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3 years ago

I got the correct answer, but I don't know what I did. Why is the answer E?

Physics

1 answer:

MariettaO [177]3 years ago

6 0

Explanation:

Displacement is the straight line distance from the starting position to the final position.

Person X walks halfway around circle. So her displacement is 100 m.

Person Y walks 3/4 of the way around. So his displacement is 50√2 m ≈ 70.7 m.

Person Z walks completely around the circle, so their displacement is 0 m.

Therefore:

Z < Y < X

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A 12-kg projectile is launched with an initial vertical speed of 20 m/s. It rises to a maximum height of 18 m above the launch p

kirill115 [55]

Answer:

Change in mechanical energy, $\Delta E=283.2\ J$

Explanation:

It is given that,

Mass of the projectile, m = 12 kg

Speed of the projectile, v = 20 m/s

Maximum height, h = 18 m

Initially, the projectile have only kinetic energy. it is given by :

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}\times 12\ kg\times (20\ m/s)^2$

K = 2400 J

Finally, it have only potential energy. it is given by :

P = mgh

$P=12\ kg\times 9.8\ m/s^2\times 18\ m$

P =2116.8 J

The change in mechanical energy is given by :

$\Delta E=K-P$

$\Delta E=2400-2116.8$

$\Delta E=283.2\ J$

So, the change in mechanical energy is 283.2 J. Hence, this is the required solution.

8 0

3 years ago

PLEASE PLEASE HELP ME!!!!!!!!!!

marysya [2.9K]

373 kelvin = 99.9 Celsius. Round makes it 100. 373 kelvin also equals 212 Fahrenheit so the correct answer is A.
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8 0

3 years ago

If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan

ioda

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

$h=ut+\frac{1}{2}gt^2$

here u=0 and t=time taken

$h=\frac{1}{2}gt^2$

for $t=1 s$

$h_1=\frac{1}{2}g$

for $t=2 s$

$h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g$

distance traveled in 2 nd sec $=2g-\frac{1}{2}g=\frac{3}{2}g$

for $t=3 s$

$h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g$

distance traveled in 3 rd sec $=\frac{9}{2}g-2g=\frac{5}{2}g$

so we can see that distance traveled in each successive second is increasing

5 0

3 years ago

How are the electric field lines around a positive charge affected when a second positive charge is near it?

Lesechka [4]

Answer: The field lines bend away from the second positive charge

Explanation: opposite attracts, same repulse

4 0

3 years ago

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What is not changed when work is done by a machine?

Irina18 [472]

B) The amount of work done

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