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SashulF [63]
3 years ago
7

A basketball player can jump 1.6 m off the hardwood floor. With what upward velocity did he leave the floor?

Physics
2 answers:
just olya [345]3 years ago
6 0

Answer:

The upward velocity is 5.6 m/s.

(5) is correct option.

Explanation:

Given that,

Height = 1.6 m

We need to calculate the the upward velocity

Using equation of motion

v^2=u^2-2gh

u=\sqrt{2gh}

Here, g = acceleration due to gravity

v = final velocity

u = initial velocity

h = height

Put the value into the formula

u=\sqrt{2\times9.8\times1.6}

u=5.6\ m/s

Hence,  The upward velocity is 5.6 m/s.

irinina [24]3 years ago
3 0
First do 1.6 m (how far he jumps) 9.8 m/s (what gravity is measured at) then times 2

= 31.36

Sq root = 5.6
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The lowest possible temperature in outer space is 3.13 K. What is the rms speed of hydrogen molecules at this temperature? (The
Umnica [9.8K]

Answer:

v_{rms} =196.59 m/s

Given:

Temperature, T = 3.13 K

molar mass of molecular hydrogen, m = 2.02 g/mol = 2.02\times 10^{-3}kg/mol

Solution:

To calculate the root mean squarer or rms speed of hydrogen molecule, we use the given formula:

v_{rms} = \sqrt{\frac{3TR}{m}}

where

R = rydberg's constant = 8.314 J/mol-K

Putting the values in the above formula:

v_{rms} = \sqrt{\frac{3\times 3.13\times 8.314}{2.02\times 10^{-3}}}

v_{rms} =196.59 m/s

5 0
3 years ago
Which is an example of a solution?
vladimir1956 [14]

Answer:

C

Explanation:

plato

6 0
3 years ago
You have a length of tubing that is closed at one end. You cut the tubing into two pieces of unequal length, giving you a tube o
umka21 [38]

Answer:

The funda mental frequency of the original tube is 182Hz.

Explanation:

See the attachment for the calculation steps.

In order to calculate the fundamental frequency of the original closed tube we need to find the length of the tube which is equal to the sum of the lengths of the two new tubes.

For closed tubes

f = nv/4L (n = 1, 3, 5,...n)

f = nv/2L (n = 1, 2, 3,...n)

The details of calculation can be found below in the attachment.

4 0
3 years ago
(a) If two sound waves, one in a gas medium and one in a liquid medium, are equal in intensity, what is the ratio of the pressur
GarryVolchara [31]

Answer:

(a) The ratio of the pressure amplitude of the waves is 43.21

(b) The ratio of the intensities of the waves is 0.000535

Explanation:

Given;

density of gas, \rho _g = 2.27 kg/m³

density of liquid, \rho _l = 972 kg/m³

speed of sound in gas, C_g = 376 m/s

speed of sound in liquid, C_l = 1640 m/s

The of the sound wave is given by;

I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}

Where;

P_o is the pressure amplitude

P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535

3 0
3 years ago
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