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gladu [14]
3 years ago
6

A car starts from a stopped position at a red light. At the end of 30 seconds, its speed is 20 meters per second. What is the ac

celeration of the car?
A. 1.5 m/s
B. 0.7 m/s
C. 0.7 m/s 2
NEED help please HEELLPP
Physics
1 answer:
Gnoma [55]3 years ago
5 0
A=deltav/t
Vf=20m/s
Vi=0m/s
(20-0)/30
=0.667 m/s^2
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A box, initially at rest, has 36.7 N of force exerted on it for 2.81 s. If the box has a mass of 7.41 kg, what was its velocity
ElenaW [278]

Answer:

13.91 m/s

Explanation:

First we need to find the acceleration:

Acceleration = Force/mass

Acceleration = 36.7N/7.41 kg

Acceleration = 4.95 m/s² (rounded to two decimal places)

Then we find the velocity:

Velocity = Acceleration * Time

Velocity = 4.95 m/s² * 2.81 s

Velocity = 13.91 m/s (rounded to two decimal places)

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2 years ago
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Nesterboy [21]

Answer:

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Explanation:

4 0
3 years ago
A train of mass 5.6 × 10^5kg is at rest in a station.at time t=0s, a resultant force acts on the train and it starts to accelera
lana66690 [7]

Answer:

420000N

Explanation:

Given parameters:

Mass of the train  = 5.6 x 10⁵kg

Acceleration  = 0.75m/s²

Unknown:

Resultant force = ?

Solution:

According to newton's second law, force is the product of mass and acceleration;

   Force  = mass x acceleration

Resultant force that acts on the train is given below;

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7 0
3 years ago
A 325 N force applied to an object for 5.5 s. What’s the impulse?
vazorg [7]
According to Newton second law of motion, the resultant force is directly proportional to the rate of change in momentum while maintaining other factors constant. Therefore, F = (mv-mu)/t where F is the resultant force , m is the mass of the object, v is the final velocity and u is the initial velocity.
Hence, Ft = mv-mu, but impulse is given by force multiplied by time, thus, impulse is equivalent to the change in momentum.
Impulse = Ft
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4 0
3 years ago
Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud
Levart [38]

Explanation:

Given that,

Charge 1, q_1=-5.45\times 10^{-6}\ C

Charge 2, q_2=4.39\times 10^{-6}\ C

Distance between charges, r = 0.0209 m

1. The electric force is given by :

F=k\dfrac{q_1q_2}{r^2}

F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}

F = -492.95 N

2. Distance between two identical charges, r=0.0209\ m

Electric force is given by :

F=\dfrac{kq_3^2}{r^2}

q_3=\sqrt{\dfrac{Fr^2}{k}}

q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}

q_3=4.89\times 10^{-6}\ C

Hence, this is the required solution.

3 0
3 years ago
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