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WARRIOR [948]
1 year ago
11

A particle moving at speed 0. 32 c has momentum p0. the speed of the particle is increased to 0. 72 c. what is its momentum now?

Physics
1 answer:
Semenov [28]1 year ago
6 0

A particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c then its momentum would be 2.25P₀.

<h3>What is momentum?</h3>

It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.

Mathematically the formula of the momentum is

P = mv

where P is the momentum of the particle

m is the mass of the particle

v is the velocity by which the particle is moving

As given in the question a particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c

by using the formula of the momentum and substituting the values of the velocity

P =mv

As mentioned in the question when the particle is moving with 0.32c velocity it has a momentum of P₀

P₀ = m*(0.32c)

If the speed of the particle is increased to 0.72 c the momentum would be

P=m*(0.72c)

by dividing the second equation from the first one

P₀/P = 0.32c/0.72c

P₀/P = 0.44

P =2.25P₀

Thus, the increased momentum of the particle would be 2.25P₀

Learn more about momentum from here

brainly.com/question/17662202

#SPJ4

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malfutka [58]

The pressure at a certain depth underwater is:

P = ρgh

P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth

The pressure exerted on the submarine window is:

P = F/A

P = pressure, F = force, A = area

The area of the circular submarine window is:

A = π(d/2)²

A = area, d = diameter

Set the expressions for the pressure equal to each other:

F/A = ρgh

Substitute A:

F/(π(d/2)²) = ρgh

Isolate h:

h = F/(ρgπ(d/2)²)

Given values:

F = 1.1×10⁶N

ρ = 1030kg/m³ (pulled from a Google search)

g = 9.81m/s²

d = 30×10⁻²m

Plug in and solve for h:

h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)

h = 1540m

5 0
3 years ago
Master of physics needed
Delicious77 [7]
Hey JayDilla, I get 1/3.  Here's how:
Kinetic energy due to linear motion is:
E_{linear}= \frac{1}{2}mv^2
where
v=r \omega
giving
E_{linear}= \frac{1}{2}mr^2 \omega ^2

The rotational part requires the moment of inertia of a solid cylinder
I_{cyl} =  \frac{1}{2}mr^2
Then the rotational kinetic energy is
E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2
Adding the two types of energy and factoring out common terms gives
\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part.  Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

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4 years ago
An object moves from point A to point C along the rectangle shown in the figure below.
Naily [24]

Answer:

Hello friend where is the figure of the question

4 0
3 years ago
What is a central idea
Mice21 [21]

Answer:

The definitive and unifying theme or idea of a story or article. It encompasses all the aspects necessary to create a coherent main idea. The central idea is typically expressed as a universal truth or theme that is built and supported by the setting and characters in a story.

Explanation:

This is the correct answer to your question.

Hope this helps!!!

Kyle.

5 0
3 years ago
A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
4 years ago
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