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WITCHER [35]
2 years ago
11

an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco

nd. The radius of the hammer's path is 2.0 meters.
Physics
1 answer:
Semenov [28]2 years ago
6 0

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

a_c = \frac{v^2}{r}

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

a_c = \frac{(12\ m/s)^2}{2\ m}

<u>ac = 72 m/s²</u>

<u></u>

Now, the centripetal force applied by the athlete on the hammer will be:

F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)

<u>Fc = 504 N</u>

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A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit
likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density \mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 120=753.6rad/sec

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt

From the relation we can see that power P\ \propto\ A^2

So if amplitude is halved then power will be \frac{1}{4} times

So power will be equal to \frac{0.2023}{2}=0.0505watt

4 0
3 years ago
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea
Kipish [7]

Answer:

6.5 m/s

Explanation:

We are given that

Distance, s=100 m

Initial speed, u=1.4 m/s

Acceleration, a=0.20 m/s^2

We have to find the final velocity at the end of the 100.0 m.

We know that

v^2-u^2=2as

Using the formula

v^2-(1.4)^2=2\times 0.20\times 100

v^2-1.96=40

v^2=40+1.96

v^2=41.96

v=\sqrt{41.96}

v=6.5 m/s

Hence, her final velocity at the end of the 100.0 m=6.5 m/s

5 0
3 years ago
The tin can with water in its bottom is heated to boil water and the steam is allowed to escape for some time. The open mouth is
Norma-Jean [14]

Explanation:

Water does expand with heat (and contract with cooling), but the amount of expansion is pretty small. So when you boil a can filled with water and seal it, the water will contract slightly as it cools. The can may kink slightly, but that will be it. Actually, most likely the only things you will be able to see is then top and bottom will be sucked in and go concave. Just like a commercial can of beans.

Now if you have a can with a little water and a big air space, things are completely different.

As the water boils, water vapour is given off. Steam. Let it boils for a minute just to make sure (nearly) all the air is expelled and the can is filled with steam.

Now when you put the lid on and cool the can, that steam condenses back to water, and goes from filling the can to a few drops of water. The can is now filled (if that is the right word) with a near vacuum, The air pressure, 15 lbs/square inch, will be pressing on every surface of the can, with nothing inside the can to resist it.

The can will crumple before your eyes.

6 0
3 years ago
On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with a block of mass 0.400 kg. Immediatel
Akimi4 [234]

Answer:48.2 Joules

Explanation:

Given

two masses of 0.2 kg and 0.4 kg collide with each other

after collision 0.2 kg deflect 30 north of east and 0.4 kg deflects 53.1 south of east

Velocity of 0.2 kg mass is

v_{0.2}=12\cos (30)\hat{i}+12\sin (30)\hat{j}

|v_{0.2}|=11.99 m/s\approx 12 m/s

Velocity of 0.4 kg mass

v_{0.4}=13\cos (53.1)\hat{i}-12\sin (53.1)\hat{j}

|v_{0.4}|=12.99 m/s\approx 13 m/s

Thus total Kinetic energy =\frac{0.2\times 12^2}{2}+\frac{0.4\times 13^2}{2}

Kinetic energy=48.2 J

8 0
3 years ago
Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net
kotykmax [81]

Answer:

0.29D

Explanation:

Given that

F = G M m / r2

F = GM(6m) / (D-r)2

G Mm/r2 = GM(6m) / (D-r)2

1/r2 = 6 / (D-r)2

r = D / (Ö6 + 1)

r = 0.29 D

See diagram in attached file

7 0
3 years ago
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