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bekas [8.4K]
4 years ago
5

An object, which is initially at rest, accelerates at a rate of 10 m/s2 . Its final position is 85 m from its initial position a

t the end of that acceleration. For how much time did it accelerate?
Physics
1 answer:
vesna_86 [32]4 years ago
4 0

Answer:

4.12 seconds

Explanation:

From the kinetic equation

s=ut+0.5at^{2} where s is the displacement, u is initial velocity, t is time of travel and a is acceleration.

Since the object is initially at rest, the initial velocity is zero hence the equation can be re-written as

s=0.5at^{2} and making t the subject of the formula

t=\sqrt {\frac {2s}{a}}

Substituting s for 85 m and a for 10 m/s2

t=\sqrt {\frac {2*85 m}{10}}=4.123105626 s

Therefore, time is approximately 4.12 seconds

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The net external force acting on an object is zero. Is it possible for the object to be traveling with a velocity that is not ze
Rzqust [24]

Answer:

Yes, this is according to the Newton's first law of motion.

Neither its direction nor its velocity changes during this course of motion.

Explanation:

Yes, it is very well in accordance with Newton's first law of motion for a body with no force acting on it and it travels with a non-zero velocity.

During such a condition the object will have a constant velocity in a certain direction throughout its motion. Neither its direction nor its velocity changes during this course of motion.

4 0
3 years ago
The slope on a distance-time graph shows a ball moving from 20 m to 40 m in 1 s and from 40 m to 80 m in the next second
eduard

The ball is accelerating

Explanation:

On a distance-time graph, the slope of the graph represents the speed of the object represented.

Let's therefore calculate the slope (so, the speed of the ball) in the two intervals given.

In the first second, we have:

\Delta y = 40 - 20 = 20 m\\\Delta x = 1 s

So the average speed is

v_1 = \frac{\Delta y}{\Delta x}=\frac{20}{1}=20 m/s

In the next second, we have:

\Delta y = 80 - 40 = 40 m\\\Delta x = 1 s

So the average speed is

v_2 = \frac{\Delta y}{\Delta x}=\frac{40}{1}=40 m/s

We notice that the speed of the ball has increased from 20 m/s in the first second to 40 m/s in the next second: this means that the speed of the ball is increasing, and therefore, the ball is accelerating.

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
Two balloons (m = 0.012 kg) are separated by a distance of d = 15 m. They are released from rest and observed to have an instant
Evgesh-ka [11]

Answer:

1.492*10^14 electrons

Explanation:

Since we know the mass of each balloon and the acceleration, let’s use the following equation to determine the total force of attraction for each balloon.

F = m * a = 0.012 * 1.9 = 0.0228 N

Gravitational forces are negligible

Charge force = 9 * 10^9 * q * q ÷ 225

= 9 * 10^9 * q^2 ÷ 225 = 0.0228

q^2 = 5.13 ÷ 9 * 10^9

q = 2.387 *10^-5

This is approximately 2.387 *10^-5 coulomb of charge. The charge of one electron is 1.6 * 10^-19 C

To determine the number of electrons, divide the charge by this number.

N =2.387 *10^-5  ÷ 1.6 * 10^-19 = 1.492*10^14 electrons

3 0
3 years ago
Earth Science help needed
denis23 [38]
D it is D the answer is D

6 0
4 years ago
Read 2 more answers
1. Two astronauts are 2.00 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via ele
Veronika [31]

Answer:

1.6949*10^6 m

Explanation:

Our values are

d=2m

v=354m/s

We can find the time through

t=\frac{d}{v}

t=\frac{2}{354}

t=5.64*10^{-3}s

The expression for the distance between the Earth and the spaceship is as follow:

d=ct

Where c is Light speed, and t our previous time.

d= (3*10^8)(5.64*10^{-3})

d= 1.6949*10^6m

Therefore the distance between the Eath and the Spaceship is 1.6949*10^6 m

4 0
4 years ago
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