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andrew11 [14]
3 years ago
13

you are building a race car. Your goal is to have a car that can go from 0 miles per hour to 80 miles per hour in 2 seconds. You

are focusing on the car's
Physics
2 answers:
Vitek1552 [10]3 years ago
6 0
<span>You want to focus on the car's acceleration.
(Hope this helps you! :) Have a nice day!)</span>
vodka [1.7K]3 years ago
3 0

Answer : Acceleration of the car is 17.89\ m/s^2

Explanation :

Initial velocity of the car, u = 0 miles

Final velocity of the car, v = 80 miles/hour = 35.77 m/s

Time taken, t = 2 s

So, from the above data we are focusing on the velocity of the car w.r.t to time i.e. acceleration of the car.

a=\dfrac{v-u}{t}

a=\dfrac{35.77\ m/s-0}{2\ s}

a=17.89\ m/s^2

So, the car of the car is 17.89\ m/s^2

Hence, this is the required solution.

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A concrete slab of mass 400 kg accelerates down a concrete slope inclined at 35°. The kinetic between the slab and slope is 0.60
belka [17]

I got you

Explanation:

normal force = 400 g cos 35

friction force up slope = .6 (400 g) cos 35

weight component down slope = 400 g sin 35

400 a = 400 g sin 35 - .6 (400 g cos 35)

a = g (sin 35 - .6 cos 35) = .082 g

I hope this helps you

8 0
2 years ago
Someone please help with this
SSSSS [86.1K]

Answer:

<em>The new force is 2/3 of the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}

Factoring out 2/3:

\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}

Substituting the original force:

F'=\frac{2}{3}F

The new force is 2/3 of the original force

7 0
3 years ago
An Earth satellite is orbiting at a distance from the Earth’s surface equal to one Earth radius (4 000 miles). At this location,
trasher [3.6K]

Answer:

The acceleration due to gravity is \dfrac{1}{4} times the value of g at the Earth’s surface.

(D) is correct option.

Explanation:

Given that,

Radius = 4000 miles

We need to calculate the gravitational force at surface

Gravitational force on the mass m on the surface of the earth

At r = R

F=mg

mg=\dfrac{GmM}{R^2}....(I)

We need to calculate the gravitational force at height

Gravitational force on a mass m from the center of the earth,

At r = R + R = 2 R

F'=mg'

mg'=\dfrac{GmM}{4R^2}....(II)

Dividing equation (II) by equation (I)

\dfrac{mg'}{mg}=\dfrac{\dfrac{GmM}{4R^2}}{\dfrac{GmM}{R^2}}

\dfrac{g'}{g}=\dfrac{1}{4}

Hence, The acceleration due to gravity is \dfrac{1}{4} times the value of g at the Earth’s surface.

4 0
3 years ago
If a ball is tossed straight up into the air, at what position is its potential energy the greatest? Question 2 options: a) when
yarga [219]
Q1: At the highest point
3 0
3 years ago
Which of the following most directly shows how physics affects society?
Nataliya [291]

the answer is definitely A.

5 0
3 years ago
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