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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 nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
2 years ago
What does the magnitude of centripetal acceleration depend on ?
MariettaO [177]

Explanation:

Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity v and has the magnitude ac=v2r;ac=rω2 a c = v 2 r ; a c = r ω 2 .

3 0
3 years ago
After a long night of cramming for a test, your college room-mate hits his head forcefully against the wall of your room in desp
zhannawk [14.2K]

Answer: the wall contracts the force exerted by his head. The wall produces the opposite force which is equal to the force his head bangs the wall with.

Explanation: if his head exerts a much greater force than the wall can counteract the wall will be destroyed, if the wall exerts a much greater force than his head exerts he will be pushed far back and might even suffer a broken head.

The wall in this case provides the opposite reactive force.

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Solar System Model Comparison and Contrast Chart
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What are you trying to here?
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3 years ago
Name the physical quantity which changes contenously during uniform-circular motion.
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The direction of the motion is constantly changing during
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