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ziro4ka [17]
3 years ago
7

A 975-kg elevator accelerates upward at 0.754 m/s2, pulled by a cable of negligible mass. Find the tension force in the cable.

Physics
1 answer:
Zina [86]3 years ago
7 0

To solve this problem we will apply the concepts of equilibrium and Newton's second law.

According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So

\sum F = ma

T-mg = ma

Here,

T = Tension

m = Mass

g = Gravitational Acceleration

a = Acceleration (upward)

Rearranging to find T,

T = m(g+a)

T = (975)(9.8+0.754)

T= 10290.15N

Therefore the tension force in the cable is 10290.15N

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Karolina [17]
I’m pretty sure the answer would be D, sorry if it’s not correct!
5 0
3 years ago
use the pendulum equation to calcuate the period of a 1.50 pendulum. Remeber that the vaule of "g" is 9.81 m/s² please help ​
kati45 [8]

Answer:

L = 3.51

Explanation:

Pendulum equation is T = 2pi\sqrt{L/9.81}

T = 1.5 and we are solving for L

1.5=2\pi \sqrt{L/9.81}

square both sides to get 2.25 = 2\pi ( L/9.81)

multiply both sides by 9.81 then divide by 2 and 3.14 as a substitue for pi. The answer should be about 3.51 in length

L = 3.51

If this helps, mark me brainliest pls

3 0
3 years ago
A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe
joja [24]

Answer:

The Doppler Effect is given by the following relation;

f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

v_o = The velocity of the observer

v_s = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(ii) When the source is receding from the observer, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

(1) Given that (v + v_s) > v, therefore, v < (v + v_s), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(1) Given that (v - v_s) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

7 0
3 years ago
Not a question but need a lot of tips
skad [1K]

DibujaZos da buenos tips e incluso el pinta con colores faber castel son económicos y el dibujo queda epico

8 0
3 years ago
A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(
Valentin [98]

Answer:

The temperature change of the copper is greater than the temperature change of the water.

Explanation:

deltaQ = mc(deltaT)

Where,

delta T = change in the temperature

m =mass

c = heat capacity

\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}

The temperature change in the copper is nearly 11 times the temperature change in the water.

So, the correct option is,

The temperature change of the copper is greater than the temperature change of the water.

Hope this helps!

7 0
3 years ago
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