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3 years ago

What are the physics

Physics

1 answer:

Yanka [14]3 years ago

4 0

Answer:

Explanation:

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Suppose air resistance does not exist, roughly what would be the velocity of a 10 lb object 5-seconds into free fall off of a 50

ira [324]

Answer:

Correct answer : v = 50 m/s

Explanation:

The movement described is a free fall with no initial velocity and we will use the following formula:

v = g · t we take that g = 10 m/s²

v = 10 m/s²· 5 s = 50 m/s

v = 50 m/s

God is with you!!!

4 0

3 years ago

_______________ appears on the ecg as having no p wave, a wide qrs complex, and t waves that deflect in the opposite direction f

melomori [17]

It would probably be Premature Ventricular Contractions.

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3 years ago

The index of refraction for red light in water is 1.331 and that for blue light is 1.340. A ray of white light enters the water

Alex Ar [27]

Answer:

(a) 47.08°

(b) 47.50°

Explanation:

Angle of incidence = 78.9°

For blue light :

Using Snell's law as:

$\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}$

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₂ is the refractive index for blue light which is 1.340

n₁ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{78.9}^0}=\frac {1}{1.340}$

${sin\theta_2}=0.7323$

Angle of refraction for blue light = sin⁻¹ 0.7323 = 47.08°.

For red light :

Using Snell's law as:

$\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}$

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₂ is the refractive index for red light which is 1.331

n₁ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{78.9}^0}=\frac {1}{1.331}$

${sin\theta_2}=0.7373$

Angle of refraction for red light = sin⁻¹ 0.7373 = 47.50°.

5 0

3 years ago

Should I apply for the Trusted Helpers Program? I want a real answer, if you don’t understand it don’t answer.

Svetradugi [14.3K]

Yes you should if you will like to. It is your opinion so follow your dreams if they are your dreams.

8 0

3 years ago

Read 2 more answers

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the drawstring ba

Alex

Answer:

v=39.05 m/s

Explanation:

Given that

x= 56 cm

F= 158 N

m= 58 g = 0.058 kg

Lets take spring constant = k

At the initial position,before releasing the arrow

F= k x

By putting the values

F= k x

158= 0.56 k

k=282.14 N/m

Now from energy conservation

Lets take final speed of the arrow after releasing

$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2$

k x²=mv²

282.14 x 0.56² = 0.058 v²

v=39.05 m/s

3 0

3 years ago