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SCORPION-xisa [38]
4 years ago
6

Impulse is the product of a force and the blank during which that force acts

Physics
1 answer:
vovikov84 [41]4 years ago
5 0
The product of the net force and the time interval for which the the force acts is Law of Momentum
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Why won't a very bright beam of red light impart more energy to an ejected electron than a feeble beam of violet light?
bearhunter [10]
This is related to the energy carried by photons of light the energy of each photon is proportional to the frequency of the light since red light has a lower frequency then violet light and photons of red light carry less energy than the photons of violet light as a result the red protons eject electrons that have less energy than the ejected electrons by Violet photons
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What action force is used when sitting down
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When sitting down there is gravity , when sat down the chair is pushing back at a equal but opposite force
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4 years ago
Can someone please help me with these physics problems? I just don’t even know where to start.
KIM [24]

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

3 0
4 years ago
What characteristics are shared by both heating and cooling curves? choose 4
Luda [366]

The characteristics shared by both heating and cooling curves is that phases changes when the temperature stays the same.

<h3>What is heating and cooling?</h3>

Heating is the process by which the temperature of a place or substances or object is increased.

Cooling Is the process of lowering the temperature of a substance or object by using conditioners or by freezing .

Therefore, The characteristics shared by both heating and cooling curves is that phases changes when the temperature stays the same.

Learn more about heating and cooling below.

brainly.com/question/9938370

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6 0
2 years ago
a proton moves at a speed of 2.0 x 10^7 m/s at right angles to a magnetic field that is directed into the page with a magnitude
nydimaria [60]
The Lorentz force acting on the proton is equal to:
F=qvB \sin \theta
where
q is the proton charge
v is the proton speed
B is the intensity of the magnetic field
\theta is the angle between the direction of v and B

since the proton travels perpendicularly to the magnetic field, in this case \theta=90^{\circ}, so the Lorentz force in this case is simply
F=qvB

The magnetic force provides also the centripetal force that keeps the proton in circular motion:
m \frac{v^2}{r}=qvB
where 
m is the proton mas
r is the radius of the orbit

If we re-arrange this equation and we use the data of the problem, we can find the radius of the orbit:
r= \frac{mv}{qB}= \frac{(1.67 \cdot 10^{-27} kg)(2.0 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19} C)(0.1 T)} =2.09 m
7 0
3 years ago
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