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3 years ago

A 3 kilogram block is sliding across the ground at a velocity of 8 meters per second. What is the momentum of the block?

Physics

1 answer:

oksano4ka [1.4K]3 years ago

4 0

Answer:

24 kgm/s

Explanation:

Momentum (p) = mass x velocity

Mass = 3kg , velocity 8m/s

momentum (p) = 3kg x 8m/s

= 24kgm/s

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For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci

MAXImum [283]

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V

maximum value of current = 3.20 A

now by ohm's law we can find the impedence as

$z = \frac{V_o}{i_o}$

now we will have

$z = \frac{125}{3.20} = 39.06 ohm$

Part b)

Now we also know that

$\frac{R}{z} = cos\theta$

$\theta = 0.982 rad = 56.3 degree$

now we have

$\frac{R}{39.06} = cos56.3$

$R = 21.7 ohm$

5 0

3 years ago

Tom rides his motorcycle at a speed of 15 meters/second for an hour.

Dahasolnce [82]

Answer:

1 hour to ride his motorcycle

4 0

3 years ago

A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.

ArbitrLikvidat [17]

Answer:

L = 499 m

Explanation:

If we assume that the speed of sound is constant, that travels along a straight line, and that the echo is instantaneous, we can find the total distance travelled by the sound, as follows, just applying the definition of average velocity:

$\Delta x = v_{s} * t = 343 m/s* 2.91 s = 998 m$

If we assume that the time needed to reach to the cliff, is the same used for the return travel, the length of the lake will be exactly half of the total distance calculated:

$l_{lake} = \frac{\Delta x}{2} = \frac{998m}{2} = 499 m$

The length of the lake is 499 m.

8 0

3 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

Zina [86]

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:

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:

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

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:

$P\propto f$

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

$f_0 = 60 MHz$

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 %.

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

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