(b) The refractive index of the glass of the prism is 1.49. The ray EF is refracted at F. Use

An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

FromTheMoon [43]

Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

$F_{drag}=12v+4v^2$

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

$F_{drag}=mg$

$12v+4v^2=80\times 9.8$

$4v^2+12v=784$

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.

6 0

1 year ago

A 63.0 kg astronaut is on a spacewalk when the tether line to the shuttle breaks. the astronaut is able to throw a spare 10.0 kg

Llana [10]

There are other forces at work here nevertheless we will imagine it is just a conservation of momentum exercise. Also the given mass of the astronaut is light astronaut.

The solution for this problem is using the formula: m1V1=m2V2 but we need to get V1:

V1= (m2/m1) V2

V1= (10/63) 12 = 1.9 m/s will be the final speed of the astronaut after throwing the tank.

6 0

2 years ago

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A 16-Ω loudspeaker, an 8.0-Ω loudspeaker, and a 4.0-Ω loudspeaker are connected in parallel across the terminals of an amplifier

kolbaska11 [484]

Answer:

2.286 ohm

Explanation:

R1 = 16 ohm

R2 = 8 ohm

R3 = 4 ohm

They all are connected in parallel combination

Let the equivalent resistance is R.

1/R = 1/R1 + 1/R2 + 1/R3

1/R = 1/16 + 1/8 + 1/4

1/R = (1 + 2 + 4) / 16

1/R = 7 / 16

R = 16/7 = 2.286 ohm

6 0

2 years ago

Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p

worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

$U1/U2=6$
$U1/U2=6$

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy is mathematically given as

$U=\frac{-k*qQ}{r}$

Therefore

The Equations of U1 and U2 is

For U1

$U1=\frac{-k*q_1Q}{r}$

For U2

$U2=\frac{-k*q_1Q}{3*2r}$

Since

U is a function of q and q2=q1/3

Therefore

$U1/U2=6$

For Question 2

For U1

$U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}$

Therefore

$U1/U2=6$

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7 0

1 year ago

Why do charges build up on clothing in an electric dryer?

Rzqust [24]

Charges build up when you have dry air and friction ,the heat to clothes which dry it out and causes friction.

4 0

2 years ago

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