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3 years ago

10

Calculate the velocity of a 80kg man that has a momentum of 720 kg m/s

1 answer:

kotykmax [81]3 years ago

8 0

V = 720 / 80 = 9 m/s

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Ilia_Sergeevich [38]

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2 years ago

13) The condenser pressure is 417.4 psig for r-410A and the condenser outlet temperature is 108f. how much subcooling is there i

Tanzania [10]

Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

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2 years ago

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Tom [10]

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3 years ago

An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.80 kW. A receiving antenna 7

Alecsey [184]

Answer:0.1759 v

Explanation:

Intensity of wave at receiver end is

I= $\frac{P_{avg}}{A}$

I= $\frac{3.80\times 10^3}{4\times \pi \times \left ( 4\times 1609.34\right )^2}$

I= $7.296\times 10^{-6} W/m^2$

Amplitude of electric field at receiver end

$E_{max}=\sqrt{2I\mu _0c}$

Amplitude of induced emf

= $E_{max}d$

= $\sqrt{2\times 7.29\times 10-6\times 4\pi \times 3\times 10^8}\times 0.75$

= $17.591\times 10^{-2}=0.1759 v$

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3 years ago

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

6 0

2 years ago