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erastova [34]
3 years ago
7

Ciara is swinging a 0.015 kg ball tied to a string around her head in a flat, horizontal circle. The radius of the circle is 0.7

0 m. It takes the ball 0.60 seconds to complete one full circle. Calculate the tension in the string and its direction that provides the centripetal force acting on the ball to keep it in the circular path.
A) 0.0077 N, toward the center of the circle
B) 1.2 N, toward the center of the circle
C) 0.0077 N, along the line tangent to the circle
D) 1.2 N, along the line tangent to the circle
Physics
2 answers:
balu736 [363]3 years ago
5 0

Answer:

1.2 N, toward the center of the circle

Explanation:

It is given that,

Mass of the ball, m = 0.015 kg

The radius of the circle, r = 0.7 m

Time taken by the ball to complete complete circle, t = 0.6 s

We need to find the tension in the string and its direction that provides the centripetal force acting on the ball to keep it in the circular path. Here, tension in the string balances the centripetal force so that the ball moves in circular path. So,

T=\dfrac{mv^2}{r}

Since, v=\dfrac{2\pi r}{t}=\dfrac{2\pi \times 0.7}{0.6}=7.33\ m/s

So, T=\dfrac{0.015\times (7.33)^2}{0.7}

T = 1.15 N

or

T = 1.2 N

The direction of centripetal force is toward the center of circle. So, the correct option is (b).

Sophie [7]3 years ago
4 0

Answer:

B) 1.2 N, toward the center of the circle

Explanation:

The circumference of the circle is:

C = 2πr

C = 2π (0.70 m)

C = 4.40 m

So the velocity of the ball is:

v = C/t

v = 4.40 m / 0.60 s

v = 7.33 m/s

Sum of the forces in the radial direction:

∑F = ma

T = m v² / r

T = (0.015 kg) (7.33 m/s)² / (0.70 m)

T = 1.2 N

The tension force is 1.2 N towards the center of the circle.

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During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approxima
vova2212 [387]

Answer:

The value  F  =  0.1396 \ N

Explanation:

From the question we are told that

   The volume blood  ejected is  V  =  65 \ cm^3  = 65*10^{-6} \  m^3

    The velocity of the blood ejected is  v  = 103 \  cm/s  = \frac{103}{100} = 1.03 \ m/s

    The density of blood is  \rho = 1060 \  kg/m^3

     The heart beat is R = 59 \  bpm(beats \  per \  minute) = \frac{59}{60}= 0.9833\ bps

The average force exerted by the blood on the wall of the aorta is mathematically represented as

      F  =  2 * \rho  *  V  *  R *  v

=>    F  =  2 * 1060  *  65*10^{-6}  *  0.9833 *  1.03

=>    F  =  0.1396 \ N

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What's the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmi
Murrr4er [49]

Complete Question:

The Voyager 1 spacecraft is now beyond the outer reaches of our solar system, but earthbound scientists still receive data from the spacecraft s 20-W radio transmitter. Voyager is expected to continue transmitting until about 2025, when it will be some 25 billion km from Earth.

What s the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmitter on Voyager transmits equally in all directions(isotropically).  In fact, the antenna on Voyager focuses the signal in a beam aimed at the earth, so this problem over-estimates the size of the receiving dish needed.

Answer:

d = 2,236 m.

Explanation:

The received power on Earth, can be calculated as the product of the intensity (or power density) times the area that intercepts the power radiated.

As we assume that  the transmitter antenna is ominidirectional, power is spreading out over a sphere with a radius equal to the distance to the source.

So, we can get the power density as follows:

I = P /A = P / 4*π*r², where P = 20 W, and r= 25 billion km = 25*10¹² m.

⇒ I = 20 W / 4*π* (25*10¹²)² m²

The received power, is just the product of this value times the area of the receiver antenna, which we assumed be a circle of diameter d:

Pr = I. Ar =( 20W / 4*π*(25*10¹²)² m²) * π * (d²/4) = 10⁻²⁰ W

Simplifying common terms, we can solve for d:

d= √(16*(25)²*10⁴/20) = 2,236 m.

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Explanation:

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Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.
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Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
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