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ss7ja [257]
3 years ago
11

What is the acceleration of a 45 kg mass pushed with 15 N of force? What is the acceleration if the force changes to 25 N? What

if the mass changes to 70 kg?
Physics
2 answers:
Rama09 [41]3 years ago
8 0
F = ma
Rearrange this so acceleration is the subject:
A = f/m

Now input your values into this equation for the first question :
A = f/m
A = 15/45
A = 0.33333333 m/s

Then using this same equation change the force of 15 to 25 :
A = f/m
A = 25/45
A = 0.55555556 m/s

For the last question keep the force of 15N but change the mass to 70kg into the same equation :
A = f/m
A = 15/70
A = 0.21 m/s (rounded)
Contact [7]3 years ago
6 0

Answer:

Part a)

a = \frac{1}{3} m/s^2

Part b)

a = \frac{5}{9} m/s^2

Part c)

a = \frac{3}{14} m/s^2

Explanation:

As per Newton's II law we know that

F = ma

Now we know that

m = 45 kg

F = 15 N

so we will have

15 = 45 a

a = \frac{15}{45} = \frac{1}{3} m/s^2

Now we have

m = 45 kg

F = 25 N

now again from Newton's Law

25 = 45 a

a = \frac{25}{45} m/s^2

a = \frac{5}{9} m/s^2

Now if mass is changed to 70 kg

again we have

15 = 70 a

a = \frac{15}{70}

a = \frac{3}{14} m/s^2

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Aneli [31]

Answer:

point in the same direction as the change in position.

require knowing the initial and final position, along with the elapsed time.

Explanation:

As we know that average velocity is defined as the ratio of change in position and the time interval of the object

It is given as

v_{avg} = \frac{\Delta x}{\Delta t}

so we will have

\Delta x = displacement

\Delta t = time interval

the direction is this average velocity is always along the direction of the displacement

so correct answer will be

point in the same direction as the change in position.

require knowing the initial and final position, along with the elapsed time.

3 0
3 years ago
Two particles, each having a mass of 3.0 mg and having equal but opposite charges of magnitude of 6.0 nC, are released simultane
algol13

Answer:

The distance that separates the two particles is 7.42 cm.

Explanation:

Given;

the mass of each particle, m = 3 mg = 3 x 10⁻⁶ kg

the magnitude of charge of each particle, q = 6.0 nC

speed of each particle, v = 5.0 m/s

F = ma = \frac{kq^2}{r^2}

a = v/t

where;

a is the acceleration of the two particles

v is the final velocity

t is time

v = u + gt

5 = 0 + 9.8t

5 = 9.8t

t = 5/9.8

t = 0.51 s

a = v/t

a = 5/0.51 = 9.8 m/s²

Total force on the two particles = (2m)a = (2* 3 x 10⁻⁶)9.8

F = 5.88 x 10⁻⁵ N

Substitute in the value of F in the above equation and calculate r

F = \frac{kq^2}{r^2}

where;

k is coulomb's constant = 8.99 x 10⁹ Nm²/c²

r is the distance of separation between the two particles

F = \frac{kq^2}{r^2} \\\\r^2 = \frac{kq^2}{F}\\\\r = \sqrt{ \frac{kq^2}{F}} = \sqrt{ \frac{8.99*10^9*(6*10^{-9})^2}{5.88*10^{-5}}} = 0.0742 \ m

Therefore, the distance that separates the two particles at the instant when each has a speed of 5.0 m/s, is 7.42 cm.

7 0
3 years ago
When fat comes in contact with sodium hydroxide, it produces soap and glycerin. Determine whether this is a physical change or a
DochEvi [55]

A quick, easy way to decide whether there was a chemical change

is to look and see whether there are NEW substances after the

event, that weren't there when it started.


This particular scenario started out with fat and sodium hydroxide (lye).

And then, suddenly, POOF ! Soap and glycerin showed up. Where did

THOSE come from ? They came from the molecules in fat and lye,

getting broken up and recombined to make different substances.

THAT's exactly a chemical change.

6 0
3 years ago
Read 2 more answers
A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of
olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through 90^{o} then position of two mirrors will be changed. The resultant path difference will be as follows.

      \frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}

Formula for change in fringe shift is as follows.

          n = \frac{2lv^{2}}{\lambda c^{2}}

       v^{2} = \frac{n \lambda c^{2}}{2l}

             v = \sqrt{\frac{n \lambda c^{2}}{2l}}

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

             l = 11 m

    \lambda = 5.9 \times 10^{-7} m

           c = 3.0 \times 10^{8} m/s

Hence, putting the given values into the above formula as follows.

            v = \sqrt{\frac{n \lambda c^{2}}{2l}}

               = \sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}

               = 2.41363 \times 10^{9} m/s

Thus, we can conclude that velocity deduced is 2.41363 \times 10^{9} m/s.

3 0
3 years ago
A sho has a length of 11 inches.which unit conversion fraction should you use to find the length in centimeters
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Answer:

2.54 cm to 1in

Explanation:

6 0
3 years ago
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