2. What makes Newton's 3rd Law an unbalanced force?

A powerful motorcycle can produce an acceleration of 3.50 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting

IrinaK [193]

Answer:1265 N

Explanation:

Given

acceleration of motorcycle $\left ( a\right )=3.5 m/s^2$

Velocity $\left ( v\right )=90 km/h\approx 25m/s$

Air friction and Friction $\left ( f\right )=425 N$

mass of the motorcycle with rider $\left ( m\right )=240 Kg$

Applying Forces on motorcycle

$F_{engine}-f=ma$

$F_{engine}=f+ma$

$F_{engine}=425+240\left ( 3.5\right )$

$F_{engine}=425+840=1265 N$

5 0

3 years ago

A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the e

IgorLugansk [536]

Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

F = kΔx

where,

F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

F = 50 N

Δx = change in length = 25 cm - 20 cm = 5 cm = 0.05 m

Therefore,

50 N = k(0.05 m)

k = 50 N/0.05 m

k = 1000 N/m

Now, we find the change in its length for F = 100 N:

100 N = (1000 N/m)Δx

Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

<u>Final Length = 30 cm</u>

6 0

3 years ago

A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l

Lelechka [254]

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>

Consider a body of mass m kept on a weighing machine in a lift.
The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
The reaction we get as the weight recorded by the machine, and it is called the apparent weight.

<h3>How to solve the question?</h3>

Here we have given with the actual weight of the body as 100lbs.
This 100lb child is standing on the scale or the weighing machine, when it is riding .
During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
There is also<em> mg </em>downwards and a normal reaction in the upward direction.
when we equate both the upward force and downward force, we get,

$ma=mg-N\\N=mg-ma$ i.e. during riding the scale reads a weight less than that of actual weight.

When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

#SPJ4

7 0

1 year ago

Which property of sugar makes it different from white crystalline sand?

yaroslaw [1]

Answer:

The property of sugar which makes it different from sand is that sugar can dissolve in water. Explanation: however the power of dissolving of sand is not present. This is due to the presence of interaction between sugar and water.

Explanation:

I hope that was useful

5 0

3 years ago

Light of wavelength 633 nm from a He-Ne laser passes through a circular aperture and is observed on a screen 4.0 m behind the ap

Verizon [17]

Answer:

The answer is " $1.144 \times 10^{-4} \ m$ ".