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Elina [12.6K]
3 years ago
9

An object with a mass of 1500kg accelerates 10.0m/s when unknown force is applied to it. What is the amount of force?

Physics
2 answers:
Hatshy [7]3 years ago
5 0

Answer:

<h3>The answer is 15000 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 1500 × 10

We have the final answer as

<h3>15,000 N</h3>

Hope this helps you

Sergio039 [100]3 years ago
5 0

Answer:

15000 N

Explanation:

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CHAPTER 6: KINETICS OF A PARTICLE
vekshin1

Explanation:

Work done by winch = kinetic energy of car

∫ T ds = ½ mv²

∫ 225s ds = ½ mv²

225/2 s² = ½ mv²

225 s² = mv²

v = 15s / √m

Given s = 10 m and m = 2500 kg:

v = 15 (10) / √2500

v = 3 m/s

8 0
3 years ago
A sound source is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners
galina1969 [7]

Answer:

Explanation:

As the source is situated on x - axis , it must be situated in between the two listeners .

So the x coordinate of source is

(-7 + 3 )/2

= - 2 m

The equation of the wave- front will be that o a circle having centre at (-2,0)

and radius = distance between -2 and 3 , that is 5 m

equation of circle

=( x+2 )² + y² = 25

It cuts y axis when x = 0

Putting x = 0

4 + y² = 25

y² = 21

y = + √21 , or - √21

7 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
1 - It is okay to use slang on a record as long as the auditor can interpret it.
vesna_86 [32]
1. false 2. false 3. true 4. not sure 5. b 6. b or d 7. not sure 8.not sure 9. not sure 10. c

lol sorry if i’m wrong on any i’m just using common sense
6 0
3 years ago
An atom has 5 protons, 6 neutrons, and 5 electrons. what is the atomic mass?​
ozzi

Answer:

the atomic mass is 11

Explanation:

the atomic mass is basically how many protons and neutrons there are so for this all you have to do is some simple math:

5 + 6 = 11

and boom, ur atomic mass is equal to 11!

4 0
2 years ago
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