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2 years ago

You decide to do an experiment to investigate Newton's Third Law. Below are several

Physics

1 answer:

Svetllana [295]2 years ago

8 0

Answer:

Explanation:

You stand on the bathroom scale and then decide to jump off and observe the

change in the reading.

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If the ambient temp. is 75 F, what is it Centigrade?

strojnjashka [21]

Explanation:

We know

C/5 = F -32/9
C/5 = 75 -32/9
C/5 = 43/9
C ≈ 25°C

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2 years ago

An 11.0 kg object traveling south with momentum of 50.0 kg-m/s collides and sticks to a 9.50 kg object traveling directly west w

Sindrei [870]

Answer:

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3 years ago

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J. J. Thomson s experiments provided evidence that an atom

gulaghasi [49]

J.J. Thomson's experiments provided evidence that an atom B.contains negatively charged particles.

<span>♡♡Hope I helped!!! :)♡♡
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3 years ago

A 2290 kg car traveling to the west at 22.3 m/s slows down uniformly. How long would it take the car to come to a stop if the fo

Elina [12.6K]

Answer:

5.72 s

Explanation:

From Newton's law, F = ma

The East is +ve direction, Hence,

F = +8930 N

m = 2290 kg

a = ?

8930 = 2290 × a

a = 8930/2290 = 3.90 m/s²

So, we will find the time it takes the car to stop using the equations of motion

a = 3.90 m/s²

u = initial velocity of the car = - 22.3 m/s (the velocity is to the west)

v = final velocity of the car = 0 m/s (since the car comes to rest)

t = time taken for the car to come to rest = ?

v = u + at

0 = - 22.3 + (3.90)(t)

3.9t = 22.3

t = 5.72 s

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2 years ago

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A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.

alisha [4.7K]

A)

It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

$S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s$

Through Definition of Velocity, comes:

$\Delta v= \frac{\Delta S}{\Delta t} \\ v_x= \frac{36}{2} \\ \boxed {v_{x}=18m/s}$

B)

Using the Velocity Hourly Equation in vertical direction, we have:

$v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}$

The angle of impact is given by:

$cos(\theta) =\frac{v_{x}}{v_{y}} \\ cos(\theta) = \frac{18}{20} \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}$

If you notice any mistake in my english, please let me know, because i am not native.

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2 years ago

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