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Murljashka [212]
2 years ago
5

Explain constant and changing velocity.

Physics
2 answers:
Igoryamba2 years ago
7 0

Answer:

To summarize, an object moving in a uniform circular motion is moving around the perimeter of the circle with a constant speed. While the speed of the object is constant, its velocity is changing. Velocity, being a vector, has a constant magnitude but a changing direction.

alina1380 [7]2 years ago
3 0

Answer:

In order for the velocity to be constant, the magnitude of the velocity (or speed) and the direction of the velocity must not change. Therefore, an object that travels with constant velocity covers the same distance during each time interval and moves in the same direction during the same time interval.

Explanation:

<em>Credit to the person who wrote the comment on your problem, i wrote it here so it is easier seen</em>

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allochka39001 [22]

Answer:

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Explanation:

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2 years ago
The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a
s344n2d4d5 [400]

Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = 0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg

r = Radius of orbit of planet = 0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m

v = Orbital speed

The kinetic and potential energy balance is given by

\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days  

The orbital period is 14.45 days

5 0
2 years ago
What are the adaptions of the tundra
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2 years ago
3. According to the Guinness Book of Records the heaviest baby ever born weighed 29 lbs 4 oz. (29.25 lbs).
Alexxandr [17]

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6 0
3 years ago
The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it
Lilit [14]

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, m=0.02 kg

Kinetic energy imparted, K=1200 J

Length of rifle barrel, d=1 m

(a)

Let the speed of bullet when it leaves the barrel is v.

Kinetic energy, K=\frac{1}{2} mv^{2}

v=\sqrt{\frac{2K}{m} }

=\sqrt{\frac{2\times1200}{0.02} }

=346.4m/s

Initial speed of bullet, u=0

The average speed in the barrel, v_a_v_g=\frac{u+v}{2}

=\frac{0+346.4}{2} \\=173.2 m/s

Time taken by bullet to cross the barrel, t=\frac{d}{v}

=\frac{1}{173.2}\\ =0.00577 second

Power, P_a_v_g=\frac{W}{t}

=\frac{1200}{0.00577} \\=207.97kW

(b)

In projectile motion,

Maximum height, H_m=\frac{v^2\sin^2\theta}{2g} \\

Range, R=\frac{v^2\sin2\theta}{g}

given that, H_m=R

then, \frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter

5 0
2 years ago
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