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3 years ago

A rock of mass m is twirled on a string in a horizontal plane. The work done by the tension in the string on the rock is:

Physics

1 answer:

Varvara68 [4.7K]3 years ago

7 0

Answer: c) 0

Explanation:

Work tension is zero because in any infinitesimal time interval, the tension force is directly perpendicular to the displacement.

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You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While

ANTONII [103]

To solve this problem it is necessary to apply the concepts related to the Rotational Force described from the equilibrium and Newton's second law.

When there is equilibrium, the Force generated by the tension is equivalent to the Force of the Weight. However in rotation, the Weight must be equivalent to the Centrifugal Force and the tension, in other words:

$W = F_T + m\omega^2r_E$

Where

$\omega = \frac{2\pi}{T} \rightarrow$ Angular velocity is equal to the Period, at this case Earth's period

$r_E = 6.371*10^6m \rightarrow$ Radius of the Earth

m = mass

$F_T$ = Force of Tension

$W = mg \rightarrow$ Newton's second law

Replacing and re-arrange to find the Tension we have,

$F_T = W- \frac{W}{g} (\frac{2\pi}{T})^2r_E$

$F_T = W(1-(\frac{2\pi}{T})^2\frac{r_E}{g})$

$F_T = (505)(1-(\frac{2\pi}{24hours})^2\frac{6.371*10^6}{9.8})$

$F_T = (505)(1-(\frac{2\pi}{24hours(\frac{3600s}{1hour})})^2\frac{6.371*10^6}{9.8})$

$F_T = (505)(1-(\frac{2\pi}{86400})^2\frac{6.371*10^6}{9.8})$

$F_T = 503.26N$

Therefore when Sneezy is on the equator he is in a circular orbit with a Force of tension of 503.26N

7 0

3 years ago

a sleepy atudent drops a calculator out of a window yhata 20.7 m off the geound. we can ignore air resistance. what js the veloc

frez [133]

The acceleration of gravity is

9.8 m/s^2 down.

When an object falls out of a hand, its speed after 1.8s is

(9.8)x(1.8) = 17.6 m/s down.

It doesn't matter what it is, how much it weighs, or how high it was dropped from.

If it's more than 17.6 m/s, then this happened on a different, bigger planet.

If it's less than 17.6 m/s, then it must have hit something on the way down, like some air or something.

6 0

3 years ago

If the velocity of blood flow in the aorta is normally about 0.32 m/s, what beat frequency would you expect if 4.40-MHz ultrasou

dusya [7]

Answer:

The beat frequency is 0.0019 MHz.

Explanation:

Given that,

Velocity = 0.32 m/s

Frequency = 4.40 MHz

Speed of wave = 1540 m/s

We need to calculate the frequency

Case (I),

Observer is moving away from the source

Using Doppler's effect

$f'=\dfrac{v-v'}{v}f$

Where, v' = speed of observer

Put the value into the formula

$f'=\dfrac{1540-0.32}{1540}\times4.40$

$f'=4.399\ MHz$

Case (II),

Cell is as the source of sound of frequency f' and it moving away from the observer.

Using formula of frequency

$f''=\dfrac{v-v_{s}}{v+v_{s}}\times f$

$f''=\dfrac{1540-0.32}{1540+0.32}\times4.399$

$f''=4.3971\ MHz$

We need to calculate the beat frequency

$\Delta f= f'-f''$

$\Delta f=4.399-4.3971=0.0019\ MHz$

Hence, The beat frequency is 0.0019 MHz.

4 0

3 years ago

PROVE YOUR INTELLIGNCE HERE!!!! (IMAGE ATTATCHED FOR YOUR CONVENIENCE)

gregori [183]

Answer:

There's an image... 〒▽〒 where?

Explanation:

8 0

3 years ago

Estimate the number of photons emitted per second from 1.0 cm2 of a person's skin if a typical emitted photon has a wavelength o

Diano4ka-milaya [45]

Answer:

2.63 x 10^18

Explanation:

A = 1 cm^2 = 1 x 10^-4 m^2

λ = 10,000 nm = 10,000 x 10^-9 m = 10^-5 m

T = 37 degree C = 37 + 273 = 310 k

Energy of each photon = h c / λ

where, h is the Plank's constant and c be the velocity of light

Energy of each photon = (6.63 x 10^-34 x 3 x 10^8) / 10^-5 = 1.989 x 10^-20 J

Energy radiated per unit time = σ A T^4

Where, σ is Stefan's constant

Energy radiated per unit time = 5.67 x 10^-8 x 10^-4 x 310^4 = 0.05236 J

Number of photons per second = Energy radiated per unit time / Energy of

each photon

Number of photons per second = 0.05236 / (1.989 x 10^-20) = 2.63 x 10^18

4 0

3 years ago