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Alexandra [31]
3 years ago
11

Riding in a car, you suddenly put on the brakes. As you experience it inside the car, do Newton's law apply? Do they apply as se

en by someone outside the car? Please explain reasoning, thank you!
Physics
1 answer:
alisha [4.7K]3 years ago
3 0

Answer with Explanation:

Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.

For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as  pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference

While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.  

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A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k
Luba_88 [7]

Answer:

The frictional force  F_{fri} = 6.446 N

The acceleration of the block a = 6.04 \frac{m}{s^{2} }

Explanation:

Mass of the block = 3.9 kg

\theta = 40°

\mu = 0.22

(a). The frictional force is given by

F_{fri} = \mu R_{N}

R_{N} = mg \cos \theta

R_{N} = 3.9 × 9.81 × \cos 40

R_{N} = 29.3 N

Therefore the frictional force

F_{fri} = 0.22 × 29.3

F_{fri} = 6.446 N

(b). Block acceleration is given by

F_{net} = F - F_{fri}

F = 30 N

F_{fri} = 6.446 N

F_{net} = 30 - 6.446

F_{net} = 23.554 N

The net force acting on the block is given by

F_{net}  = ma

23.554 = 3.9 × a

a = 6.04 \frac{m}{s^{2} }

This is the acceleration of the block.

8 0
3 years ago
A grindstone in the shape of a solid disk with a diameter of 1.0 m and a mass of 50.0 kg, is rotating at 900 rev/min. A tool is
monitta

Answer:

oh for real?

Explanation:

The solubility of glucose at 30°C is

125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.

8 0
3 years ago
At what distance will a 80 W lightbulb have the same apparent brightness as a 120 W bulb viewed from a distance of 40 m
liq [111]

Answer:

32.6mm

Explanation:

Using area of a sphere(bulb) = 4πr²

So A is proportional to radius²

So the Energy will be proportional to r²

But 120/80 = 1.5 is the energy factor so

Using

1.5/d² = 1/r²

1.5/40²= 1/r^2

r = √( 40²/ 1.5)

r = 32.6m

4 0
3 years ago
7. Which law describes when a person lands on a
marissa [1.9K]

Answer:

Newton's Third Law

Explanation:

Newton's third law

Newton's third law: “for every action, there is an equal and opposite reaction.” This is where you get the bounce. When you push down on the trampoline (or fall downward onto the trampoline bed), Newton's third law says that an equal and opposite reaction pushes back.

:)

6 0
3 years ago
Read 2 more answers
The mass of the car?
Nonamiya [84]

Answer:

1050 kg

Explanation:

The formula for kinetic energy is:

KE (kinetic energy) = 1/2 × m × v² where <em>m</em> is the <em>mass in kg </em>and <em>v</em> is the velocity or <em>speed</em> of the object <em>in m/s</em>.

We can now substitute the values we know into this equation.

KE = 472 500 J and v = 30 m/s:

472 500 = 1/2 × m × 30²

Next, we can rearrange the equation to make m the subject and solve for m:

m = 472 500 ÷ (1/2 × 30²)

m = 472 500 ÷ 450

m = 1050 kg

Hope this helps!

3 0
2 years ago
Read 2 more answers
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