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slavikrds [6]
3 years ago
13

A red racecar accelrates at a constant rate of 5 m/s2. How much time does it take to increase its speed from 50 m/s to 60 m/s?

Physics
1 answer:
natulia [17]3 years ago
7 0
A red racecar accelrates at a constant rate of 5 m/s2. How much time does it take to increase its speed from 50 m/s to 60 m/s?


A.17 s


B.0.058 s


C.0.25 s


D.2.0 s
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Read 2 more answers
Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x directi
pogonyaev

Answer:

V2 = -25.93 m/s

Explanation:

First of all, we know that;

Momentum = mass x change in velocity

Thus;

Momentum = m(v2 - v1)

Also, we know that;

Impulse = force x time

And also that;

Momentum = Impulse

From the question, m = 0.145kg and V1 = 32m/s while impulse = -8.4 N.s

Thus;

0.145(v2 - 32) = -8.4

Now,

0.145v2 - 4.64 = -8.4

0.145v2 = -8.4 + 4.64

0.145v2 = -3.76

v2 = -3.76/0.145

= -25.931 m/s and to two significant figures gives -25.93 m/s

7 0
3 years ago
Two 1.50-V batteries-with their positive terminals in the same direction-are inserted in series into a flashlight. One battery h
DIA [1.3K]

Based on the voltage of the batteries and the internal resistance, the bulb's resistance is 4.595 Ω

<h3>What is the resistance?</h3>

First, find the total emf of the circuit:

= 2 x 1.5V

= 3.0V

The internal resistance is:

= 0.255 + 0.15

= 0.405 Ω

The resistance in the bulb is:

= 3.0 / 0.600 - 0.405

= 4.595 Ω

Find out more on resistance at brainly.com/question/24974191

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8 0
2 years ago
A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at
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Answer:

\Delta U = 0.2072 J

Explanation:

Potential difference between two points in constant electric field is given by the formula

\Delta V = E.\Delta x

here we know that

E = 370 N/C

also we know that

\Delta x = 2.1 - 1.9 = 0.2 m

now we have

\Delta V = 370 (0.2) = 74 V

now change in potential energy is given as

\Delta U = Q\Delta V

\Delta U = (2.80 \times 10^{-3})(74)

\Delta U = 0.2072 J

3 0
4 years ago
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