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3 years ago

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You serve a volley ball with a mass of 2.1 kg. The ball leaves your hand with a speed of 30 m/s. The ball has_______ energy. Can

you show me how to calculate this too?

1 answer:

Greeley [361]3 years ago

5 0

Kinetic energy because the ball is in motion or moving with energy behind it... kinda like when you shoot a gun, the bullet is fired out of the muzzle with kinetic energy ( Punch ) and the bullet goes through a wall or something. Sorry but my math skills aren't very good to give complex calculations but I would recommend that you maybe talk to some of the top ranking math guys on the website. Maybe they can give you better help...

Anyways, I hope I have been helpful to you.

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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

2 years ago

A skier of mass 60 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 75 m up a 30° slop

postnew [5]

Answer:

Explanation:

Given

mass of skier=60 kg

distance traveled by skier=75 m

inclination $(\theta )=30^{\circ}$

speed (v)=2.4 m/s

as the skier is moving up with a constant velocity therefore net force is zero

$F_{net}=0$

Force applied by cable $=mg\sin \theta$

$F=60\times 9.8\times \sin (30)=294 N$

work done $=F\cdot x$

$W=294\cdot 75=22.125 J$

(b)Power $=F\cdot v$

$P=294\cdot 2.4=705.6 W\approx 0.946 hp$

5 0

2 years ago

A professor's office door is 0.91 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg; and pivots on frictionless hinges.

olasank [31]

Answer:

F= 5.71 N

Explanation:

width of door= 0.91 m

door closer torque on door= 5.2 Nm

In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.

so wee need to exert 5.2 Nm torque on the door.

If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.

T= r x F

T= r F sin∅

F= T/ (r * sin∅)

F= 5.2/ (0.91 * 1)

F= 5.71 N

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3 years ago

Which is better, ramen, or spaghetti

Sergio039 [100]

Answer:

ramen

Explanation:

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2 years ago

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A student is getting ready to go on a bike ride, but her bike back tire is flat! She connects a pump to the valve and adds air t

Mariana [72]

Wheres the answer below?

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2 years ago