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3 years ago

Which is the best example of Newton's first law of motion

Physics

2 answers:

Svet_ta [14]3 years ago

4 0

Object will remain without motion unless some kind of force forces it to change. Hope this helps

Black_prince [1.1K]3 years ago

3 0

I believe the best example of Newton's First Law of motion would be the example or illustration with the basketball player. An object will move in a straight line or a given direction at a constant speed unless or until another force acts upon the object, causing a change in speed and or direction.

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Arrange the stars based on their temperature. Begin with the coolest star, and end with the hottest star.

Slav-nsk [51]

Answer:

Uranus, Pluto, Neptune, Saturn , Jupiter, mars, Venus ,mercury and sun

4 0

2 years ago

HELP PLEASE!! GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted! (20pts)

anyanavicka [17]

Explanation:

potential energy =360800J

mass(m)=?

height (h)=25m

g=9.8m/s²

we have

potential energy =360800J

mgh=360800J

m×9.8×25=360800

m=360800/(9.8×25)=1472.653061kg

8 0

3 years ago

Read 2 more answers

The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a

jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A $V_a=382\ V$

Potential at $V_c=785\ V$

when particle starts from A it reaches with velocity $v_b$ at Point while when it starts from C it reaches at point B with velocity $2v_b$

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

$q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1$

Change in Kinetic Energy of particle moving under the Potential From C to B

$q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2$

Divide 1 and 2 we get

$\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}$

on solving we get

$V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c$

$V_b=\frac{743}{3}=247.67\ V$

4 0

3 years ago

Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect

Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

Where Vo = initial voltage

e = natural logarithm = 2.718

For the first capacitor 2.50 µF,

V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

4 0

3 years ago

Distinguishing between Speed and Velocity.

koban [17]

Answer:

Speed: Distance per time, 400 km/h, and a scalar quantity.

Velocity: Displacement per time, 20 m/s south, and a vector quantity.

Explanation:

Hope this helps! Please mark as brainliest.

Thanks!

8 0

3 years ago