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STALIN [3.7K]
3 years ago
7

What is the centripetal acceleration of a ball on a string that is whirled horizontally over a boy’s head? The radius of the cir

cle is 1.8 meters, and the ball is traveling at 9.2 meters/second.
Physics
2 answers:
Mazyrski [523]3 years ago
6 0

Answer:

Centripetal acceleration of boy on a string is 47.02 m/s²

Explanation:

It is given that,

Radius of the circle, r = 1.8 m

Speed of the ball, v = 9.2 m/s

Centripetal acceleration is given by :

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

a_c=\dfrac{(9.2)^2}{1.8}

a_c=47.02\ m/s^2

Hence, the centripetal acceleration of a ball on a string that is whirled horizontally over a boy’s head is 47.02 m/s²

Leya [2.2K]3 years ago
4 0
Well the centripetal acceleration would be 9.82 squared. That is the correct answer. I hope this helped you. Have a great day! :)
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Oduvanchick [21]

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E=225000 J  

P=9000 W  

P=22500 W

Ew=Fd=E  

P=E/t

5 0
2 years ago
If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is
leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

  • \rm y = 0.02\sin(30x-200 t).

The vertical displacement of a wave is given in generalized form as

\rm y = A\sin(kx -\omega t).

<em>where</em>,

  • A = amplitude of the displacement of the wave.
  • k = wave number of the wave = \dfrac{2\pi }{\lambda}.
  • \lambda = wavelength of the wave.
  • x = horizontal displacement of the wave.
  • \omega = angular frequency of the wave = \rm 2\pi f.
  • f = frequency of the wave.
  • t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

\rm A=0.02.\\k=30.\\\omega =200.\\

therefore,

\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.

It is the required frequency of the wave.

3 0
2 years ago
Present day glaciers are found primarily in _______________.
Mazyrski [523]
<h3><u>Answers;</u></h3>

Antarctica and Greenland

Present day glaciers are found primarily in <em><u>Antarctica and Greenland</u></em>.

<h3><u>Explanation;</u></h3>
  • <em><u>The two major ice sheets that exists today are found primarily in Antarctica and Greenland. Ice sheets are large masses of glacial ice that are also known as continental glaciers.</u></em>
  • Most ice in Antarctica and Greenland spill out into the ocean from a few spots. The Antarctica and Greenland ice sheets combined comprise more than 99 percent of freshwater ice found on Earth.
8 0
3 years ago
Read 2 more answers
A 400g sample of water absorbs 500j of energy. how did the water temperature change if the specific heat of water is 4.18j/g©. S
Mashutka [201]

In general, the quantity of heat energy, Q, required to raise a mass m kg of a substance with a specific heat capacity of <span>c </span>J/(kg °C), from temperature t1 °C to t2 °C is given by:

<span>Q </span>= <span>mc(t</span><span>2 </span><span>– t</span>1<span>) joules</span>

<span>So:</span>

(t2-t1) =Q / mc

<span>As we know:
Q = 500 J </span>

<span>m = 0.4 kg</span>

<span>c = 4180 J/Kg </span>°c

<span>We can take t1 to be 0</span>°c

t2 - 0 = 500 / ( 0.4 * 4180 )

t2 - 0 = 0.30°c

6 0
3 years ago
Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea
melisa1 [442]

Answer:

m = 1.26*10²⁵ kg.

Explanation:

Assuming that the mass of the stone is much smaller than the mass of the planet, we can get the mass, applying the Universal Law of  Gravitation to both masses, as follows:

Fg = G* ms* mp / rp²

Now, if we apply Newton's 2nd Law to the mass of the stone, we can get the gravitational acceleration, as follows:

Fg = ms*a = ms*g ⇒ g = G*mp / rp²

First of all, we need to get the value of g.

Assuming that this acceleration is constant, we can appy the kinematic equations to this situation.

We know that the stone is thrown upward with an initial velocity vo = 15 m/s.

At the highest point in the trajectory, just before of changing direction, the stone comes momentarily to a stop.

At this point, applying the definition of acceleration, we can write:

vf = vo -g*t ⇒ 0 = vo -gt ⇒ g = vo/t (1)

We have the total time since the stone was thrown upwards, not the one used for the upward trajectory.

It can be showed, using the expression for the displacement (which is the same in both directions) that the time used for going up, it's the same used to go down, so the time that we need to put in (1). is just the half of the total time.

So, replacing in (1) we get the value of g, as follows:

g = 15 m/s / 4.5 s = 3.33 m/s²

Now, we can replace this value in the equation that gives us g based in the Universal Law of Gravitation, as follows:

g=G*mp / rp² (2)

Before solving for mp, however, we need to get the value of the radius of the planet.

Assuming that it's a perfect sphere, we can get this value from the value of the circumference at the planet's equator:

rp = 2*π*rp / 2*π ⇒ rp = 1.0*10⁵ km / 2*π = 15,915 km.

With this value for  rp, we can solve (2) for mp, as follows:

mp= g*rp² / G = 3.33 m/s² * (15,915 km)² / 6,67*10⁻¹¹ N.m²/kg²

mp = 1.26*10²⁵ kg.

8 0
3 years ago
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