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solong [7]
3 years ago
13

A 1500 kg car is pushing a 4000 kg truck. The car and truck are accelerating at 2.0 m/s^2. Assuming that the frictional force on

the truck is negligible, what force is the truck exerting on the car?
Physics
1 answer:
Crazy boy [7]3 years ago
7 0

To solve this problem we will use the Force equation according to the definition given in Newton's second law. There we have that the Force is equal to

F =ma

Where,

m = mass

a = acceleration

Our values are given as

m_1 = 1500 kg

m_2 = 4000 kg

a = 2.0 m/s^2

Considering that both mass are equal to one, we have that:

F = (m_1+m_2)* a

F = (1500 + 4000)(2.0)

F = 11000 N = 11kN

Therefore the truck exert a force on the car of 11kN

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CAN YOU PLS CHECK IF ITS CORRECT I'LL MARK YOU BRAINLIST
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i think it looks good

yea it correct

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5 0
3 years ago
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a runner starts from rest and has an acceleration of 3 m/s^2. How fast is she running after 1.1 seconds
grigory [225]
Acceleration x time = velocity 

Since you're given acceleration and time, just plug the values into the equation. 

3\frac{m}{s^{2}} x 1.1 s = ? 

Solve that equation, and remember your velocity should be in m/s.
8 0
3 years ago
A billiards ball B rests on a horizontal surface and is struck by another billiards ball A of the same mass m = 0.2 kg. Ball A i
SOVA2 [1]

Answer:

v1 = 15.90 m/s

v2 = 8.46 m/s

mechanical energy before collision = 32.4 J

mechanical energy after collision = 32.433 J

Explanation:

given data

mass m = 0.2 kg

speed = 18 m/s

angle =  28°

to find out

final velocity and  mechanical energy both before and after the collision

solution

we know that conservation of momentum remain same so in x direction

mv = mv1 cosθ + mv2cosθ

put here value

0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)

3.6 =  0.1765 V1 + 0.09389 v2    ................1

and

in y axis

mv = mv1 sinθ - mv2sinθ

0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)

0 = 0.09389 v1 - 0.1768 v2   .......................2

from equation 1 and 2

v1 = 15.90 m/s

v2 = 8.46 m/s

so

mechanical energy  before collision = 1/2 mv1² + 1/2 mv2²

mechanical energy before collision = 1/2 (0.2)(18)² + 0

mechanical energy before collision = 32.4 J

and

mechanical energy after collision = 1/2 (0.2)(15.90)² + 1/2 (0.2)(8.46)²

mechanical energy after collision = 32.433 J

7 0
3 years ago
A boy is pulling a wagon with a force of 70.0 N directed at an angle of 30 degrees above the horizontal. What are the x and y co
bogdanovich [222]

Answer:

x component is 60.60 N and y component is 35 N ⇒ answer A

Explanation:

A boy is pulling a wagon with a force of 70.0 N directed at an angle

of 30 degrees above the horizontal then,

F_{x} is the horizontal component of the force in the direction

of positive part of x-axis

F_{y} is the vertical component of the force force in the direction

of positive part of y-axis

The x component of the force F_{x} is the force multiplied by

cos(30)°

The y component of the force F_{y} is the force multiplied by

sin(30)°

→ The x component F_{x} =  F cos(30)

→ The x component F_{x} = 70 cos(30) =  60.60 N

→ The y component F_{y} = F sin(30)

→ The y component F_{y} = 70 sin(3) = 35 N

<em>x component is 60.60 N and y component is 35 N</em>

7 0
4 years ago
Does anyone know what the answers are?
baherus [9]

Answer:

Option (C) is the answer

Explanation:

may be it is possible if that we stand so far

4 0
3 years ago
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