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Dmitriy789 [7]
3 years ago
12

A 20 g ball is fired horizontally with speed v0 toward a 100 g ball hanging motionless from a 1.0-m-long string. the balls under

go a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle umax = 50. what was v0 ?
Physics
1 answer:
Novay_Z [31]3 years ago
7 0

Sorry but i dont know i was doing this for some point sorry

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A physics student throws a ball straight up. The student catches the ball in exactly the same place from which it was released.
mr Goodwill [35]

Answer:

The correct answer is H ÷ ¹/₂T

Explanation:

The formula for velocity is distance covered ÷ time.

Neglecting air resistance;

If the ball's time of overall time flight is T, the time it will take for the second half/return trip is ¹/₂T.

If the ball's maximum height above its released point is H, the height will also be the distance it covered for the second part of the trip since the student caught the ball in the exact same place the ball was thrown. Hence, the distance for the second half of the trip will be H.

Since velocity = distance/time

The average velocity during the second half of the trip will be = H ÷ ¹/₂T

4 0
4 years ago
David is driving a steady 28.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady
rodikova [14]

Answer:

Explanation:

Let t represent the time for Tina to catch David.

Hence, considering the equation of linear motion S = ut + 1/2at^2..... 1

For David u = 28.0 m/s where 'a' is set to nought

S = ut

S = 28t.......2

For Tina consider equation 1

Where acceleration = 2.90m/s^2 and u is set at nought

S = 1/2×2.90 m/s×t^2.......3

Equate 2 and 3

28t = 1.45t^2

Divide through by t

28 = 1.45t

t = 28/1.45

t = 19.31seconds

Now put the value of t into equation 3

S = 1/2×2.90 m/s×t^2.......3

= 1.45×20×20

= 580m

Tina must have driven 580meters before passing David

Considering the equation of linear motion : V^2 = U^2+2as

Where u is set at nought

V^2 = 2as

V^2 = 2×2.9×580

V^2 = 3364

V = √3364

V = 58m/s

Her speed will be 58m/s

7 0
3 years ago
One student bangs two bricks together
Romashka-Z-Leto [24]

Answer:

This could be done if a stop watch is used to calculate the time taken to hear the echo and a rule should be used to calculate the distance between the bricks and the wall. Then divide distance by time

Explanation:

I hope this is what you need

PLEASE MAKE ME BRAINLIEST

7 0
4 years ago
A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A
zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

V_{orbital} = \sqrt{\frac{GM_E}{R}}

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}

V_{orbital} = 5591.62m/s

V_{orbital} = 5.591*10^3m/s

PART B) The period of satellite is given as,

T = 2\pi \sqrt{\frac{r^3}{Gm_E}}

T = \frac{2\pi r}{V_{orbital}}

T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}

T = 238.61min

PART C) The gravitational force on the satellite is given by,

F = ma

F = \frac{1}{4} mg

F = \frac{270*9.8}{4}

F = 661.5N

5 0
3 years ago
How does a sound wave transfer energy to your ears?
inessss [21]

Answer:

I think it's C!

Explanation:

Sound waves travel at 343 m/s through the air and faster through liquids and solids. The waves transfer energy from the source of the sound, e.g. a drum, to its surroundings. Your ear detects sound waves when vibrating air particles cause your eardrum to vibrate. The bigger the vibrations the louder the sound.

Hope this helps!

4 0
3 years ago
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