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Ivahew [28]
3 years ago
14

How long is a football field?

Physics
2 answers:
Nana76 [90]3 years ago
7 0

Answer:

91.44 meters or 300 feet

ycow [4]3 years ago
6 0

Answer:

100 yards

Explanation:

If comparing from actual playing field between the goal posts. Add another 10 yards for end lines. Making it 120 yards in length.

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If a 50 W light bulb and a 90 W light bulb operate from 120 V, which bulb has a greater current in it?
PIT_PIT [208]
The 50 W bulb would have a greater current than the 90 W bulb
8 0
3 years ago
Kinetic energy of an object is equal to
Arte-miy333 [17]
Choice-'b' says the formula for kinetic energy in words.

     KE = (1/2) · (M) · (S²)
7 0
3 years ago
Read 2 more answers
Which would take a longer time to warm up, a piece of brass or a piece of marble, given that both
diamong [38]

Answer:

Explanation:

specific heat of marble is 880 J/kg °C

specific heat of brass is 920 J/kg °C

If Q heat is given to a body of mass m and specific heat s , and there is a rise of temperature t

Q = m s t

t = Q / m s

if Q and m is constant

t is inversely proportional to specific heat .

so rise in temperature will be inversely proportional to specific heat .

lower the specific heat , higher  will be the rise in temperature . Since marble has lower specific heat , rise in temperature in marble will be higher and quicker .

So brass will take longer time to warm up.

7 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
Most of the bright stars in our galaxy are located in the galactic
GenaCL600 [577]

False. Most of the bright stars in our galaxy are located in the disk.

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3 years ago
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