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Slav-nsk [51]
3 years ago
9

A ball is kicked from the ground with an initial speed v0 at an angle θ above the horizontal. Which of the following best descri

bes the magnitudes of the velocity and acceleration of the ball when it reaches the highest point of its trajectory?
Physics
1 answer:
adoni [48]3 years ago
5 0

Answer:

Explanation:

velocity of projection = vo

angle of projection = θ

When the ball is at maximum height, its vertical component of velocity is zero, So it has only the horizontal component of velocity at highest point.

velocity at highest point = vo Cos θ

The acceleration acting on the ball is only the acceleration due to gravity which always acts vertically downward.

So, teh acceleration of the ball at the highest point is g.

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Is being social media famous worth real-life relationships?
Maru [420]

Answer:

No

Explanation:

Social media fame isnt worth it, because youre not really able to connect with people. Real life relationships with friends, family, and everything else is way more important. having people notice you or be attracted to you through the internet is nothing compared to real life social interactions. Dont leave people just to be online, youre going to regret it.

3 0
3 years ago
A geosynchronous satellite moves in a circular orbit around the Earth and completes one circle in the same time T during which t
andreyandreev [35.5K]

Answer:

Explanation:

The time period of geosynchronous satellite must be equal to T .

The radius of its orbit will be (  R+ h )

orbital velocity  V₀ =  \sqrt{\frac{GM}{( R+h)} }

Time period T = 2π( R + h ) / V₀

= 2π( R + h ) x \sqrt{\frac{( R+h)}{GM } }

\frac{T^\frac{2}{3}(GM)^\frac{1}{3}  }{(2\pi )^\frac{2}{3} } = R +h

h = \frac{T^\frac{2}{3}(GM)^\frac{1}{3}  }{(2\pi )^\frac{2}{3} } - R.

7 0
3 years ago
Read 2 more answers
The distance between the earth and sun is 1.5 x 108 kilometers and the speed of light is 3.00 x 108 meters per second. Calculate
butalik [34]

Answer:

time = 8.3333 minutes.

Explanation:

distance between earth and sun = 1.5 * 10^{8}km

speed of light = 3* 10^{8}m/s

convert the distance unit from km to m so we can have uniform units.

distance between earth and sun = 1.5 *10^{8}*1000m

distance between earth and sun = 1.5 * 10^{11}m

speed = distance /time

time = distance / speed

time = \frac{1.5*10^{11} }{3*10^{8} }

= 0.5*10^{3}

time =500 sec

time = 500/60 minutes

time = 8.3333 minutes.

3 0
3 years ago
A charged sphere with 1 × 10 8 units of negative charge is brought near a neutral metal rod. The half of the rod closer to the s
jeyben [28]

Answer:

-4*10⁴ units.

Explanation:

As the metal rod was initially neutral (which means that it has the same quantity of positive and negative charges), after being close to the charged sphere, as charge must be conserved, the total charge of the metal rod must  still remain to be zero.

So, if due to the influence of the negative charge in the sphere, the half of the road closer to the sphere has a surplus charge of +4*10⁴ units, the charge on the half of the rod farther from the sphere must be the same in magnitude but of the opposite sign, i.e., -4*10⁴ units.

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