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3 years ago

6

A football player kicks a ball horizontally off a hill with an initial velocity of 42.0 m/s. It travels a horizontal distance of

59.2m. How y'all was the hill he kicked the ball off of?

1 answer:

damaskus [11]3 years ago

7 0

Explanation:

.........................

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A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 358

aleksandr82 [10.1K]

Answer:

Period is 86811.5 seconds.

Explanation:

${ \boxed{ \bf{T {}^{2} = (\frac{4 {\pi}^{2} }{GM}) {r}^{3} }}}$

${ \tt{T {}^{2} = \frac{4 {(3.14)}^{2} }{(6.6 \times {10}^{ - 11} ) \times (5.98 \times {10}^{24} )} \times {((35880\times {10}^{3}) } + (6370 \times {10}^{3} )) {}^{3} }} \\ \\ { \tt{T {}^{2} = 7.54 \times {10}^{9} }} \\ { \tt{T = \sqrt{7.54 \times {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}$

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3 years ago

Write the first equation of motion. Under what condition(s) is this equation valid?

Zepler [3.9K]

Explanation:

The first equation of motion in kinematics is given by :

$v=u+at$ .....(1)

u is initial speed

a is acceleration

v is final speed

t is time

Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.

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3 years ago

A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?

garri49 [273]

<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>

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4 years ago

What are the units for mass and weight?

Luda [366]

The kilogram is the SI unit of mass and it is the almost universally used standard mass unit. The associated SI unit of force and weight is the Newton, with 1 kilogram weighing 9.8 Newtons under standard conditions on the Earth's surface.

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3 years ago

Read 2 more answers

The pressure P (in kilopascals), volume V (in liters), and temperature T (in kelvins) of a mole of an ideal gas are related by t

stich3 [128]

Answer:

$\dfrac{dV}{dt} = -0.466 m^3/s$

Explanation:

given,

P (in kilo pascals), volume V (in liters), temperature T (in kelvins)

P V = 8.31 T

Rate of increase the temperature = 0.1 K/s

temperature = 285 K

Pressure = 18 kPa

increasing at the rate of 0.07 k Pa/s

Rate at which volume is changing = ?

$V = 8.31 \dfrac{T}{P}$

$\dfrac{dV}{dt} = 8.31 \dfrac{P\dfrac{dT}{dt}-T\dfrac{dP}{dt}}{P^2}$

$\dfrac{dV}{dt} = 8.31 \dfrac{18 \times 0.1-285\times 0.07}{18^2}$

$\dfrac{dV}{dt} = 8.31 \dfrac{-18.15}{324}$

$\dfrac{dV}{dt} = -0.466 m^3/s$

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3 years ago