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MrMuchimi
3 years ago
6

Will give brainliest!!

Physics
1 answer:
ruslelena [56]3 years ago
8 0

Answer:

iam not sure but I think its NaNO3

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To describe velocity you need to know _______ and _________
Crazy boy [7]
To get the velocity you need to know the displacement and the time.

I hope that helps!
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Can electromagnetic waves travel in a vacuum? <br><br> Yes <br> Or<br> No
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Yes

Explanation:

Yes because Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.

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3 years ago
samantha a high school teacher wants to spend more time with her children and is looking for a job in her field. what would be a
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She could look for an online job or a job she can work at from home.

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A stuntman drives a car with a mass of 1500 kg on a drawbridge. The car accelerates with a constant force of 10,000 N. While he
Viktor [21]

Answer:

1.77 m/s^2

Explanation:

There are two forces acting on the car along the direction parallel to the incline:

- The driving force of 10,000 N, which pushes forward

- The component of the weigth of the car parallel to the incline, which pulls backward

The component of the weight of the car parallel to the incline is:

W_p = mg sin \theta=(1500 kg)(9.8 m/s^2)( sin 30^{\circ})=7350 N

So now we can apply Newton's second law to find the acceleration of the car:

F-W_p = ma\\a=\frac{F-W_p}{m}=\frac{10000 N-7350 N}{1500 kg}=1.77 m/s^2

7 0
3 years ago
What must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between t
vlabodo [156]

Answer:

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² ...................................... Equation 1

Where F = electrostatic force  between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

6 0
3 years ago
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