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Alenkasestr [34]

3 years ago

15

Two balls are kicked with the same initial speeds. Ball A was kicked at the angle 20° above horizontal and ball B was kicked at

the angle 75° above horizontal. What ball will have bigger speed at the highest point of their trajectory? O Ball A O Ball B O Their will have equal speeds O Impossible to answer without knowing their actual initial speeds.

1 answer:

GREYUIT [131]3 years ago

5 0

Answer:

Ball A

Explanation:

Let the initial speed of the balls be u .

Angle of projection for ball A = 20°

Angle of projection for ball B = 75°

As we know that at highest point, the ball has only horizontal speed which always remains constant throughout the motion because the acceleration in horizontal direction is zero.

Speed of ball A at highest point = u Cos 20° = 0.94 u

Speed of ball B at highest point = u Cos 75° = 0.26 u

So, the ball A has bigger speed than B.

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The volume of water in the Pacific Ocean is about 7.00 × 108 km3. The density of seawater is about 1030 kg/m3. For the sake of t

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The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.

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First we calculate the mass through the ratio given by density.

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$m = 7.210*10^{11}Kg$

PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon

Now we define the radius at the most distant point

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

Then the potential energy at this point would be,

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$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

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PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.

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$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

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3 years ago

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4 years ago

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3 years ago

A child and sled with a combined mass of 58.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed o

erma4kov [3.2K]

Answer:

The height is $h = 0.5224 \ m$

Explanation:

From the question we are told that

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The speed of the sled is $u = 3.20 \ m/s$

Generally applying SOHCAHTOA on the slope which the combined mass is down from

Here the length of the slope(L) where the combined mass slides through is the hypotenuses

while the height(h) of the height of the slope is the opposite

Hence from SOHCAHTOA

$sin (\theta) = \frac{h}{L}$

=> $Lsin(\theta) = h$

Generally from the kinematic equation we have that

$v^2 = u^2 + 2aL$

Here the u is the initial velocity of the combined mass which is zero since it started from rest

and a is the acceleration of the combined mass which is mathematically evaluated as

$a = g * sin (\theta )$

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=> $2Lsin(\theta ) = \frac{v^2 - u^2 }{g}$

=> $h = \frac{ v^2 - u^2}{2g}$

=> $h = \frac{ 3.20^2 - 0^2}{2 * 9.8 }$

=> $h = 0.5224 \ m$

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3 years ago