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kotegsom [21]
3 years ago
9

A student compared some soccer players to the atoms in the gaseous state. Which of the following activities were the soccer play

ers most likely performing?
A. Lining up next to each other
B. Huddling close together for the ball
C. Running randomly in all directions
D. Slowly jumping up and down in their places
Physics
2 answers:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

The answer will be <em>D</em>

Explanation:

I choose answer <em>C</em> and got it wrong on the test. Also Liquid atoms vibrate fast and slide past each other. what answer <em>C </em>shows is a gas.

Elan Coil [88]3 years ago
6 0

Answer:

D

Explanation:

Slowly jumping up and down in their places

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mario62 [17]

Velocity of vw bug is given as

v_1 = 54 mpH towards north

velocity of imphala is given as

v_2 = 47 mph towards south

now we need to find velocity of imphala relative to bug

v_{21} = v_2 - v_1

v_{21} = 47 - (-54) = 101 mpH

so the speed is 101 mpH towards south so correct option is C

5 0
3 years ago
A person's resting heart rate typically lessens<br> with age.
yan [13]

Answer:

Ok

Explanation:

Please mark brainliest

8 0
3 years ago
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A garden hose with a diameter of 0.64 in has water flowing in it with a speed of 0.46 m/s and a pressure of 1.9 atmospheres. At
STALIN [3.7K]

Answer:

(a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

Explanation:

Given that,

Nozzle diameter = 0.25 in = 0.00635 m

Hose pipe diameter = 0.64 in = 0.016256 m

Pressure = 1.9 atm =192518 Pa

(a). We need to calculate the speed of the water in the nozzle

Flow Speed at the inlet pipe will be given by using Continuity Equation

Q_{1}=Q_{2}

v_{1}A_{1}=v_{2}A_{2}

v_{1}=v_{2}\times(\dfrac{A_{2}}{A_{1}})

Where, A = area of pipe

A=\pi\times \dfrac{d^2}{4}

v_{1}=v_{2}\times(\dfrac{d_{2}^2}{d_{1}^2})

Put the value into the formula

v_{1}=0.46\times\dfrac{(0.016256)^2}{(0.00635)^2}

v_{1}=3.014\ m/s

The speed of the water in the nozzle is 3.014 m/s.

(b). We need to calculate the pressure in the nozzle

Using Bernoulli's Theorem,

P_{1}+\dfrac{1}{2}\rho\times v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2+\rho gh_{2}

Where, h_{1}=h_{2}

P_{1}+\dfrac{1}{2}\rho\times v_{1}^2=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2

P_{1}=P_{2}+\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)

Put the value into the formula

P_{1}=192518 +\dfrac{1}{2}\times1000\times((0.46)^2-(3.014)^2)

P_{1}=188081.702\ Pa

P=1.86\ atm

Hence, (a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

7 0
2 years ago
Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r
Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

v=u-gt

u=gt

t=\dfrac{u}{g}

t=\dfrac{13.7}{9.8}

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

h=ut+\dfrac{1}{2}at^2

Here, a = -g

h=ut-\dfrac{1}{2}gt^2

h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0
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