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Burka [1]
2 years ago
14

A football player kicks a field goal from a distance of 45 m from the goalpost. The football is launched at a 35° angle above th

e horizontal. What initial velocity is required so that the football just clears the goalpost crossbar that is 3.1 m above the ground? Ignore air resistance and the dimensions of the football.
Physics
1 answer:
daser333 [38]2 years ago
7 0

Answer:  try searching It up on go0gle

Explanation:

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A physical quantity is defined as mass per uint volume .what is its si unit.


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If one light bulb burns out in a string of lights, and the rest stay lit, it is reasonable to assume that the lights are wired i
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The lights are wired in PARALLEL.

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3 years ago
A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35m x 0.55m. The magnetic field has a
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Answer:

E = 0.38V

Explanation:

See attachment below.

4 0
3 years ago
The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the
Evgen [1.6K]

Question Continuation

Derive an expression for x in terms of m, M, and D. b) If the net force is zero a distance ⅔D from the planet, what is the ratio R of the mass of the planet to the mass of the moon, M/m?

Answer:

a. x = (D√M/m)/(√M/m + 1)

b. The ratio R of the mass of the planet to the mass of the moon=4:1

Explanation:

Given

m = Mass of moon

M = Mass of the planet

D = Distance between the centre of the planet and the moon

Net force = 0

Let Y be a point at distance x from the planet

Let mo = mass at point Y

a.

Derive an expression for x in terms of m, M and D.

Formula for Gravitational Force is

F = Gm1m2/r²

Y = D - x

Force on rest mass due to mass M (FM) =Force applied on rest mass due to m (Fm)

FM = G * mo * M/x²

Fm = G * mo * m/Y²

Fm = G * mo * m/(D - x)²

FM = Fm = 0 ------ from the question

So,

G * mo * M/x² = G * mo * m/(D - x)² ----- divide both sides by G * mo

M/x² = m/(D - x)² --- Cross Multiply

M * (D - x)² = m * x²

M/m = x²/(D - x)² ---_ Find square roots of both sides

√(M/m) = x/(D - x) ----- Multiply both sides by (D - x)

(D - x)√(M/m) = x

D√(M/m) - x√(M/m) = x

D√(M/m) = x√(M/m) + x

D√(M/m) = x(√(M/m) + 1) ------- Divide both sides by √M/m + 1

x = (D√M/m)/(√M/m + 1)

b. Here x = ⅔D

FM = G * mo * M/x²

Fm = G * mo * m/(D - x)²

FM = Fm

G * mo * M/x² = G * mo * m/(D - x)² ----- divide both sides by G * mo

M/x² = m/(D - x)² --- (Substitute ⅔D for x)

M/(⅔D)² = m/(D - ⅔D)²

M/(4D/9) = m/(⅓D)²

9M/4D = m/(D/9)

9M/4D = 9m/D ---- Divide both side by 9/D

M/4 = m

M = 4m

M/m = 4

M:m = 4:1

So, the ratio R of the mass of the planet to the mass of the moon=4:1

3 0
3 years ago
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