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3 years ago

A basketball player jumped straight up to grab a rebound. If she was in the air for 0.80 second, how high did she jump?

2 answers:

4 0

-- She went up for 0.4 sec and down for 0.4 sec.

-- The vertical distance traveled in gravity during ' t ' seconds is

   D = (1/2) x (g) x (t)²

   = (1/2) (9.8 m/s²) (0.4 sec)²

   = (4.9 m/s²) x (0.16 s²)

   = 0.784 meter ( B )

Alinara [238K]3 years ago

4 0

After 0.40 seconds, she must have reached her max height
she spends half the time going up and the other half falling down

t = 0.40
vf = 0
a = -9.8

y = vf t - 1/2 a t^2

y = 0 - 1/2 (-9.8) (0.4)^2

y = 4.9 (0.4)^2

Y=0.784

This will be your answer.

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What are possible units for impulse? Check all that apply.

Neporo4naja [7]

Units of impulse: N • s, kg • meters per second

Explanation:

Impulse is defined in two ways:

Impulse is defined as the product between the force exerted in a collision and the duration of the collision:

$I=F\Delta t$

where

F is the force

$\Delta t$ is the time interval

Since the force is measured in Newtons (N) and the time is measured in seconds (s), the units for the impulse are

$[I] = [N][s]$

So,

N • s

Impulse is also defined as the change in momentum experienced by an object:

$I=\Delta p$

where the change in momentum is given by

$\Delta p = m\Delta v$

where m is the mass and $\Delta v$ is the change in velocity.

The mass is measured in kilograms (kg) while the change in velocity is measured in metres per second (m/s), therefore the units for impulse are

$[I]=[kg][m/s]$

so,

kg • meters per second

Learn more about impulse:

brainly.com/question/9484203

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How does sound travel?

Doss [256]

Answer:

Sound vibrations travel in a wave pattern, and we call these vibrations sound waves. Sound waves move by vibrating objects and these objects vibrate other surrounding objects, carrying the sound along. ... Sound can move through the air, water, or solids, as long as there are particles to bounce off of.

Explanation:

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water flows horizontally from a hose that is 3m above the ground. the water lands 7.2m to the right of the hose . how long does

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Answer:

0.782 s

Explanation:

The water flows horizontally from the hose, so its initial vertical velocity is 0.

Given:

y₀ = 3 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

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Water vapor enters a turbine operating at steady state at 500°C, 40 bar, with a velocity of 200 m/s, and expands adiabatically t

faltersainse [42]

Answer:

W = 5701 KW

Explanation:

From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;

Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s

Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s

Volumetric flow rate = 15 m^(3)/s

Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).

So from the table,

v2 = 2.087 m^(3)/kg

Now, mass flow rate (m) = (AV) /v

Where AV is the volumetric flow rate.

Thus, the mass flow rate at exit could be calculated as;

m = 15/(2.087) = 7.17 kg/s

We also know energy equation could be defined as;

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Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;

-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}

From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg

Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;

h2 = 2665.8 KJ/Kg

So we now calculate power developed;

W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW

Since the sign is not negative but positive, it means that the power is developed from the system.

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