal as you enter the middle of the curve.
7 0
3 years ago
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