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3 years ago

Which of your groups should you NOT change anything for? (in other

Physics

1 answer:

ella [17]3 years ago

7 0

Answer:

Explanation:

The control is something that is meant to not be changed, the control is a comparison of the experimental.

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A light year is defined as the distance that light can travel in 1 year. What is the value of 1 light year in meter? Show your c

Ipatiy [6.2K]

Answer:

d=9.462×10^15 meters

Explanation:

<u>Relation between distance, temps and velocity:</u>

d=v*t

t=1year*(365days/1year)*/(24hours/1day)*(3600s/1h)=31536000s

So:

1 light year=d=3*10^8m/s*3.154*10^7s=9.462×10^15 meters

6 0

2 years ago

Three cars collide, a 1500 kg sports car, a 1750 kg family car, and a 1200 kg compact car. Which experiences the greatest change

sweet [91]

Answer:

Explanation:

This is because momentum is defined as p = mv

delta p = Force *time

neither velocity nor time is given so a conclusion cannot be made on which has the greatest momentum change.

5 0

2 years ago

Choose the true statement concerning the mall or subatomic particles:

solniwko [45]

C is true, and just one of those has as much mass as about 1,840 electrons.

4 0

3 years ago

Read 2 more answers

What is the magnitude of the force that lifts a 3-kilogram object straight upwards?

frez [133]

Any force of 29.4 Newtons or greater can do it.

4 0

3 years ago

Read 2 more answers

A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re

lbvjy [14]

Answer:

The beam of light is moving at the peed of:

$\frac{dy}{dt} = \frac{80\pi}{3}$ km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, $\omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi$ (1)

Now, in the right angle in the given fig.:

$tan\theta' = \frac{y}{3}$

Now, differentiating both the sides w.r.t t:

$\frac{dtan\theta'}{dt} = \frac{dy}{3dt}$

Applying chain rule:

$\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}$

$sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}$

Now, using $tan\theta = \frac{1}{m}$ and y = 1 in the above eqn, we get:

$(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}$

Also, using eqn (1),

$8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}$

$\frac{dy}{dt} = \frac{80\pi}{3}$

7 0

2 years ago