A little girl kicks a soccer ball, it goes 10 feet then comes back to her, how is this possible??

What will happen to plant height if the amount of available light is reduced due to global dimming? Which type of investigation

Lorico [155]

The plant will not grow. In fact it could have all the nutrients and all the water it needs, but without a sufficient amount of light, it could die because its leaves are meant for a certain minimum amount of light.

I'll come back and see if you have posted the question you wanted and edit my answer.

6 0

3 years ago

In the lab downstairs physics majors use a rotating mirror to measure the speed of light within a few percent of the actual valu

iris [78.8K]

The number of complete cycles the rotating mirror goes through before the angular velocity gets to zero is approximately 1166.8 revs

<h3>What is angular velocity?</h3>

Angular velocity is the ratio of the angle turned to the time taken.

The kinematic equation for angular velocity are presented as follows;

ω = ω₀ + α·t

θ = θ₀ + ω₀·t + 0.5·α·t²

Where;

θ₀ = The initial angle turned = 0

ω₀ = The initial angular velocity of the mirrors = 115 rad/s clockwise

α = The angular acceleration = (115 - (-115))rad/s/(85 s) = -46/17 m/s²

t = The duration of the motion;

When the angular velocity, ω is zero, we get;

0 = 115 - 46/17·t

t = 85/2

Which indicates;

θ = 0 + 115× (85/2) + 0.5×(46/17) ×(85/2)² = 7331.25

θ = 7331.25 radians

θ = 7331.25/(2×π) ≈ 1166.8 rev

The mirrors would have turned through approximately 1166.8 revolutions when the angular gets to zero

Learn more about angular velocity and acceleration here:

brainly.com/question/13014974

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7 0

1 year ago

Can any one give me a amazon code i will repay you. plz i am trying to buy my mom somin off amazon

Alexandra [31]

Answer:i have none but good luck also this is a learning site ;-;

Explanation:

4 0

3 years ago

Where does the energy come from in a nuclear fission reaction?

poizon [28]

Answer:ni

Explanation:

nnn

4 0

3 years ago

Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de G

MakcuM [25]

Answer:

The magnitude of the electric field is 1.124 X 10⁷ N/C

Explanation:

Magnitude of electric field is given as;

$E = \frac{kq}{r^2} , N/C$

where;

E is the magnitude of the electric field, N/C

q is the point charge, C

k is coulomb's constant, Nm²/C²

r is the distance of the point charge, m

Given;

q = 5mC = 5×10⁻³ C

r = 2m

k = 8.99 × 10⁹ Nm²/C²

Substitute these values and solve for magnitude of electric field E

$E = \frac{kq}{r^2} = \frac{(8.99 X10^9)(5X10^{-3})}{2^2} = 1.124 X10^7 N/C$

Therefore, the magnitude of the electric field is 1.124 X 10⁷ N/C

7 0

3 years ago