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3 years ago

15

A player kicks a soccer ball from ground level and sons at flying at an angle of 30° at a speed of 26MS how far did the ball tra

vel before it hit the ground round the answer to the nearest meter

1 answer:

RSB [31]3 years ago

7 0

Answer:

Explanation:

60 meters is he answer for this question

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A body is moving with an acceleration of 60 m/s^2(square).If the force applied to the body is 4200N.Calculate its mass.

Pavel [41]

Hello, for this use the <u>2nd law of Newton:</u>

F = ma

Replacing:

4200 N = m * 60 m/s^2

Resolving:

4200 N / 60 m/s^2 = m

70 kg = m

The mass is <u>70 kg</u>

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2 years ago

How to calculate average breaking force physics

Alexandra [31]

Answer:

Formula: Remove the zero from the speed, multiply the figure by itself and then multiply by 0.4. The figure 0.4 is taken from the fact that the braking distance from 10 km/h in dry road conditions is approximately 0.4 metres.

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3 years ago

The temperature of 20 liters of helium gas is increased from 100°K to 300°K. If the pressure was kept constant, the new volume w

slavikrds [6]

Answer:

New volume (V2) = 60 liter

Explanation:

Given:

Amount of helium (V1) = 20 Liter

Temperature (T1) = 100°K

Temperature (T2) = 300°K

Find:

New volume (V2)

Computation:

According to Gas Law:

V1 / T1 = V2 / T2

20 / 100°K = V2 / 300°K

V2 = 60 liter

New volume (V2) = 60 liter

8 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

2 years ago

What is the acceleration of a 30.0-kg go kart if the thrust of its engine is 300.0 N?

VikaD [51]

Answer : $a=10\ m/s^2$

Explanation :

It is given that,

Mass of the engine, m = 30 kg

Thrust is equivalent to the force acting perpendicularly and it is F = 300 N

According to Newton's second law of motion :

$F = m\times a$

a is the acceleration of the engine.

$a=\dfrac{F}{m}$

$a=\dfrac{300\ N}{30\ Kg}$

$a=10\ m/s^2$

So, the acceleration of the engine is $10\ m/s^2$ .

Hence, this is the required solution.

5 0

3 years ago

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