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Phantasy [73]
3 years ago
10

A ball on the end of a rope is moving in a vertical circle near the surface of the earth. Point A is at the top of the circle; C

is at the bottom. Points B and D are exactly halfway between A and C. Which one of the following statements concerning the A boy is whirling a stone at the end of a string around his head. The string makes one complete revolution every second, and the tension in the string is FT. The boy increases the speed of the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions per second. What happens to the tension in the sting?
Physics
1 answer:
mrs_skeptik [129]3 years ago
3 0

Answer:

Tension in the string will increase

Explanation:

As we know that tension in the string at any angle with the vertical is given as

T - mgcos\theta = m\omega^2 R

now we have

T = mgcos\theta + m\omega^2 R

also we know that

angular speed of the stone is directly depending on the time period of the motion

so it is given as

\omega = \frac{2\pi}{T}

since the frequency of the revolution is increased from n = 1 rev/s to 2 rev/s

so the angular speed would be doubled

So here we can say that

tension in the string will increase when we will increase the frequency of revolution.

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Asexual reproduction involves only one parent. The offspring of this type of reproduction have - F DNA identical to the DNA of t
Sedbober [7]

Answer:

DNA identical to the DNA of the parent

Explanation:

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3 years ago
The maximum current output of a 60 ω circuit is 11 A. What is the root mean square voltage of the circuit?
goldenfox [79]

Answer:

660V

Explanation:

V=IR

V=?, I=11A,R=60w

V=60 ×11

=660V

8 0
3 years ago
Does the horizontal distance d travelled by the ball depend on the height of release? If it does depend on the height, what is t
elena-s [515]

Answer:

Explanation:

Yes , the horizontal distance travelled by the ball will depend upon the height of release .

When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play .  But the horizontal range covered by the body thrown

depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .

Horizontal distance covered = t x horizontal velocity

If V be the velocity of throw and Vx be its horizontal component

Horizontal distance covered = t x Vx

Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .

If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity

from the relation

s = ut + 1/2 at²

h = - Vy t  + 1/2 at²

As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only  horizontal velocity initially , Vy = 0

h = 1/2 gt²

t = \sqrt{\frac{2h}{g} }

Horizontal distance covered  = t x Vx

= \sqrt{\frac{2h}{g} } \times  V_x

From this expression also

Horizontal distance covered is proportional to \sqrt{h} .

7 0
3 years ago
the brightest , hottest, and most massive stars are the brilliant blue stars designated as spectral class O. if a class O star w
4vir4ik [10]

The speed is 0.956 m / s.

<u>Explanation</u>:

The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.

                   K.E = 1/2 \times m \times v^{2}

where K.E represents the kinetic energy,

           m represents the mass,

            v represents the velocity.

                  K.E = 1/2 \times m \times v^{2}

    1.10 \times 10^42 = 1/2 \times 3.26 \times 10^31 \times v^{2}

                     v^{2} = (1.10 \times 10^42 \times 2) / (3.26 \times 10^31)

                     v = 0.956 m / s.

6 0
3 years ago
Recall from Chapter 1 that a watt is a unit of en- ergy per unit time, and one watt (W) is equal to one joule per second ( J·s–1
harkovskaia [24]

Answer:

Explanation:

The energy of a photon is given by the equation E_p=h f, where h is the <em>Planck constant</em> and f the frequency of the photon. Thus, N photons of frequency f will give an energy of E_N=N h f.

We also know that frequency and wavelength are related by f=\frac{c}{\lambda}, so we have E_N=\frac{N h c}{\lambda}, where c is the <em>speed of light</em>.

We will want the number of photons, so we can write

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We need to know then how much energy do we have to calculate N. The equation of power is P=E/t, so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be E_N.

Putting this paragraph in equations:

E_N=(\frac{4}{100})E=0.04Pt=(0.04)(100W)(1s)=4J.

And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

N=\frac{\lambda E_N}{h c}=\frac{(520 \times10^{-9}m) (4J)}{(6.626\times10^{-34}Js) (299792458m/s)}=1.047 \times10^{19}

3 0
3 years ago
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