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ExtremeBDS [4]
3 years ago
5

16.

Physics
2 answers:
kompoz [17]3 years ago
4 0

Answer:

The answer is D, free-body!

Explanation:

Free-body diagrams are used in Physics to demonstrate all the forces that are acting on an object (usually a rectangle for efficiency).

Airida [17]3 years ago
3 0
The answer is D hope this helps
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Calculate the acceleration of a 7,000 kg, single-engine airplane just before takeoff when the thrust of its
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The engine is 10,000 N I think. Hopefully this helps!!
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3 years ago
You and your friend Peter are putting new shingles on a roof pitched at 21∘. You're sitting on the very top of the roof when Pet
Marat540 [252]

Answer:

3.69 m/s

Explanation:

Forces :

mgsin Θ - mumgcosΘ = ma  

g x sinΘ  - mu x g x cosΘ  = a

9.8 x sin 21 - 0.53 x 9.8 x cos 21 = a

a = -1.337 m/s²

so you have final velocity = 0 m/s

initial velocity = ? m/s

Given d = 5.1 m

By kinematics

vf² = vo² + 2ad

0 = vo² + 2 x -1.337*5.1

vo = 3.69 m/s

8 0
4 years ago
A rock weighs 110 N in air and has a volume of 0.00337 m3 . What is its apparent weight when submerged in water? The acceleratio
pickupchik [31]

Explanation:

It is given that,

Weight of the rock in air, W = 110 N

Since, W = mg

m=\dfrac{W}{g}

m=\dfrac{110\ N}{9.8\ m/s^2}

m = 11.22 kg

We need to find the apparent weight of the rock when it is submerged in water. Apparent weight is equal to the weight of liquid displaced i.e.

M=d\times V

d is the density of water, d=1000\ kg/m^3

V is the volume of rock, V=0.00337\ m^3

M=1000\ kg/m^3\times 0.00337\ m^3

M = 3.37 kg

The apparent weight in water, W = m - M

W=7.85\ kg\times 9.8\ m/s^2

W = 76.93 N

So, the apparent weight of the rock is 76.93 N. Hence, this is the required solution.

4 0
3 years ago
mr. tolman believes that our universe is expanding, but with all of the gravitational force from the celestial bodies of space,
marishachu [46]
Based on the description, it will be most likely that he believes in Oscillating model (also known as Cyclic model) of the universe

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hope this helps
8 0
3 years ago
Read 2 more answers
A 17.3 eV electron has a 0.295 nm wavelength. If such electrons are passed through a double slit and have their first maximum at
4vir4ik [10]

Answer:

0.541 nm

Explanation:

The condition for maxima is,

dsin\theta=m\lambda

Here, m=0,1,2,.....

And d is the slit separation, m is the order of maxima, \lambda is the wavelength.

Given that, the 17.3 eV electron posses a wavelength of

\lambda=0.295 nm\\\\\lambda=0.295\times 10^{-9}m

And the order of maxima is m=1.

And the angle at which first order maxima occur is,  \theta=33^{\circ}.

Put these values in maxima condition while solving for d.

d=\frac{1\times 0.295\times 10^{-9}m}{sin33^{\circ}} \\d=\frac{0.295\times 10^{-9}m}{0.545} \\d=0.541\times 10^{-9}m}\\d=0.541 nm

Therefore, the slit separation is 0.541 nm.

7 0
3 years ago
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