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frosja888 [35]
2 years ago
11

The velocity of a car decreases from 28m/s to 20m/s in a time of 4 seconds. what is the average acceleration of the car in this

process?
Physics
1 answer:
Olin [163]2 years ago
3 0

The average acceleration of the car in this process is -2 m/s²

<h3>What is Acceleration ?</h3>

Acceleration can be defined as change in velocity per time taken. It is a vector quantity. That is, it has both the magnitude and direction. It is measured in m/s²

Given that a velocity of a car decreases from 28m/s to 20m/s in a time of 4 seconds. Then,

  • Initial velocity u = 28 m/s
  • Final velocity v = 20 m/s
  • Time t = 4 s
  • Acceleration a = ?

From the definition of acceleration,

a = Δv / t

a = ( 20 - 28 ) / 4

a = -8 / 4

a = - 2 m/s²

Therefore, the average acceleration of the car in this process is -2 m/s² which mean that the car is decelerating.

Learn more about Deceleration here: brainly.com/question/25311290

#SPJ1

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A lever and fulcrum are used to lift a fallen tree, which has a weight of 480N. if the lever has a mechanical advantage of 5.5,
Alisiya [41]

Answer:

<h2>The input force is 87.27N</h2>

Explanation:

Mechanical advantage (M.A)of a machine is the ratio of load to effort

<h2>Given</h2>

M.A= 5.5

load= 480N

effort= ?

Applying the formula and substituting to find the effort we have

M.A= \frac{load}{effort}

5.5= \frac{480}{effort}\\\effort= \frac{480}{5.5} \\\effort= 87.27N

8 0
3 years ago
A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of
Pie

Answer

given,

distance = 200 m

radius of curvature = 35 m

time = 30 s

centripetal acceleration = ?

speed of the runner

v = \dfrac{d}{t}

v = \dfrac{200}{35}

v = 5.71 m/s

acceleration of the runner

a_c = \dfrac{v^2}{r}

a_c = \dfrac{5.71^2}{35}

a_c = 0.931 m/s^2

5 0
3 years ago
Earth is the third planet from the Sun. This placement most affects Earth's unique _____and ______ conditions. 1.A)gravity
Scrat [10]

<u>Answer: </u>

For 1: The correct option is b.

For 2: The correct option is a.

<u>Explanation: </u>

Earth is the third planet in our solar system. It's placement in the solar system is such that, it is the only planet on which life survival is possible.

Earth lies in the Goldilock zone that defined as the zone in which there is an accurate temperature and atmospheric conditions which supports the life.

So, this placement most affects earth's unique temperature and atmospheric conditions. Hence, the correct answer is Option b for 1 and Option a for 2.

3 0
3 years ago
The resistivity of gold is 2.44 x 10-8 ohms.m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries
harkovskaia [24]

Answer:

E=0.036 V/m

Explanation:

Given that

Resistivity ,ρ=2.44 x 10⁻⁸ ohms.m

d= 0.9 mm

L= 14 cm

I = 940 m A = 0.94 A

We know that electric field E

E= V/L

V= I R

R=ρL/A

So we can say that

E= ρI/A

Now by putting the values

E=\dfrac{ 2.44\times 10^{-8}\times 0.94}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}

E=0.036 V/m

6 0
3 years ago
Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

U=k\frac{q_1q_2}{r_{1,2}}                  (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}           (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J

The electric potential energy between the three charges is 80.91 J

7 0
3 years ago
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