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2 years ago

Which unequal condition causes a sea breeze to develop?

Physics

1 answer:

pickupchik [31]2 years ago

8 0

Drier air Overland then water BR &/BR are the answers so the second one the second I’m in the first option

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A ball is dropped off a cliff and falls for 24 seconds. How far did the ball fall from the top down

DerKrebs [107]

Answer:

9266 feet

Explanation:

with Earth's gravity and long it fell that's as good as it gets if there was no other factors like wind mass weight but your welcome

4 0

2 years ago

Using energy considerations, calculate the average force (in N) a 67.0 kg sprinter exerts backward on the track to accelerate fr

Illusion [34]

Answer:

$F_{sprinter}=110.4N$

Explanation:

Given data

Mass m=67.0 kg

Final Speed vf=8.00 m/s

Initial Speed vi=2.00 m/s

Distance d=25.0 m

Force F=30.0 N

From work-energy theorem we know that the work done equals the change in kinetic energy

W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²

And

$W=F_{total}.d$

$W=1/2mv_{f}^2-1/2mv_{i}^2\\F_{total}=\frac{1/2mv_{f}^2-1/2mv_{i}^2}{d} \\F_{total}=\frac{1/2(67.0kg)(8.00m/s)^2-1/2(67.0kg)(2.00m/s)^2}{25.0m} \\F_{total}=80.4N$

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N

$F_{sprinter}=F_{total}+F_{wind}\\F_{sprinter}=80.4N+30N\\F_{sprinter}=110.4N$

7 0

2 years ago

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What is addictive about vaping?

AnnZ [28]

Answer:
The last one, nicotine from tobacco

4 0

2 years ago

Two trains are traveling side-by-side along parallel, straight tracks at the same speed. In a time t, train A doubles its speed.

sergij07 [2.7K]

Answer:

derdddrdickdd

Explanation:

3 0

2 years ago

Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that:

$\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}$

$\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83$

6 0

3 years ago

Read 2 more answers