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ioda
3 years ago
15

Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.

Physics
1 answer:
Marina86 [1]3 years ago
3 0
Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s 
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5  
Frequency f = 2 Hz
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A kettle is rated at 1 kW, 220 V. Calculate the working resistance of the kettle.
Anna [14]

Explanation:

Power of electric kettle, P = 1 kW

Voltage, V = 220 V

(a) Electric power is given by the formula as follows :

P=\dfrac{V^2}{R}

R is resistance

R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{10^3}\\\\R=48.4\ \Omega

(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.

Energy supplied is given by :

E=P\times t

P is power, P=\dfrac{V^2}{R}

E=\dfrac{V^2}{R}t\\\\E=\dfrac{(220)^2}{48.4}\times 180\\\\E=180000\ J\\\\E=180\ kJ

5 0
3 years ago
A transformer has 100 turns in its primary coil and 75 turns in its secondary coil. If the input voltage is 12.0 V, what is the
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This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broa
Montano1993 [528]

Answer:

2.77287\times 10^{15}\ m

Explanation:

P = Power = 50 kW

n = Number of photons per second

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

\nu = Frequency = 781 kHz

r = Distance at which the photon intensity is i = 1 photon/m²

Power is given by

P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s

Photon intensity is given by

i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m

The distance is 2.77287\times 10^{15}\ m

3 0
3 years ago
The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa
Sonja [21]

Answer:

m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

c_{pA} > c_{pB}

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

T_{iA}= T_{iB}

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

Q = m c_p \Delta T

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

Q_A = Q_B

And if we replace the formula for sensible heat we got:

m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B

And if we replace for the change of the temperature we got:

m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})

And since T_{iA}= T_{iB}= T we have this:

m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

4 0
3 years ago
You are running at a speed of 10km/h and hit a patch of mud. Two seconds later you speed is 8km/h. What is your acceleration in
Vlad1618 [11]

Answer:

0.28 m/s^2

Explanation:

Acceleration is given by

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time interval

In this problem:

u = 10 km/h \cdot \frac{1000 m/km}{3600 s/h}=2.78 m/s is the initial velocity

v = 8 km/h \cdot \frac{1000 m/km}{3600 s/h}= 2.22 m/s is the final velocity

t = 2 s is the time

Substituting, we find the acceleration:

a=\frac{2.78-2.22}{2}=0.28 m/s^2

5 0
3 years ago
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