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RSB [31]
3 years ago
14

A crate is to be pulled across the floor. The crate will experience a force of friction of 34 N. You pull on the crate with a fo

rce of 75 N. The mass of the crate is 8.5 kg.
What is the net force acting on the crate?

What is the crate's rate of acceleration?
Physics
1 answer:
Jobisdone [24]3 years ago
6 0
The net force would be 75N - 34N = 41N

The crates rate of acceleration would be 41N=8.5*a so a = 4.8m/s^2
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Arunner starts from rest and accelerates uniformly to a speed of 8.0 meters per
VladimirAG [237]

Answer:

2.0 m/s/s

Explanation:

The acceleration of an object is the rate of change of velocity of the object.

Mathematically, it is given by:

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

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

Acceleration is a vector, so it has both a magnitude and a direction.

For the runner in this problem, we have:

u = 0 is the initial velocity (he starts from rest)

v = 8.0 m/s is the final velocity

t = 4.0 s is the time taken

Substituting, we find

a=\frac{8.0-0}{4.0}=2.0 m/s^2

3 0
3 years ago
You guys cant answer a dam question I ask it fricking 6th grade questions
nexus9112 [7]

Answer:

calm down please its not that serious maybe no one saw it yet

Explanation:

7 0
3 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s =  2500/(185644.54)

s=0.0134


So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0
3 years ago
Plz give me the answer of qiestion ofno 31
77julia77 [94]
The resistance between A and B is 10 ohms.
5 0
3 years ago
2. A woman prevents a 3kg brick from falling by pressing it against a vertical wall. The coefficient of friction
Anettt [7]
<h3>Answer:</h3>

49 N

<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of the brick as 3 kg
  • The coefficient of friction as 0.6

We are required to determine the force that must be applied by the woman so the brick does not fall.

  • We need to importantly note that;
  • For the brick not to fall the, the force due to gravity is equal to the friction force acting on the brick.
  • That is; Friction force = Mg

But; Friction force = μ F

Therefore;

μ F = mg

0.6 F = 3 × 9.8

0.6 F = 29.4

      F = 49 N

Therefore, she must use a force of 49 N

6 0
3 years ago
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