All categories

2 years ago

What is the net force on a truck if the force of friction is 31 N and the force of the engine is 79 N?

Physics

1 answer:

Evgen [1.6K]2 years ago

3 0

Answer:

Fr = 48 [N] forward.

Explanation:

Suppose the movement is on the X axis, in this way we have the force of the engine that produces the movement to the right, while the force produced by the brake causes the vehicle to decrease its speed in this way the sign must be negative.

∑F = Fr

$F_{engine}-F_{brake} =F_{r}\\F_{r}=79-31\\F_{r}=48[N]$

The movement remains forward, since the force produced by the movement is greater than the braking force.

You might be interested in

Which set of ordered pairs represents a function?

olchik [2.2K]

Answer:

The second one is a function. {(-8, -2), (7, -2),(-9,2), (0,0)

Explanation:

Its because the y-value is repeated twice.

Hope it helps.

3 0

3 years ago

Read 2 more answers

Compare the pressure exerted by the liquid at points A, B and C. Justify your answer

zzz [600]

Answer:

Pressure is equal in A, B, C and D

Explanation:

Pressure does not depends on the shape of the container

Pressure acting all direction

5 0

2 years ago

A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one

Fittoniya [83]

I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)

8 0

3 years ago

Read 2 more answers

A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location

amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

$f=-16.0 cm$ (the focal length is negative for a diverging lens)

$p=10.0 cm$ is the distance of the object from the lens

Solvign the equation for q, we find

$\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}$

$q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm$

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0

3 years ago

As mass increases what happens to the kinetic energy

AnnyKZ [126]

As mass increases kinetic energy also increases; kinetic energy is directly proportional to mass so whatever is done to either affects the other one the same. i hope this helps :)

5 0

3 years ago