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LiRa [457]
2 years ago
14

A car travels West at 35 mph for a time period of 2.5 hours and then travels West at 55 mph for an additional time of 4.0 hours.

How far will this car be from its starting point at the end of the journey?
Pls show work
Physics
1 answer:
OlgaM077 [116]2 years ago
7 0

The total distance travelled is 307.5 miles.

<h3>What is the distance covered?</h3>

We know that the distance covered can be obtained from the formula from speed since the motion all took place in the same direction.

Speed = distance/time

distance = speed * time

In the first case;

Distance =  35 mph  *  2.5 hours = 87.5 miles

In the second case;

Distance =  55 mph * 4.0 hours = 220 miles

Total distance = 87.5 miles + 220 miles = 307.5 miles

Hence the total distance travelled is 307.5 miles.

Learn more about total distance:brainly.com/question/20330990

#SPJ1

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A device experiences a voltage drop of 5.0 V across it while a current of 10.0 mA flows through it. How much power does it dissi
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0.05 W

Explanation:

The power dissipated by a device can be written as

P=VI

where

P is the power dissipated

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In this problem, we have

V = 5.0 V is the voltage drop across the device

I = 10.0 mA = 0.01 A is the current through it

By applying the formula, we find the power dissipated:

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4 0
3 years ago
Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, t
Levart [38]

Answer:

3N/C

Explanation:

Thinking process:

Let, electric field of 4N/C is in the positive x direction, that is + 4 NC

And the the resulting electric field on the x axis at x =2m

Therefore, based on the data above, the net charge must be in the positive direction. Thus, the magnitude is given by the equation:

q = \frac{Er^{2} }{k} \\   = \frac{4 (4)}{9*10^{4} } \\   = \frac{16}{9}NC

Now considering that at x = + 4 NC, the net charge will be:

E_{net} = E_{0} + E_{q}   \\             = 4 -(\frac{9*10^{} }{16}*\frac{16}{9} * 10^{-90})\\             = 3 (N/C)

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