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Kisachek [45]
3 years ago
13

A ball on a cart is moving at a rate of 2 m/s. The cart suddenly stops and the ball continues to travel in the same direction at

the same speed. This is an example of which Newton’s Laws?
Physics
2 answers:
sukhopar [10]3 years ago
7 0

Answer:

The first

Explanation:

Cause it will continue in motion till another force is applied

Nitella [24]3 years ago
3 0

Newton's law of inertia

Explanation:

I remember reading it two years ago in the science book it's not law is called Newton's law of inertia you can also look it up on Google to make sure

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How much power will be required to force a current of 4.13 amps to flow through a conductor whose resistance is 113 ohms? Use tw
-Dominant- [34]

The power required to force the current of 4.13 A to flow through the conductor is 1927.43 watts

<h3>What is power? </h3>

This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:

Power (P) = square current (I²)× resistancet (R)

P = I²R

<h3>How to determine the power</h3>
  • Current (I) = 4.13 A
  • Resistance (R) = 113 ohms
  • Power (P) =?

P = I²R

P = 4.13² × 113

P = 1927.43 watts

Thus, the power required is 1927.43 watts

Learn more about electrical power:

brainly.com/question/64224

#SPJ1

7 0
2 years ago
How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

T_{f} =  (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

T_{f}  = 466.8 K

so change in temperature ΔT

ΔT =  466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x_{i\\}² - x_{f\\}² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q =  16 J

Therefore, the required heat energy is 16 J

5 0
3 years ago
Sonia's grandfather was not able to see clearly after returning from his walk on a bright sunny day. He was about to hit a chair
Vinil7 [7]

Answer:

Sonia's grandfather was not able to see clearly after returning from his walk on a bright sunny day. He was about to hit a chair when Sonia held him and guided him to the nearby sofa, in sunny environment grandfather's eye is set according to high brightness, his pupils become small to lower the number of light rays entering his eye, when he returned from sunny environment to his house which was having low brightness so the pupils should enlarge to absorb more light to see clearly but due to old age his Ciliary muscle of the eye would have worn out and due to this poor coordination the image was not clear.

4 0
3 years ago
Who water rocket starts from rest and roses straight up with an acceleration of 5 m/s until it runs out of water 2.5 seconds lat
Kitty [74]

Answer:

23. 4375 m

Explanation:

There are two parts of the rocket's motion

1 ) accelerating  (assume it goes upto  h1 height )

using motion equations upwards

s = ut+\frac{1}{2}*a*t^{2} \\h_1=0+\frac{1}{2}*5*2.5^{2} \\=15.625 m

Lets find the velocity after 2.5 seconds (V1)

V = U +at

V1 = 0 +5*2.5 = 12.5 m/s  

2) motion under gravity (assume it goes upto  h2 height )

now there no acceleration from the rocket. it is now subjected to the gravity

using motion equations upwards (assuming g= 10m/s² downwards)

V²= U² +2as

0 = 12.5²+2*(-10)*h2

h2 = 7.8125 m

maximum height = h1 + h2

                            = 15.625 + 7.8125

                            = 23. 4375 m

3 0
3 years ago
An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert
cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.                

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, k=4.2\times 10^5\ N/m

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

mgh=\dfrac{1}{2}kx^2

x is the compression in spring

x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m

So, the compression in the spring is 5.88 meters.                                                                                                                  

6 0
3 years ago
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