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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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Answer:

Explanation:

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3 years ago

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

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