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Radda [10]
4 years ago
6

In a generator the current changes direction each time the passes through the of a d

Physics
2 answers:
Alexus [3.1K]4 years ago
4 0

Answer:

Commutator passes the slip rings

NeX [460]4 years ago
3 0

Answer:

armature, pole

Explanation:

I just took the test

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A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.
labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

W=K_f -K_i

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

K_i = 66,120 J is the initial kinetic energy of the car

K_f is the final kinetic energy

Solving,

K_f = K_i + W=66,120+(-36,733)=29387 J

The final kinetic energy of the car can be written as

K_f = \frac{1}{2}mv^2

where

m = 661 kg is its mass

v is its final speed

Solving for v,

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s

4 0
4 years ago
1.5 x 10^3 standard notation
poizon [28]

Answer:

1500

Explanation:

3 0
3 years ago
A uniform meter stick is pivoted at the 50.00 cm mark on the meter stick. A 400.0 gram object is hung at the 20.0 cm mark on the
babunello [35]

Answer:C

Explanation:

Given

mass m_1=400\ gm is at x=20\ cm mark

mass m_2=320\ gm is at x=75\ cm mark

Scale is Pivoted at x=50\ cm mark

For scale to be in equilibrium net torque must be equal to zero

Taking ACW as positive thus

T_{net}=0.4\times g\times (0.5-0.2)-0.32\times g\times(0.75-0.50)

T_{net}=0.12g-0.08g=0.04g

Therefore a net torque of 0.04 g is required in CW sense which a mass 400\ gm can provide at a distance of x_o from pivot

0.04g=0.4\times g\times x_o

x_o=0.1\ m

therefore in meter stick it is at a distance of x=60\ cm

6 0
3 years ago
1. A plane starts from rest and aceelerates in a
larisa86 [58]

s=600 m

t=12 s

s=0.5*a*t² (initial speed V0=0)

a=(2*s)/t²

a=(2*600)/12²

a≈8.33 m/s²

L= s(t2=12s)-s(t1=11s) -> (distance during the twelfth second)

L=0.5*a*(t2²-t1²)

L=0.5*((2*s)/t²)*(t2²-t1²)

L=0.5*((2*600)/12²)*(12²-11²)

L ≈ 95.83 m

6 0
3 years ago
A car is traveling at a speed of 10m/s. A 0.5kg clump of mudis
CaHeK987 [17]

Answer:B

Explanation:

Given

speed of car v=10 m/s

mass of clump m=0.5 kg

Radius of car tire r=0.2 m

Since the tire is rotating about axle so a centripetal force is acting constantly on each particle towards the center of tire.

Centripetal force is given by

F_c=\frac{mv^2}{r}

where m=mass\ of\ element

v=speed

r=distance\ from\ center

F_c=\frac{0.5\times 10^2}{0.2}

F_c=250\ N (inward)

           

3 0
4 years ago
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