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adelina 88 [10]

3 years ago

5

A .2.-kg soccer ball is rolling at 6.0 m/s toward a player. The player kicks the ball back and gives it a velocity of 14 m/s in

the opposite direction.

1 answer:

Sergeu [11.5K]3 years ago

5 0

Answer:

25(6+14) = 5

25(6+14) = 55/.02 = 500/2 = 250 N

25(6+14) = 55/.02 = 500/2 = 250 Na = F/m = 250 N/.25 = 1000 m/s^2

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How many meters in 2.50 miles? (Use these two conversions: 1000 m = 1 km and 1.00 km = .621 mi )

Artyom0805 [142]

2.50 miles is equal to 4026 m.

<u>Explanation:</u>

As it is known that 1000 m =1 km and 1 km = 0.621 miles. So first we have to convert miles to km and then to metre as follows.

As 1 km = 0.621 miles, then

$\text { 1 miles }=\frac{1}{0.621} \mathrm{km}$

So, 2.50 miles will be equal to

$2.50 \text { miles }=\frac{2.50}{0.621} \mathrm{km}=4.026 \mathrm{km}$

Then, in order to get the answer in meters, we have to convert this km to meter by the conversion of 1000 m =1 km. So,

$1 \mathrm{km}=1000 \mathrm{m}$

Thereby,

$4.026 \mathrm{km}=4026 \mathrm{m}$

7 0

3 years ago

(8 points) Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet

trapecia [35]

Answer:

a ) $\dot m = 351.49 kg/s$

b) $\dot m_{actual} = 1046.15 kg/s$

Explanation:

given data:

pressure ration rp = 12

inlet temperature = 300 K

TURBINE inlet temperature = 1000 K

AT the end of isentropic process (compression) temperature is

$\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}$

$\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}$

$T_2' = 610.181 K$

AT the end of isentropic process (expansion) temperature is

$\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}$

$\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}$

$T_4' = 491.66 K$

isentropic work is given as

$w(compressor) = CP (T_2' -T_1)$

w = 1.005(610.18 - 300)

w = 311.73 kJ/kg

w(turbine) = 1.005( 1000 - 491.66)

w(turbine) = 510.88 kJ/kg

a) mass flow rate for isentropic process is given as

$\dot m = \frac{70000}{510.88 - 311.73}$

$\dot m = 351.49 kg/s$

b) actual mass flow rate uis given as

$\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}$

$\dot m_{actual} = 1046.15 kg/s$

6 0

3 years ago

What energy output objects work with the turbine

Angelina_Jolie [31]

Answer:

maybe alternator..generator..

4 0

3 years ago

What are all stars made of

ivolga24 [154]

Stars are huge celestial bodies made mostly of hydrogen and helium that produce light and heat from the churning nuclear forges inside their cores. Aside from our sun, the dots of light we see in the sky are all light-years from Earth. They are the building blocks of galaxies, of which there are billions in the universe. It’s impossible to know how many stars exist, but astronomers estimate that in our Milky Way galaxy alone, there are about 300 billion.

7 0

3 years ago

A box is dropped onto a conveyor belt moving at 3.2 m/s. If the coefficient of friction between the box and the belt is 0.28, ho

Lemur [1.5K]

Answer:

t = 1.16 s.

Explanation:

Given,

speed of conveyor belt, v = 3.2 m/s

coefficient of friction,f = 0.28

Using newton second law

f = ma

and we also know that frictional force

f = μ N

f = μ m g

equating both the force equation

a = μ g

a = 0.28 x 9.81

a = 2.75 m/s²

Using Kinematic equation

v = u + at

3.2 = 0 + 2.75 x t

t = 1.16 s.

Time taken by the box to move without slipping is 1.16 s.

6 0

2 years ago