2 years ago

2.2.4 Quiz: Conservation of Energy

Physics

1 answer:

dexar [7]2 years ago

7 0

Answer:

it is the judicious use of energy to prevent wastage

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If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the 1200.-kg pick-up truck behind him continues tr

Alborosie

Answer: 20.2 m/s

Explanation:

From the question above, we have the following data;

M1 = 800kg

M2 = 1200kg

V1 = 13m/s

V2 = 25m/s

U (common velocity) =?

M1V1 + M2V2 = (M1 + M2). U

(800*13) + (1200*25) = (800+1200) * U

10400 + 30000 = 2000u

40400 = 2000u

U = 40400 / 2000

U = 20.2 m/s

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2 years ago

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If an object was accelerating at 10 m/s2, and a mass of 1 kg, what was size of the force acting on the object?

ryzh [129]

Answer:

10 N

Explanation:

$f = ma$

= 10m/s^2 * 1 kg

=10N

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There isnt enough sanitidzer(alcohol)what solution to be replaced

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For me: WASH OUR HANDS REGULARLY

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Is Li Chen abiotic or biotic?

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Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

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2 years ago