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viktelen [127]
3 years ago
8

Explain your observation in terms of force and Newtons 3rd law

Physics
1 answer:
Simora [160]3 years ago
4 0

Answer:

If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A

Explanation:

the law of force show that focuses will always come in 2 or pairs and it also says that one body cannot exert a force on another without experiencing a force itself.

( hope that helps :) #pk1208 is here to help )

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pantera1 [17]

let Coefficients of Friction of Rubber on asphalt (dry) =0.7

F= Coefficients of Friction * normal force = 0.7 * 60 =42 N

so the net force of the rubber is zero, meaning it will travel at a constant speed.

When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed

6 0
3 years ago
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A ball of mass 0.500 kg is carefully balanced on a shelf that is 2.70 m above the ground. What is its gravitational potential en
ratelena [41]

Answer:

The gravitational potential energy of the ball is 13.23 J.

Explanation:

Given;

mass of the ball, m = 0.5 kg

height of the shelf, h = 2.7 m

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P.E = mgh

where;

m is mass of the ball

g is acceleration due to gravity = 9.8 m/s²

h is height of the ball

Substitute the givens and solve for gravitational potential energy;

PE = (0.5 x 9.8 x 2.7)

P.E = 13.23 J

Therefore, the gravitational potential energy of the ball is 13.23 J.

8 0
3 years ago
If youre reading this goodluck in school homie <3
bazaltina [42]
I needed this ! Thanks a lot
7 0
3 years ago
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Radio waves travel at 3.00 · 108 m/s. Calculate the wavelength of a radio wave of frequency 900 kHz. (9.00 · 105 Hz.)
Anuta_ua [19.1K]
V: velocity of wave
f: frequency 
L: wavelenght

v = fL => L = v/f => L = (3x10^8)/(900x10^3) => L = 3.33 x 10^2m
7 0
3 years ago
The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. a typical axon carries a peak cu
Gemiola [76]

Answer:

6.66\cdot 10^{-12}T

Explanation:

The magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

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\mu_0 is the vacuum permeability

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r is the distance from the wire

In this problem we have

I=0.040 \mu A=4\cdot 10^{-8}A

r = 1.2 mm = 0.0012 m

So the magnetic field strength is

B=\frac{(4\pi \cdot 10^{-7} H/m)(4\cdot 10^{-8}A)}{2\pi (0.0012 m)}=6.66\cdot 10^{-12}T

5 0
3 years ago
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