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andrew-mc [135]
3 years ago
13

A 20-kg object sitting at rest is struck elastically in a head-on collision with a 10-kg object initially moving at 3.0 m/s. Fin

d the final velocity of the 10-kg object after the collision.

Physics
1 answer:
Sati [7]3 years ago
3 0

Answer:1m/s

Explanation: As the stationary ball is hit by the moving ball ,the two moves together after collision, with a single velocity. The attached photo further explains how the answer is calculated

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What is the frequency of a radio wave with an energy of 3.686 × 10−24 j/photon?answer in units of hz?
Makovka662 [10]

To solve this problem, we have to use the formula:

E = h f

where E is total energy, h is Plancks constant 6.626x10^-34 J s, f is frequency

 

f = E / h

f = 3.686 × 10−24 J / (6.626x10^-34 J s)

<span>f = 5.56 x 10^9 Hz</span>

4 0
3 years ago
A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is
alexgriva [62]

Answer:

R=4.22*10⁴km

Explanation:

The tangential speed v of the geosynchronous satellite is given by:

v=\frac{2\pi R}{T}

Because 2\pi R is the circumference length (the distance traveled) and T is the period (the interval of time).

Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:

F_c=\frac{mv^{2} }{R}

If we substitute the expression for v in this formula, we get:

F_c=\frac{m(\frac{2\pi R}{T})^{2}}{R}=\frac{4m\pi ^{2}R}{T^{2}}

Since the centripetal force is the gravitational force F_g between the satellite and the Earth, we know that:

F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }

Where G is the gravitational constant (G=6.67*10^{-11} Nm^{2}/kg^{2}) and M is the mass of the Earth (M=5.97*10^{24}kg). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:

R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km

This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.

5 0
3 years ago
what is the gravitational potential energy of a 150-kg object that is suspended 5-m above the earth's surface
77julia77 [94]
150x5x9.7(gravity on earth)=7275
7 0
3 years ago
Which of the following statements is TRUE for high-visibility clothing? A. High-visibility clothing helps to reduce insect probl
Vlad1618 [11]

Answer:

The answer for the above statement is:

C. High-visibility clothing is important to wear in areas with moving vehicles.

because in bright clothes you are easier to see, so people driving can see you.

Explanation:

3 0
3 years ago
Read 2 more answers
INSTI
max2010maxim [7]

Answer:

B

Explanation:

8 0
3 years ago
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