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tekilochka [14]
2 years ago
13

A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car

moves 30.0 m. Neglecting friction between car and road, find v and the horizontal force exerted on the car.
(a) the speed v

______m/s

(b) the horizontal force exerted on the car (Enter the magnitude.)
_______N
Physics
1 answer:
Illusion [34]2 years ago
5 0

The horizontal force applied is  160 N while the velocity is  2.03 m/s.

<h3>What is the speed of the car?</h3>

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

Learn more about force and work:brainly.com/question/758238

#SPJ1

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Musya8 [376]

Answer:

Solution given:

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  3. <u>Used</u><u> </u><u>in</u><u> </u><u>communication</u><u>.</u>

2.

Solution given;

power [P]=100W

Current [I]=5A.

Voltage [V]=?

we have

P=IV

100=5*V

V=\frac{100}{5}=<u>2</u><u>0</u><u>V</u>

<u>the voltage of the battery used in the </u><u>car</u><u> </u><u>is</u><u> </u><u>2</u><u>0</u><u>V</u><u>.</u>

8 0
3 years ago
Read 2 more answers
A worker pushes a 200-N box (weight of box, you have to find mass) resting on a level floor with a force of 30 N. What is the ac
QveST [7]

For the problem we will apply the concepts related to Newton's second law. Recall that this is defined as the product between mass and acceleration, and that in special cases, when the acceleration is equivalent to the force of gravity, the force is equivalent to the weight of the person. From these relationships we will find the mass, and then the acceleration with the given force.

F_w = mg

Here,

F_w= Force

m = Mass

g = Gravity

Replacing and rearranging we have that the mass is,

m = \frac{F_w}{g} = \frac{200}{9.81}

m = 20.387kg

Now using the value of the force, but solving for the acceleration with the previous value of the mass we have,

F = 30N

a = \frac{F}{m}

a = \frac{30}{20.387}

a = 1.47m/s^2

Therefore the acceleration of the box is 1.47m/s^2

8 0
3 years ago
The pressure of the earth's atmosphere at sea level is 14.7 lb/in2. What is the pressure when expressed in g/m2? (2.54 cm = 1 in
miss Akunina [59]
<h2>Answer:</h2>

14.7 lb / in² = 10333018.166 g / m²

<h2>Explanation:</h2>

Given from the question;

pressure = 14.7lb/in²

<u>To convert from lb / in² to g / m²</u>

Note;

2.54 cm = 1 in       ------------(i)

But;

100 cm = 1m

=> 2.54cm = 2.54 cm x 1 m / 100 cm = 0.0254m

=>2.54cm = 0.0254m

Substitute 2.54cm = 0.0254m into equation(i)

=> 0.0254m = 1 in

<em>Note also;</em>

2.205 lb = 1 kg            ------------------(ii)

But;

1kg = 1000g

Substitute 1kg = 1000g into equation (ii)

=> 2.205 lb = 1000g   ---------------------(iii)

<em>From equation(iii);</em>

if, 2.205 lb = 1000g

then, 1 lb = 1 lb x 1000 g / 2.205 lb = 453.5g

=> 1 lb = 453.5 g

<em>Now let's convert 14.7 lb/in² to g / m²;</em>

=> 14.7 lb / in² could be written as 14.7 x 1 lb / (1 in x 1 in)

i.e

=> 14.7 lb / in² = 14.7 x 1 lb / (1 in x 1 in)     ------------------(iv)

<em>Substitute the values of 1 lb = 453.5g and 1 in = 0.0254m into equation(iv)</em>

=> 14.7 lb / in² = 14.7 x 453.5g / (0.0254m x 0.0254m)

=> 14.7 lb / in² = 6666.45 g / (0.00064516m²)

=> 14.7 lb / in² = 10333018.166 g / m²

<em>Therefore, 14.7 lb / in² = 10333018.166 g / m²</em>

5 0
3 years ago
Q1: Using the known values of the mass of the earth and the radius of the earth, calculate the
kupik [55]

Answer:

m g = G * m * M / R^2      force of attraction

g = G * M / R^2

g = 6.67E-11 * 5.96E24 / (6.37E6)^2

g = 6.67 * 5.96 / 6.37^2 * 10^1

g = .98 * 10 = 9.8 m/s^2

Q2.

d g = G M * (-2 R^-3 dR ) = -2 G M / R^3       (very small where dR = 1)

3 0
3 years ago
A body moves a speed of 20km/hr in 15secs, what is the distance covered.​
Taya2010 [7]

Answer:

aa

Explanation:

aa

5 0
3 years ago
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