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Ludmilka [50]
3 years ago
9

A particle is projected from a point on a horizontal plane and has an initial velocity of 28root 3 m/s at an angle of 60 degree

.find the greatest heights reached by the particle ​
Physics
1 answer:
Lelechka [254]3 years ago
4 0

Answer:

uujjjjjctc7tox7txr9ll8rz8lr5xl8r6l8dl85x8rl5x8rl5x8rl5xrx8l58rk5xr8l5xr6l8xr68lc

You might be interested in
Which condition is necessary for total internal reflection? A. The refracted ray should lie along the boundary of the two media.
g100num [7]

There are two conditions necessary for total internal reflection, which is when light hits the boundary between two mediums and reflects back into its original medium:

Light is about to pass from a more optically dense medium (slower) to a less optically dense medium (faster).

The angle of incidence is greater than the defined critical angle for the two mediums, which is given by:

θ = sin⁻¹(n_{fast}/n_{slow})

Where θ = critical angle, n_{fast} = refractive index of faster medium, n_{slow} = refractive index of slower medium.

Choice C gives one of the above necessary conditions.

6 0
3 years ago
On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan
frutty [35]

Answer:

The distance in kilometers is 4012 ×10^{-6} km.

Explanation:

We know that the conversion of 1 millimeters is equal to 10^{-3} meter. And then the conversion of 1 meter is equal to 10^{-3} km. Then the conversion of 1 millimeter to km will be

1 mm = 10^{-3} m

1 m = 10^{-3} km

So, 1 mm = 10^{-3}×10^{-3} km = 10^{-6} km.

As here the the distance is 4012 mm, then the distance in km will be

4012 mm = 4012 ×10^{-6} km.

So the distance is 4012 ×10^{-6} km.

5 0
3 years ago
I need help i don’t want to go to summer school
Sonbull [250]

Answer:

makes you hungry is not a science based benefit of meditation

7 0
3 years ago
A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground.
Nadusha1986 [10]
<h2>Thus the force of friction is 235 N</h2>

Explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E =  27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting = \frac{1}{2} m v²

=   \frac{1}{2} x 27 x ( 6.1 )²  = 490 J

The force of friction f = P.E - K.E = 3290 J

Because the work done = Force x Distance

Thus frictional force = \frac{3290}{14} = 235 N

3 0
4 years ago
A 20.0-kg rock is dropped and hits the ground at a speed of 90.0 m/s. Calculate the rock’s gravitational potential energy before
KonstantinChe [14]
Answer:

<span>GPE=81000J or 81kJ</span>

Explanation

Potential Energy = mgh = 20 x 9.8 x ? 

<span>To find H use one of the equation of motion </span>

<span>= [(90)^2 - 0 ] / 2(9.8) </span>

<span>Potential Energy = mgh = 20 x 9.8 x 8100 /2(9.8) = 81000 J</span>

5 0
3 years ago
Read 2 more answers
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