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3 years ago

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Two ice skaters stand facing each other at rest on a frozen pond. They push off against one another and the 48 kg skater acquire

s a speed of 0.67 m/s. If the other skater acquires a speed of 0.83 m/s, what is her mass in kilograms?

2 answers:

BaLLatris [955]3 years ago

5 0

Answer:

The mass from the second skater is 40.2Kg

Explanation:

To solve this problem it is necessary to apply the concepts related to conservation of the moment.

By definition for the particular case we have to,

$m_1u_1 = m_2v_2$

Where,

$m_1$ = Mass first skater

$u_1$ = Speed of first skater

$v_2$ = Velocity of second skater

Our values are given as,

$m_1= 48Kgu_1 = 0.67m/sv_2 = 0.8m/s$

Replacing in our equation and re-arrange to find $m_2,$

$m_2 = \frac{m_1u_1}{v_2}m_2 = \frac{(48)(0.67)}{0.8}m_2 = 40.2Kg$

The mass from the second skater is 40.2Kg

Papessa [141]3 years ago

5 0

Answer:

38.75 kg

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = mv+m'v'................... Equation 1

Note: Since both skater were initially at rest, u = u' = 0 m/s.

Also Note that both skater moves in opposite direction after collision, and let the direction of the first skater be positive and that of the second skater be negative.

mv = -m'v'.......................... Equation 2

make m' the subject of the equation above,

m' = -mv/v'..................... Equation 3

Given: m =48 kg, v = 0.67 m/s, v = -0.83 m/s (opposite direction to the first skater)

substitute into equation 3

m' = -48(0.67)/-0.83

m' = -32.16/-0.83

m' = 38.75 kg

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A string under a tension of 36 N is used to whirl a rock in a horizontal circle of radius 3.6 m at a speed of 16.12 m/s. The str

ExtremeBDS [4]

Answer:

6010.457N

Explanation:

Centripetal acceleration = a= V²/R

At a radius of 3.6m and velocity of 16.12m/s,

Acceleration is

a = 16.12²/ 3.6 = 72.182 m/s²

Force = Mass (m) * Acceleration (a)

36 = m * 72.182

m = 36/72.182

At breaking point

Radius = 0.468 m and Velocity = 75.1 m/s

a = V²/R = 75.1²/0.468

a = 12051.3 m/s

F = Mass(m) * Acceleration (a)

F = m * 12051.3

m = F/ 12051.3

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F = 12051.3 * 36/72.182

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3 years ago

A phonograph record 0.15 m in its radius rotates 18 times per 90 seconds what is the frequency?

ioda

Answer:

The frequency of the phonograph record is 0.2 Hz

Explanation:

The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period

The given parameters of the phonograph record are;

The radius of the record = 0.15 m

The number of times the phonograph record rotates, n = 18 times

The time it takes the phonograph record to rotate the 18 times, t = 90 seconds

The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)

∴ The frequency of the phonograph record, f = n/t = 18/(90 s) = 0.2 Hz

The frequency of the phonograph record = 0.2 Hz.

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2 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

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a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

a)

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

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Here the frequency increases by 1 MHz: the original frequency was

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so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

c)

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$

Now we can find the radiation intensity with the equation

$I=\frac{P}{A}$

Where the area is

$A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2$

And substituting,

$I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2$

Learn more about electromagnetic radiation:

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2 years ago

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3 years ago

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Doss [256]

Explanation:

The total energy of an aircraft flying in the atmosphere can be calculated using equation 1. [2]

E = ½ m v2 + mgh

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3 years ago