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Evgen [1.6K]
3 years ago
8

A car’s momentum is p when it is traveling with a velocity of v. If the velocity of that car doubles, what is the new momentum o

f the car?
Physics
2 answers:
abruzzese [7]3 years ago
7 0

Answer:

c. 2p

Explanation:

edge 2021

lubasha [3.4K]3 years ago
5 0

Answer:

∆p=(m2v)kg.m/s

Explanation:

∆p=mv where v=2v. hence ∆p=m2v

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Which statement describes characteristics of a concave lens?
stiv31 [10]
I dont know what the statements are but concave lens are thinner in the middle which cause light to diverge or scatter
5 0
3 years ago
Read 2 more answers
Why do the passengers in high-altitude jet planes feel the sensation of weight while passengers in an orbiting space vehicle, su
kykrilka [37]

Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.

In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.

Remember that the Force of Gravity is given under the principle

F_g = \frac{GMm}{r^2}

Where,

G = Gravitational Universal constant

M = Mass of the planet

m = mass of the object

r = Distance from center of the planet

When the radius grows considerably the gravitational force begins to decrease.

7 0
2 years ago
A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul
GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that for spherical shell and pure rolling conditions

v = R \omega

I = \frac{2}{3}mR^2

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh

\frac{5}{6}mv^2 = mgh

h = \frac{5v^2}{6g}

h = \frac{5(8^2)}{6(9.81)} = 5.44 m

Part b)

If ball is not rolling and just sliding over the hill then in that case

\frac{1}{2}mv^2 = mgh

h = \frac{v^2}{2g}

h = \frac{8^2}{2(9.81)} = 3.16 m

3 0
3 years ago
Explain electric circuits. Be sure to include the features, the roles, and the types. Help PLEASEEEE!!
maksim [4K]
If there was any way to do that, then your teacher wouldn't
need to keep you coming into class every day and doing
homework every night.  She could just give you the 3 or 4
paragraphs and a few pictures that you're asking me for,
and bada-bing ! you'd know it !

The time it takes, and the amount of homework it takes, is
EXACTLY the time you spent hearing about it in class.

(Unless you're some kind of genius savant prodigy, which
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5 0
3 years ago
A ball is tossed vertically upward. When it reaches its highest point (before falling back downward) Group of answer choices the
nignag [31]

Answer:

the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward

Explanation:

Is this exercise in kinematics

          v = v₀ - g t

where g is the acceleration of the ball, which is created by the attraction of the ball to the Earth.

At the highest point

velocity must be zero.

The acceleration depends on the Earth therefore it is constant at this point and with a downward direction.

The force of the earth on the ball is towards the center of the Earth, that is, down

all other alternatives are wrong

7 0
3 years ago
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