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Paraphin [41]
3 years ago
12

A 3-kg mass is in free fall. What is the velocity of the mass after 7 seconds??

Physics
1 answer:
dmitriy555 [2]3 years ago
6 0
69 ms. I hope that helps
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It has been proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of f
Anestetic [448]

Answer:

A = 8.34 x 10^(5) m²

Explanation:

The intensity of the solar radiation is the average solar power per unit area. Thus,

I = P/4A = P/(4(πr²))

Where;

P is average solar power.

r is it's distance from centre of sun

A is area

Now, The rate at which the Sun emits energy has a standard value of 3.90 × 10^(26) W

Thus;

I = [3.90 × 10^(26)]/(4πr²)

I = [3 x 10^(25)]/r²

Now, the formula for radiation pressure is;

P_rad = 2I/c

Where c is speed of light and has a value of 3 x 10^(8) m/s

Thus,

P_rad = 2([3 x 10^(25)]/r²)/(3 x 10^(8))

P_rad = [2.07 x 10^(17)]/r² N/m²

Also, Radiation pressure on ship; P_rad = F/A

Where Force on ship and A is area.

Thus;

F = P_rad x A

So, F = [2.07 x 10^(17)•A]/r²

Now,

F_grav = GMm/r²

Where;

G is gravitational constant with a value = 6.67 x 10^(-11) Nm²/kg²

M is mass of sun with a value of 1.99 x 10^(30)

m is mass of ship and sail = 1300 kg

Thus, plugging in the relevant values to obtain;

F_grav = (6.67×10^(-11) × 1.99 x 10^(30) × 1300)/r²

F_grav = [17.255 x 10^(22)]/r²

Now, equating F to F_grav, we get;

[2.07 x 10^(17)•A]/r² = [17.255 x 10^(22)]/r²

r² will cancel out to give;

2.07 x 10^(17)•A= [17.255 x 10^(22)]

A = [17.255 x 10^(22)]/2.07 x 10^(17)

A = 8.34 x 10^(5) m²

7 0
3 years ago
A basketball player has made 60​% of his foul shots during the season. Assuming the shots are​ independent, find the probability
marissa [1.9K]

Answer:

a) 0.0864

b) 0.0384

c) 0.936

Explanation:

Probability that he makes his shot, P(A) = 0.6

Probability that he doesn't make the shot, P(A') = 1 - P(A) = 1 - 0.6 = 0.4

a) Probability that he Misses for the first time on his fourth attempt

P(A) × P(A) × P(A) × P(A') = 0.6 × 0.6 × 0.6 × 0.4 = 0.0864

b) Probability that he Makes his first basket on his fourth shot

P(A') × P(A') × P(A') × P(A) = 0.4 × 0.4 × 0.4 × 0.6 = 0.0384

c) Probability that he Makes his first basket on one of his first 3 shots

Sum of the probabilities that he makes all three first shots, two of the first three shots and one of the first three shots with the order irrelevant.

- Probability that he makes all first three shots = P(A) × P(A) × P(A) = 0.6 × 0.6 × 0.6 = 0.216

- Probability that he makes two out of the first three shots = (P(A) × P(A) × P(A')) + (P(A) × P(A') × P(A)) + (P(A') × P(A) × P(A)) = 3(0.6 × 0.6 × 0.4) = 0.432 (it's multipled by 3 because the probability is the same regardless of order)

- Probability of making only one of the first three shots = (P(A) × P(A') × P(A')) + (P(A') × P(A) × P(A')) + (P(A') × P(A') × P(A)) = 3(0.6 × 0.4 × 0.4) = 0.288 (It's multiplied by 3 too because the probabilities are the same too, regardless of order).

Probability that he Makes his first basket on one of his first 3 shots = 0.216 + 0.432 + 0.288 = 0.936

5 0
3 years ago
If the box weighs 40 newtons and is lifted a distance of 2 meters, how much work is done on the box?
Zarrin [17]

Answer:

The work done on the box is 80 J.

Explanation:

Given that,

Weight of box = 40 N

Distance = 2 meter

We need to calculate the work done

Using formula of work done

W=F\times x

W=mg\times x

Where, x = distance

mg = weight

Put the value into the formula

W=40\times2

W= 80\ Nm

W=80\ J

Hence, The work done on the box is 80 J.

5 0
3 years ago
The red lines on this
tino4ka555 [31]
The circular lines you see on the chart are isobars, which join areas of the same barometric pressure.
8 0
3 years ago
A raised plaster cover (often called a plaster ring or mud ring) is permitted to increase the maximum number of conductors permi
andreev551 [17]

Answer:

When it is marked with its cubic-inch volume

Explanation:

Because this allows for best and efficient identification

6 0
3 years ago
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