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lora16 [44]
3 years ago
15

A mass m = 4.7 kg hangs on the end of a massless rope L = 2.06 m long. The pendulum is held horizontal and released from rest. 1

)How fast is the mass moving at the bottom of its path?
Physics
1 answer:
AysviL [449]3 years ago
6 0

Answer:

PE = KE

mgh = ½mv²

v = √(2gh)

v = √(2(9.81)2.07)

v = 6.37 m/s

Explanation:

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Motor 1 and motor 2 each lift objects of equal weight. They both have the same power rating but motor 2 lifts the weight faster.
Oksanka [162]

Answer:

motor 2

Explanation:

because of strenth

5 0
3 years ago
Why does a car slide when it rolls over an icy patch on a road?
Rainbow [258]
Hello. The answer, though I don't know what options you need to be answered, is that there is less friction between the car tires and the ice, than when they are on the road. Causing the car to slide.
7 0
3 years ago
A mass of 15 kg of air in a piston-cylinder device is heated from 25 o C to 77 o C by passing current through a resistance heate
Sophie [7]

Answer:

The electrical energy supplied is 0.233 kWh

Explanation:

Given :

Mass of air m = 15 kg

Initial temperature T _{1} = 298 K

Final temperature T_{2} = 350 K

Heat loss Q _{out} = 60 kJ

From the first law of thermodynamics,

     \Delta U = W_{in} - W_{out} - Q_{out}

We know that internal energy is proportional to difference of temperature,

    W_{in} = m (T_{2} - T_{1}  ) + 60

    W_{in} = 15 \times  (350-298) + 60 }

Now we need answer into kWh so ( 1 kWh = 3600 kJ ).

So we need to multiply ( \frac{1}{3600} )

     W_{in} = [15 \times  (350-298) + 60 ] \times \frac{1}{3600} kWh

     W_{in} = 0.233 kWh

Therefore, the electrical energy supplied 0.233 kWh

3 0
3 years ago
If E1 = 13.0 V and E2 = 5.0 V , calculate the current I2 flowing in emf source E2.
swat32

The solution would be like this for this specific problem:<span>

KCL at Junction a. </span><span>
<span>+ I1 + I2 + I3 = 0 (1) </span>

<span>KVL
<span>+ 13 V - 0.2 R I1 - 0.025 R I1 - 5 V + 0.02 R I2 = 0 (2) </span>
<span>8 + 0.02 I2 = 0.225 I1 </span>
<span>I1 = 35.6 + 0.0889 I2 (2A) </span></span></span>

 

<span>KVL (bottom loop - CCW direction) </span><span>
<span>- 0.02 R I2 + 5 V + 0.5 R I3 = 0 (3) </span>
<span>0.5 I3 = -5 + 0.02 I2 </span>
<span>I3 = -10 + 0.04 I2 (3A) </span></span>

 

<span>Replace 2A and 3A into 1. </span><span>

<span>+ I1 + I2 + I3 = 0 </span>
<span>( 35.6 + 0.0889 I2 ) + I2 + ( -10 + 0.04 I2 ) = 0 </span>
<span>1.129 I2 = -25.6 </span>
<span>I2 = -22.6A </span>

<span>Solve 2A and 3A for other currents. </span>

<span>I1 = 35.6 + 0.0889 I2 = 35.6 + 0.0889 * -22.6 = 33.5A </span>
<span>I3 = -10 + 0.04 I2 = -10 + 0.04 * -22.6 = -10.9A

So the answer is letter D.</span></span>

3 0
3 years ago
A catapult fires a 0.015kg stone at 70m/s.The spring constant of the catapult is 150N/M. Find the exrtention of the catapult.
adelina 88 [10]

Answer:

0.495m

Explanation:

The kinetic energy of the stone will be equal to the work done by the spring

Mathematically;

W = KE

1/2ke² = 1/2mv²

ke² = mv²

k is the spring constant = 150N/m

e is the extension

m is the mass = 0.015kg

v is the velocity = 70m/s

Substitute

150e² = 1/2 * 0.015 * 70²

150e² = 36.75

e² = 36.75/150

e² =  0.245

e = √0.245

e = 0.495m

Hence the extension of the catapult is 0.495m

7 0
3 years ago
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