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Ivahew [28]
3 years ago
5

When a container decreases in size, what will happen to an amount of gas in the container??

Physics
1 answer:
alexgriva [62]3 years ago
7 0
The gas would also decrease in size since the container lost gas to decrease the size of the container.
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A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car
Illusion [34]

The horizontal force applied is  160 N while the velocity is  2.03 m/s.

<h3>What is the speed of the car?</h3>

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

Learn more about force and work:brainly.com/question/758238

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5 0
2 years ago
The following are possible ways to express the quantity 0.391 (Give ALL correct answers, i.e., B, AC, BCD...) Note: 3.45E-8 is a
laiz [17]
To answer, evaluate the power of 10 in the given choices. If it is positve, move the decimal n places to the right. If it is negative, move the decimal n corresponding places to the left. From all the choices given, only the choices D, E, and F will give us the correct answer. 
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3 years ago
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p
Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0
2 years ago
In a tape recorder, the tape is pulled past the read-and-write heads at a constant speed by the drive mechanism. Consider the re
Irina-Kira [14]

Answer:

Torque decreases .

Explanation:

The tape is pulled at constant speed , speed v is constant , so there is

v = ω r where ω is angular speed and r is radius , As radius decreases , angular speed ω increases , So there is angular acceleration .

Let it be α . Let I be moment of inertia of reel .

Reel is in the form of disc

I = 1/2 m r²

I x α = torque

1/2 m r² x α =  torque

As the reel is untapped , its mass decreases , r also decreases , so torques also decreases .

6 0
3 years ago
A node is a point on a standing wave that has no displacement from the rest position. At the nodes, _____.
Kitty [74]

Answer: (B) There is complete destructive interference between the incoming and reflected waves

Explanation:

For example, if you pluck a guitar the waves will travel back and forth. They consist of nodes and anti-nodes. It is created, when the wave traveling to one side and bounces of the other end and comes back. As it travels to the other side, it is reflected thus, comes back. So standing waves occurs when there is interference.

When the wave is produced, the points where the string is not moving are called nodes and where they are moving are called anti-nodes. The positions where nodes are produced, destructive interference occurs and where anti-nodes are produced, constructive interference occurs

8 0
3 years ago
Read 2 more answers
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