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Vinil7 [7]
3 years ago
10

A motorcycle is travelling at a constant velocity of 90 km/h (no acceleration). How many

Physics
1 answer:
Mazyrski [523]3 years ago
8 0
2km*1000=2000m
S=ut+1/2at^2
a=0,thus s=ut
2000=90t
t=2000/90
=22.22s(to 4 s.f)
Please mark as brainliest
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A race car starting from rest accelerates uniformly at 4.9 m/s^2. What is the car's speed after it has traveled 200 meters?
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The equation v²=u²+2as will help with solving this equation
v is what we are trying to find
u=0
a=4.9
s=200
v²=0²+2*4.9*200
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v=√1960
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5 0
4 years ago
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A woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 50.2 s.
lorasvet [3.4K]

Explanation:

(a) The period of a wave is the time required for one complete cycle. In this case, we have the time of five cycles. So:

T=\frac{t}{n}\\\\T=\frac{50.2s}{5}\\T=10.04s

(b) The frequency of a wave is inversely proportional to its period:

f=\frac{1}{T}\\f=\frac{1}{10.04s}\\f=0.01Hz

(c) The wavelength is the distance between two successive crests, so:

\lambda=30.2m

(d) The speed of a wave is defined as:

v=f\lambda\\v=(0.1Hz)(30.2m)\\v=3.02\frac{m}{s}

8 0
3 years ago
A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. the wire ful
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1.875*10^-5

Explanation:

8 0
3 years ago
Which of the following is a contact force?
3241004551 [841]

Answer:

b friction

Explanation:

Contact forces

Contact forces are forces that act between two objects that are physically touching each other. Examples of contact forces include:

Reaction force

An object at rest on a surface experiences reaction force. For example, a book on a table.

A box rests on a table. There are two arrows, equal in size but going in opposite directions, up and down, from the point where the box meets the table.

Tension

An object that is being stretched experiences a tension force. For example, a cable holding a ceiling lamp.

A box hangs from a rope. Two arrows which are equal in size act upwards and dowards from the top and bottom of the rope.

Friction

Two objects sliding past each other experience friction forces. For example, a box sliding down a slope.

A box rests on an incline. There are three arrows; one acting vertically downwards from the centre of the box’s base. One arrow acts perpendicular to the incline. One arrow acts up the incline.

Air resistance

An object moving through the air experiences air resistance. For example, a skydiver falling through the air.

A box falls from the sky. Two arrows, equal in size and opposite in direction act upwards from the box and downwards from the box

When a contact force acts between two objects, both objects experience the same size force, but in opposite directions. This is Newton's Third Law of Motion.

3 0
3 years ago
A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou
Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

<u>Mathematical expression for the Newton's second law of motion is given as:</u>

F=\frac{dp}{dt} ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

p=m.v

Now, equation (1) becomes:

F=\frac{d(m.v)}{dt}

<em>∵mass is constant at speeds v << c (speed of light)</em>

\therefore F=m.\frac{dv}{dt}

and, \frac{dv}{dt} =a

where: a = acceleration

\Rightarrow F=m.a

also

F\propto \frac{1}{dt}

so, more the time, lesser the force.

<em>& </em><u><em>Impulse:</em></u>

I=F.dt

I=m.a.dt

I=m.\frac{dv}{dt}.dt

I=m.dv=dp

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0
3 years ago
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