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Andrei [34K]
2 years ago
15

A coin is dropped. What is its displacement after 0.30 s.

Physics
1 answer:
Morgarella [4.7K]2 years ago
3 0
Assuming this coin is on earth and that it wasn’t dropped forcefully:
Use the formula d = 1/2at^2. Rewriting using a=g and solving for height h gets us h = 1/2(9.8)t^2.
In this case that would get that the change in height h is 0.5(9.8)(0.3^2) = 0.441 m.
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Sleep is a difficult problem for dolphins because dolphins __________. A. need more sleep than their predators B. must be awake
Alex17521 [72]
Need more sleep than their predators
5 0
2 years ago
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A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?
Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

H_m = \frac{u^2sin^2 \theta}{2g}

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta  = sin^{-1}(0.2123)\\\\\theta  = 12.26^0

Therefore, the angle of projection is 12.26⁰.

6 0
2 years ago
Pleaseee HElpp!!!!!!!
lutik1710 [3]

Coulomb's Law

Given:

F = 3.0 x 10^-3 Newton

d = 6.0 x 10^2 meters

Q1 = 3.3x 10^-8 Coulombs

k = 9.0 x 10^9 Newton*m^2/Coulombs^2

Required:

Q2 =?

Formula:

F = k • Q1 • Q2 / d²

Solution:

So, to solve for Q2

 

Q2 = F • d²/ k • Q1

Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9 Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)

Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)

Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)

Then, take the reciprocal of the denominator and start multiplying

Q2 = 1080 • 1 Coulombs/297

Q2 = 1080 Coulombs / 297

Q2 = 3.63636363636 Coulombs

Q2 = 3.64 Coulumbs

6 0
3 years ago
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Water drips from a shower head (the sprayer at the top of the shower) and falls onto the floor 2.3 m below. The droplets are fal
Maurinko [17]

Answer:

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Explanation:

We are given that the time interval between each droplet is equal.

We are also given that the fourth drop is just dripping from the shower when the first hits the floor.

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The distance between each drop has to be the same given that they fall at equal time intervals.

Let this distance be x.

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The answer is 300 km

To the destination
4 0
3 years ago
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