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Cloud [144]
3 years ago
14

In 1.71 minutes, a ski lift raises four skiers at constant speed to a height of 148 m. The average mass of each skier is 62.9 kg

. What is the average power provided by the tension in the cable pulling the lift?
Physics
1 answer:
Vitek1552 [10]3 years ago
6 0

Answer:

3560.36 Watts

Explanation:

Power, P=\frac {nΔW}{Δt) where P is power, n is the number of skiers, t is time in seconds and Δt is change in time, ΔW is given by mgh where m is mass, g is gravitational constant, h is height

Substituting n for 4 skiers, m for 62.9 Kg, g for 9.81, h for 148 m and t for 1.71*60=102.6 seconds

P=\frac {4*62.9*9.81*148}{1.71*60}=3560.360702  Watts

Average power is approximately 3560.36 Watts

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D. gravity

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Gravity keeps the atmosphere from escaping into space.

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If the elephant were then allowed to fall straight down, how fast would it be moving when it landed back on the ground?
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2 years ago
A small plastic bead has been charged to -15nC.
marishachu [46]

Answer:

Explanation:

q = - 15 n C = - 15 x 10^-9 c

(a) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on proton, F = charge on proton x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (1.67 x 10^-27)

a = 1.6 x 10^14 m/s²

(B) the direction of acceleration is towards the bead, as the force is attractive.

(C) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on electron, F = charge on electron x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (9.1 x 10^-31)

a = 3 x 10^17 m/s²

(D) the direction of acceleration is away from the bead, as the force is repulsive.  

8 0
3 years ago
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
3 years ago
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