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3 years ago

11

A space craft is traveling in interplanetary space at a constant velocity. what is the estimated distance traveled by the space

craft in 9.50 seconds?
A. 2.30 x 10^2 kil
B. 2.42 x 10^2 kil
C. 2.51 x 10^2 kil
D. 3.02 x 10^2 kil

1 answer:

mina [271]3 years ago

7 0

Answer:

A. 2.30 x 10^2 kil

Explanation:

3/2 = 1.5 and 0.72/2 = 0.36

1.93 + 0.36 = 2.29

2.29 = about 2.30

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Read 2 more answers

A generator produces 38 mwmw of power and sends it to town at an rms voltage of 78 kvkv. part a what is the rms current in the t

4vir4ik [10]

The rms current in the transmission lines is I = 487.18 A.

The root-imply-rectangular (rms) voltage of a sinusoidal supply of electromotive force is used to represent the source. it is the rectangular root of the time average of the voltage squared.

Alternating-present day circuits. the root-imply-square (rms) voltage of a sinusoidal source of electromotive force is used to symbolize the supply. it's far the square root of the time average of the voltage squared.

Electric power is by using present day or the waft of electric fee and voltage or the capacity of rate to deliver electricity. A given cost of power can be produced by using any combination of contemporary and voltage values

power = 38 M watt

rms voltage = 78 K v

power = IV

I = power/V

I = (38 * 1000000)/78*1000

I = 487.18 A.

Learn more about rms current here:-brainly.com/question/20913680

#SPJ4

7 0

2 years ago

10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when

SVEN [57.7K]

Answer:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

Explanation:

For this case we can use the second law of Newton given by:

$\sum F = ma$

The friction force on this case is defined as :

$F_f = \mu_k N = \mu_k mg$

Where N represent the normal force, $\mu_k$ the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

$F_f = ma$

Replacing the friction force we got:

$\mu_k mg = ma$

We can cancel the mass and we have:

$a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}$

And now we can use the following kinematic formula in order to find the distance travelled:

$v^2_f = v^2_i - 2ad$

Assuming the final velocity is 0 we can find the distance like this:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

5 0

3 years ago