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Rus_ich [418]
3 years ago
14

An electric iron of resistance 20Ω takes a current of 5A. Calculate the heat developed in 30seconds?

Physics
1 answer:
Komok [63]3 years ago
6 0
<h2>ANSWER:-</h2>

The amount of heat (H) produced is given by the joule’s law of heating as H= Vlt

Where,

Current, I = 5 A

Time, t = 30 s

Voltage, V = Current x Resistance = 5 x 20 = 100V

H= 100 x 5 x 30 = 1.5 x 10⁴ J.

Therefore, the amount of heat developed in the electric iron is 1.5 x 10⁴J.

Please mark me as a BRAINLIST

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Answer:

See below

Explanation:

At point A    the PE = mgh = 2 * 10 * 1 = 20 J

  at point B, all of the PE , 20 J , is converted to Kinetic Energy

KE = 1/2 m v^2

20 = 1/2 (2)(v^2 )

20 = v^2     v = sqrt 20 = 4.47 m/s

for the friction part

 vf = vo t  + 1/2 a t^2      vf = final velocity = 0 (stopped)

                                       vo = original velocity = 4.47 m/s  

                                         a = -1 m/s^2

  0 = 4.47 t + 1/2 (-1) t^2

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6 0
2 years ago
Far out in space, a 100,000-kg rocket and a 200,000-kg rocket are docked at opposite ends of a motionless 90m-long tunnel. The t
postnew [5]

Answer:

a) X_{cm} = 60m

b) I = 5.4*10^8Kg*m^2

c) T= 4.5*10^6 N*m

d) a = 8.3*10^{-3}rad/s^2

e) W= 0.25 rad/s

Explanation:

a) we know that:

X_{cm} = \frac{m_1x_1+m_2x_2}{m1+m2}

where X_cm is the ubicaton of the center of mass, m_1 the mass of the first rocket, x_1 its distances with the rocket 1, m_2 the mass of the second rocket and x_2 its distance with the rocket 1. So, replacing values, we get:

X_{cm} = \frac{(100,000kg)(0)+(200,000kg)(90m)}{100,000kg+200,000kg}

X_{cm} = 60m

So, the center of mass is at 60m from the rocket 1.

b) we know that:

I = M_1R_1^2 +M_2R_2^2

where I is the moment of inertia, M_1 is the mass of the rocket 1, R_1 its distance from the center of mass, M_2 the mass of the second rocket and R_2 the distance between the rocket 2 and the center of mass. So, replacing values, we get:

I = (100,000kg)(60m)^2 +(200,000kg)(30m)^2

I = 5.4*10^8Kg*m^2

c) We know that:

T = Fr

where T is the net torque, F is the force and r is the distance between the rocket and the radius. Then:

FR_1+FR_2 = T

Replacing values, we get:

50,000N(60m)+50,000N(30m) = T

T= 4.5*10^6 N*m

d) We know that:

T = Ia

where T is the net torque, I the moment of inertia and a is the angular aceleration. So, replacing values, we get:

4.5*10^6Nm = 5.4*10^8Kg*m^2a

solving for a:

a = 8.3*10^{-3}rad/s^2

e) Finally, using:

W = at

where W is the angular velocity, a is the angular aceleration and t is the time.

Then, replacing values. we get:

W = (8.3*10^{-3}rad/s^2)(30s)

W = 0.25 rad/s

3 0
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