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lilavasa [31]
2 years ago
14

What relation does distance have with force and mass​

Physics
1 answer:
Reika [66]2 years ago
4 0

This relationship was first published by Sir Issac Newton. His law of universal gravitation says that the force (F) of gravitational attraction between two objects with Mass1 and Mass2 at distance D is: F = G(mass1*mass2)/D squared

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A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5
svlad2 [7]
The magnetic force experienced by the proton is given by
F=qvB \sin \theta
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and \theta the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so \sin \theta=1 and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:
F=ma

So we have
ma=qvB
from which we can find the magnitude of the field:
B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T
4 0
2 years ago
when an incandescent light bulb is turned on, its thin wire filament heats up quickly. as the temperature of this wire filament
Leno4ka [110]

Well, we usually assume that the resistance of a circuit component
is constant and doesn't change.  But the truth is that for anything
that conducts current, its resistance always increases somewhat
when it warms up. 

For things like light bulbs, electric toasters, space heaters, electric
stove burners, the heat coils in a blow-dryer ... anything that's
designed to be really hot when it's doing its job ... the resistance
of those things increases significantly when they come up to their
operating temperatures.
6 0
3 years ago
Which of the following is closest to 2cm?
Sloan [31]
C because the the smallest thing then the other ones because it never said what kind of size of it
4 0
3 years ago
Two forces are applied on a body. One produces a force of 480-N directly forward while the other gives a 513-N force at 32.4-deg
n200080 [17]

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N

Thus, the net force over the body is:

F=(913.14N)\hat{i}+(274.87N)\hat{j}

Next, you calculate the magnitude of the force:

F=\sqrt{(913.14N)+(274.87N)^2}=953.61N

and the direction is:

\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°

7 0
3 years ago
A football punter accelerates a .55 kg football
Ronch [10]

Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

F=\frac{\Delta p}{\Delta t}

where

\Delta p is the change in momentum of the football

\Delta t=0.25 s is the time elapsed

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.55 kg is the mass of the football

u = 0 is the initial  velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we  find the force exerted on the ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N

5 0
3 years ago
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