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3 years ago

Some machines do not multiply the force that is applied to them. Question 20 options: True False

Physics

1 answer:

olga55 [171]3 years ago

3 0

Explanation:

it is true that some machines do not multiply the force that is applied on them

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Please help

natali 33 [55]

Answer:

Thermosphere

temperature continue to rise up to 1500*C

Ionosphere

Mesosphere

temperature falls to -93*C

Stratosphere

protective ozone layer

UV radiation causes temperature to rise

Troposphere

78% nitrogen, 21% oxygen

temperature drops from 17 to -52*C

Explanation:

3 0

3 years ago

A particle beam is made up of many protons, each with a kinetic energy of 3.25times 10-15 J. A proton has a mass of 1.673 times

ArbitrLikvidat [17]

Answer:

The magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 1.01 x $10^{-4}$ N/C

Explanation:

given information,

kinetic energy, KE = 3.25 x $10^{-15}$ J

proton's mass, m = 1.673 x $10^{-27}$ kg

charge, q = 1.602 x $10^{-19}$ C

distance, d = 2 m

to find the electric field that will stop the proton, we can use the following equation:

E = F/q

= (KE/d) / q , KE = Fd --> F = KE/d

= KE/qd

= (3.25 x $10^{-15}$ J) / (1.602 x $10^{-19}$ C)(2 m)

= 1.01 x $10^{-4}$ N/C

3 0

3 years ago

The parents of a young child carefully placed plastic plug protectors in all the electrical outlets in their house. Which statem

DanielleElmas [232]

I think the answer is c or b but i p ick c

4 0

3 years ago

Read 2 more answers

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

$r = 7.08 \times 10^{-3} m$

so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

6 0

3 years ago

A 50.0-kg wolf is running at 10.0 m/sec. What is the wolfs kinetic energy

Fiesta28 [93]

Kinetic energy = mass time squared speed divided by 2
W=mv^2/2 = 50*10*10/2 = 2500 J

4 0

3 years ago