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3 years ago

Calculate the intensity of current flowing through a computer that consumes 180W and operates at 120 V.

Physics

1 answer:

padilas [110]3 years ago

5 0

Answer:

C) 1.5 A

Explanation:

P = IV

180 W = I (120 V)

I = 1.5 A

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In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6

stealth61 [152]

Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

r₁ = 5 cm

diameter of pipe outlet, d₂ = 15 cm

r₂= 7.5 cm

head upto water level is to rise = 60 + 5

= 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q

π r₁² v₁ = π r₂² v₂ = 0.015

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{7.5^2}{5^2} v_2$

$v_1= 2.25 v_2$

$v_2 = \dfrac{0.015}{\pi r_2^2}$

$v_2 = \dfrac{0.015}{\pi 0.075^2}$

v₂ = 0.848 m/s

v₁ = 1.908 m/s

Applying Bernoulli's equation

$P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h$

$P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65$

$P_p= 635539.32 Pa$

P_p is the pump pressure

Power of the pump

W = P_p x Q

W = 635539.32 x 0.015

W = 9533.09 Watt

6 0

4 years ago

A ball rolls down an inclined plane with a constant acceleration of 2.5 m/s/s. How fast is the ball traveling after 3 seconds

AveGali [126]

7.5 m/s just multiply the two values

8 0

3 years ago

I NEED THE ANSWER TO PASS SO I CAN GET MY PHONE BACK HELPPPPPPPPPPPPPP

lianna [129]

Answer:

it would be c.

Explanation:

starts at zero and isn't a straight shot up

6 0

3 years ago

a driver brings a car to a full stop in 2.0 s. if the car was initially traveling at 22 m/s, what is the acceleration

Nuetrik [128]

Answer: what is it?

Explanation:

4 0

4 years ago

Read 2 more answers

A ball is thrown vertically upward with a speed of 18.0 m/s. (a) How high does it rise? (b) How long does it take to reach its h

Mrrafil [7]

Answer:

a) $y=16.53m$

b) $t_{up}=1.83s$

c) $t_{down}=1.84s$

d) $v=-18m/s$

Explanation:

a) To find the highest point of the ball we need to know that at that point the ball stops going up and its velocity become 0

$v^{2} =v^{2} _{o} +2g(y-y_{o})$

$0=(18)^{2} -2(9.8)(y-0)$

Solving for y

$y=\frac{(18)^{2} }{2(9.8)}=16.53m$

b) To find how long does it take to reach that point:

$v=v_{o}+at$

$0=18-9.8t$

Solving for t

$t_{up} =\frac{18m/s}{9.8m/s^{2} }= 1.83s$

c) To find how long does it take to hit the ground after it reaches its highest point we need to find how long does it take to do the whole motion and then subtract the time that takes to go up

$y=y_{o}+v_{o}t+\frac{1}{2}gt^{2}$

$0=0+18t-\frac{1}{2}(9.8)t^{2}$

Solving for t

$t=0 s$ or $t=3.67s$

Since time can not be negative, we choose the second option

$t_{down}=t-t_{up}=3.67s-1.83s=1.84s$

d) To find the velocity when it returns to the level from which it started we need to use the following formula:

$v=v_{o}+at$

$v=18m/s-(9.8m/s^{2} )(3.67s)=-18m/s$

The sign means the ball is going down

8 0

4 years ago