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IrinaVladis [17]

3 years ago

7

A hunter stood at about 60 m away from a tree. He used the bow to release the arrow in order to shoot a coconut held by a monkey

in the tree. The coconut was at a height of 25 m from the ground. The hunter aimed directly at the coconut and the arrow left the bow at 45 m/s making an angle of 20 degree to the horizontal. At the moment thebhunter released the arrow, the monkey dropped the coconut such that it falls vertically from rest. Neglect air resistance and the arrow's size. What is the time taken for the arrow to hit the coconut?

1 answer:

Sergeeva-Olga [200]3 years ago

5 0

Answer:

$v = u + at (upward)\\ 0 = 45 \sin(20) - 9.81t \\ t = 1.56s$

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A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer

777dan777 [17]

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

$K = \frac{1}{2}mv^{2}$

By substituting the values

$1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}$

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, $v = \left (\frac{4.9}{1000} \right )\times \left ( \frac{3600}{1} \right )$

v = 17.64 km/h

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3 years ago

The gamma ray released by each decay carries 140kev of energy. find the total energy e released by decays in the 2 hours.

olya-2409 [2.1K]

First, we would need to know the decaying isotope.
Next, we use the decay formula
A = Ao e^(-kt)
After determining the remaining amount after two hours, the decay reaction can be used to determine the number of gamma rays released. If the given is in terms of mole, then the total energy is
E = 140n KeV where n is the number of moles of gamma rays released

8 0

3 years ago

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Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun

Dominik [7]

Answer:

a. $K_{Axis}=2.574x10^{29}J$

b. $K_{Orbit}=2.6577x10^{33}J$

Explanation:

$K_{Axis}=\frac{1}{2}I*w^2$

$I_{Sphere}=\frac{2}{5}*m*r^2$

$w=\frac{2\pi }{T}$ , $T=24hrs*\frac{3600s}{1hr} =86400s$

radius earth = 6371 km

mass earth = 5,972*10^24 kg

a.

$K_{Axis}=\frac{1}{2}*\frac{2}{5}*m*r^2*(\frac{2\pi}{T})^2$

$K_{axis}=\frac{4\pi^2}{5}*5.98x10^{24}kg*(6.38x10^6m)^2*(\frac{1}{86400s})^2$

$K_{Axis}=2.574x10^{29}J$

b.

$T=1year*\frac{365day}{1year}*\frac{24hr}{1day}*\frac{3600s}{1hr}=31536000s$

$K_{Orbit}=\frac{1}{2}*I*w$

$I=m*r^2$

$K_{Orbit}=\frac{1}{2}*m*r^2*(\frac{2\pi}{T})^2$

$K_{Orbit}=\frac{4\pi^2}{5}*5.98x10^{24}*6.38x10^6m*(\frac{1}{31.536x10^6s})^2$

$K_{Orbit}=2.6577x10^{33}J$

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4 years ago

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6 0

3 years ago

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A spherically spreading EM wave comes from an 1800-W source. At a distance of 5.0 m, what is the intensity, and what is the rms

Aleonysh [2.5K]

Explanation:

It is given that,

Power of EM waves, P = 1800 W

We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.

Intensity,

$I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2$

The formula that is used to find the rms value of the electric field is as follows :

$I=\epsilon_o cE^2_{rms}$

c is speed of light and $\epsilon_o$ is permittivity of free space

So,

$E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m$

Hence, this is the required solution.

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3 years ago