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motikmotik
3 years ago
13

A mover pushes a 30 kg crate across a level florr with a force of 250 N, but the crate accelerates at a rate of only 5.83 m/s sq

uared because a friction force opposes the crate's motion. What is the magnitude of the force of friction?
Physics
1 answer:
arlik [135]3 years ago
7 0

Force of friction=75 N

Explanation:

we use Newton's second law of motion

F- Ff= ma

F = applied force=250 N

Ff= force of friction

a= acceleration= 5.83 m/s²

m= mass=30 kg

250- Ff= 30 (5.83)

Ff=250-30(5.83)

Ff=250-175

Ff=75 N

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(1-dimension) A fish has a mass of 6 kg and is moving at a speed of 4m/s to the right. What is its momentum?
artcher [175]

Answer:

<h2>24 kg m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 6 × 4

We have the final answer as

<h3>24 kg m/s</h3>

Hope this helps you

4 0
3 years ago
A 26.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m th
frozen [14]

Answer:

18 m/s

Explanation:

Force is given by mass × acceleration

<em>F = ma</em>

<em>a = F/m</em>

For the first 11.0 m, there is no friction. Hence, the constant force is applied fully on the crate. The acceleration is

<em>a = </em>225/26.0<em> </em>m/s²

By the equation of motion, <em>v² = u² + 2as</em>

where <em>v</em> is the final velocity, <em>u</em> is the initial velocity, <em>a</em> is the acceleration and <em>s</em> is the distance travelled.

Using the values for the first part of the motion,

<em>v² = </em>0² + 2 × 8.65 × 11.0

<em>v = </em>13.8 m/s

This is the initial velocity for the second part of the motion. This part has friction of coefficient 0.20.

The frictional force = 0.20 × weight = 0.20 × 26.0 × 9.8 = 50.96 N

The effective force moving the crate in the second motion = 225 - 50.96 N = 174.04 N

This gives an acceleration of 174.04/26.0 = 6.69 m/s².

Using parameters for the equation of motion, <em>v² = u² + 2as</em>

<em>v² = </em>13.8² + 2 × 6.69 × 10 = 324.24

<em>v = </em>18<em> </em>m/s

8 0
4 years ago
What is question 22?
Fynjy0 [20]

the answer is a!! its pretty simple I just read the graph.
4 0
3 years ago
Read 2 more answers
A resistor converts 350 J of electrical energy to other forms of energy. What is the amount of
Helga [31]

Answer:

50C

Explanation:

Given parameters:

Electrical energy  = 350J

Potential difference  = 7V

Unknown:

Amount of charge = ?

Solution:

To solve this problem, use the expression below;

  E  = q x v

E is the electrical energy

q is the quantity of charge

v is the voltage

Insert the parameters and solve for q;

            350 = q x 7

            q = \frac{350}{7}

             q = 50C

5 0
3 years ago
The Indianapolis speedway consists of a 2.5 mile track having four turns, each 0.25 mile long and banked at 9 12'
blsea [12.9K]

Answer: Your question is missing below is the question

Question : What is the no-friction needed speed (in m/s ) for these turns?

answer:

20.1 m/s

Explanation:

2.5 mile track

number of turns = 4

length of each turn = 0.25 mile

banked at 9 12'

<u>Determine the no-friction needed speed </u>

First step : calculate the value of R

2πR / 4 =  πR / 2

note : πR / 2 = 0.25 mile

∴ R = ( 0.25 * 2 ) / π

      = 0.159 mile ≈ 256 m

Finally no-friction needed speed

tan θ = v^2 / gR

∴ v^2 = gR * tan θ

 v = √9.81 * 256 * tan(9.2°)  = 20.1 m/s

8 0
3 years ago
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