3 years ago

What force must be overcome to set an object in motion

Physics

2 answers:

forsale [732]3 years ago

8 0

Newtons Law of motion
HOPE IT HELPS:)

podryga [215]3 years ago

7 0

Friction must be overcome

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if a car is traveling at an average speed of 60 kilometers per hour, how long does it take to travel 12 kilometers?

VashaNatasha [74]

Im sure its twelve minutes

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3 years ago

Please help on this one

mixas84 [53]

Ep=mgh
h= Ep/mg
h=57÷(3.3×9.8)
h= 57÷32.34
h= 1.8m
So; the answer is B. 1.8m

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3 years ago

Which of the following is an example of a conductor?

nordsb [41]

Answer: B. a gold chain

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2 years ago

A circuit contains two devices that are connected in parallel. If the resistance of one of these devices is 12 ohms and the resi

finlep [7]

<u>Answer</u>

3 Ohms

<u>Explanation</u>

when the resistors are in series, the resistance in the circuit increases. For example, if two resistors, R1 and R2 are in series, the combined resistance is R1+R2.

When connected in parallel, the total resistance is the reciprocal of (1/R1 + 1/R2)

In this case the resistors are in parallel.

Total resistance = (1/12 + 1/4)⁻¹

= (1/3)⁻¹

= 3 Ohms

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2 years ago

Read 2 more answers

The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a

vampirchik [111]

Answer:

$F = 1.5 \times 10^{-16} N$

this force is $1.68 \times 10^{13}$ times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

$KE = 1 keV$

$KE = 1 \times 10^3 (1.6 \times 10^{-19}) J$

$KE = 1.6 \times 10^{-16} J$

now the speed of electron is given as

$KE = \frac{1}{2}mv^2$

now we have

$v = \sqrt{\frac{2 KE}{m}}$

$v = 1.87 \times 10^7 m/s$

now the maximum force due to magnetic field is given as

$F = qvB$

$F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})$

$F = 1.5 \times 10^{-16} N$

Now if this force is compared by the gravitational force on the electron then it is

$\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}$

$\frac{F}{F_g} = 1.68 \times 10^{13}$

so this force is $1.68 \times 10^{13}$ times more than the gravitational force

4 0

3 years ago