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katen-ka-za [31]
3 years ago
15

Dave throws a 10 kg bowling ball straight up in the air. At the very tippy-top of its path, what is it's momentum?

Physics
1 answer:
svetoff [14.1K]3 years ago
8 0

Answer:Zero

Explanation:

Given

mass of ball m=10\ kg

If the ball is thrown upward then at maximum point velocity of ball is zero because ball is no longer able to move upward

Momentum(P) of a particle is given by

P=mass\times velocity

P=10\times 0

P=0

Therefore at the highest point momentum is zero .

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A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
A 18.0-kg rock is sliding on a rough, horizontal surface at 7.10 m/s and eventually stops due to friction. the coefficient of ki
Bond [772]
A = .3*g = 2.94 m/s² 

<span>t = v/a = 9/2.94 = 3.061 sec </span>

<span>W = E/t = ½mv²/t = ½*40*9²/3.061 = 529.2 watts</span>
4 0
3 years ago
You have a mass of 60-kg, and you are facing your friend who has a mass of 100 kg. You skate towards
White raven [17]

Answer: My initial velocity is 5 m/s.

Explanation:

In this case, momentum can be conserved.

initial momentum = final momentum

Since both the bodies come to rest after collision,

Final momentum = 0

Let my velocity be v, and mass, m1 = 60 kg

Friend's mass, m2 = 100 kg

Friend's velocity, v2 = 3 m/s

Intial momentum = m1v + m2v2

= 60v + 300

Conserving momentum,

60v + 300 = 0

v= -5 m/s

( Negative sign indicates that me and my friend are moving in opposite directions that is towards each other)

4 0
3 years ago
Which picture best demonstrates the phenomenon of diffraction?
timama [110]

Answer:

The image to the left (with the disks on it)

Explanation:

Interference in any type of wave can be gotten in two forms, constructive interference, and destructive interference.

The constructive interference is between two waves with the same phase, that is, each crest and trough correspond with the crest and trough of the another getting as result a wave with twice the amplitude of the original one.

The destructive interference is between two waves out of phase, in which the crest of one cancels with the trough of another.

If light passes for a slit it will get a diffraction pattern in a screen, at which each bright pattern corresponds to a crest and a dark pattern to a trough, as a consequence of constructive interference and destructive interference in different points of its propagation to the screen.

The circular shape of the disks can be a representation of the wavefront and how the overlaps make constructive and destructive interference in order to get the diffraction pattern.

5 0
3 years ago
Self-registering thermometers are used for:
Angelina_Jolie [31]

Self-registering thermometers are used for temperature or depth profile. It registers the maximum and minimum temperature of the substance occurring in the interval of time between two consecutive settings of the instrument.

5 0
3 years ago
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