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kondaur [170]
4 years ago
5

The elevator accelerates upward (in the positive direction) from rest at a rate of 1.95 m/s2 for 2.15 s. Calculate the tension i

n the cable supporting the elevator in newtons.
Physics
1 answer:
sammy [17]4 years ago
3 0

The mass is missing. The mass of the elevator is 1650 kg.

Answer:

The tension in the cable is 19387.5 N.

Explanation:

Given:

Initial velocity of the elevator (u) = 0 m/s

Acceleration in the upward direction (a) = 1.95 m/s²

Time taken by the elevator (t) = 2.15 s

Mass of the elevator and persons (m) = 1650 kg

Let the tension in the cable wire be 'T' Newtons.

Now, there are 2 forces acting in the vertical direction. One is tension in the upward direction and the other the weight of the elevator in the downward direction.

As the elevator is accelerating upward, the net force acts in the upward direction.

So, net force on the elevator is given as:

F_{net}=T-mg

Now, from Newton's second law, net force equals mass times acceleration.

F_{net}=ma\\\\T-mg=ma\\\\T=m(g+a)

Plug in the given values and solve for 'T'. This gives,

T=1650\ kg(9.8+1.95)\ m/s^2\\\\T=1650\times11.75\ N\\\\T=19387.5\ N

Therefore, the tension in the cable is 19387.5 N.

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0.5*0.8*45^2. =810J

Explanation:

KE=1/2 mv^2

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When operated on a household 110.0-V line, typical hair dryers draw about 1650 W of power. We can model the current as a long st
Dimas [21]

Answer:

Current = 15 A

Resistance = 7.33 ohm

Magnetic field = 1.62 x 10^-4 Tesla

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V = 110 V, P = 1650 W, r = 1.85 cm,

(a) Let i be the current

P = V x i

i = 1650 / 110 = 15 A

(b) Let R be the resistance

V = i R

R = 110 / 15 = 7.33 Ohm

(c) Let B be the magnetic field

B = μ0 / 4π x 2i / r

B = 10^-7 x 2 x 15 / 0.0185 = 1.62 x 10^-4 Tesla

5 0
3 years ago
The compressor of an air conditioner draws an electric current of 23.7 A when it starts up. If the start-up time is 2.35 s long,
defon

Answer:

Electric charge, Q = 55.69 C

Explanation:

It is given that,

Electric current drawn by the compressor, I = 23.7 A

Time taken, t = 2.35 s

We need to find the electric charge passes through the circuit during this period. The definition of electric current is given by total charge divided by total time taken.

I=\dfrac{q}{t}

Where,

q is the electric charge

q=I\times t

q=23.7\ A\times 2.35\ s

q = 55.69 C

So, the electric charge passes through the circuit during this period is 55.69 C. Hence, this is the required solution.

6 0
3 years ago
a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string. The string is hung over a pulley as shown in t
svet-max [94.6K]

The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately

<h3>What are we to consider in equilibrium ?</h3>

Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.

Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.

The acceleration at which they will both move will be;

a = (m_{1} - m_{2}) / (m_{1} + m_{2})

a = (5.75 - 3.53) / (5.75 + 3.53)

a = 2.22 / 9.28

a = 0.24 m/s²

Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m

We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.

v² = u² + 2as

since u =0

v² = 2 × 0.24 × 2.47

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v = √1.19

v = 1.0888 m/s

Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately

Learn more about Equilibrium here: brainly.com/question/517289

#SPJ1

5 0
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