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2 years ago

Why do scientists build models.

Physics

2 answers:

Lisa [10]2 years ago

5 0

D all of the above all of them fit the question

zvonat [6]2 years ago

3 0

Answer: D. All of the Above

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A student buys a 5000 Btu window air conditioner for his apartment bedroom. He monitors it for one hour on a hot day and determi

pochemuha

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

A small ball is attached to one end of a spring that has an unstrained length of 0.169 m. The spring is held by the other end, a

Nadya [2.5K]

The ball moves around in circular motion. The centripetal force keeping it in circular motion is given by:

F = mv²/r

F = centripetal force, m = mass of ball, v = velocity of ball, r = radius of motion

The force the spring exerts on the ball is given by:

F = kΔx

F = spring force, k = spring constant, Δx = change of spring length

The spring provides the centripetal force that keeps the ball in circular motion, so set the spring force equal to the centripetal force:

kΔx = mv²/r

Let's do some algebra:

m/k = rΔx/v²

Now if we attach the same spring to the ceiling with the ball still fixed to one end and let the ball hang in static equilibrium, the spring force would balance the ball's weight:

kΔx' = mg

k = spring constant, Δx' = new change in spring length, m = mass, g = gravitational acceleration

Let's do some more algebra:

Δx' = gm/k

Substitute m/k with our previous result, rΔx/v²:

Δx' = grΔx/v²

Given values:

g = 9.81m/s²

r = 0.169m + 0.0131m = 0.1821m

Δx = 0.0131m

v = 3.15m/s

Plug in the values and solve for Δx':

Δx' = 9.81(0.1821)(0.0131)/3.15²

Δx' = 0.0024m

5 0

2 years ago

PLEASE HELP ILL MARK BRAINLIEST!!! A ball is initially thrown downwards with an initial speed of 20 m/s from the top of a 300 m

Sholpan [36]

Using the 3rd equation of motion:

= v² - u² = 2gs ------ [g = Acceleration due to gravity]

= v² - 20² = 2 × 10 × 300

= v² - 400 = 6000

= v² = 6000 - 400

= v = √5600

= v = 74.83 m/s

And yeah it's done :)

8 0

3 years ago

A pendulum consisting of a 0.5 kg mass tied to a 0.3 m string is set into oscillation at the same moment that a stone is dropped

lara31 [8.8K]

Answer:

2.72 cycles

Explanation:

First of all, let's find the time that the stone takes to reaches the ground. The stone moves by uniform accelerated motion with constant acceleration g=9.8 m/s^2, and it covers a distance of S=44.1 m, so the time taken is

$S=\frac{1}{2}at^2\\t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(44.1m)}{9.8 m/s^2}}=3 s$

The period of the pendulum instead is given by:

$T=2 \pi \sqrt{\frac{L}{g}}=2 \pi \sqrt{\frac{0.3 m}{9.8 m/s^2}}=1.10 s$

Therefore, the number of oscillations that the pendulum goes through before the stone hits the ground is given by the time the stone takes to hit the ground divided by the period of the pendulum:

$N=\frac{t}{T}=\frac{3 s}{1.10 s}=2.72$

6 0

2 years ago

A 45 N girl sits on a bench 0.6 meters off the ground. How much work is done on the bench?

ycow [4]

Answer: 27 joules

Explanation:

Work is done when force is applied on the bench over a distance. it is measured in joules.

Workdone = force x distance

= 45 N x 0.6 metres

= 27 joules

Thus, 27 joules of work is done on the bench.

6 0

3 years ago