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VikaD [51]
3 years ago
5

What is the free-throw line used for?

Physics
1 answer:
Genrish500 [490]3 years ago
5 0

Answer:

it be the second one lad

Explanation:

i have done the playing pf the basketball

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By what factor would your weight increase if you could stand on the sun? (never mind that you can't.)
Fofino [41]

Gravity on the surface of sun is given as

g = \frac{GM}{R^2}

here we know that

M = 1.98 \times 10^{30} kg

R = 6.95 \times 10^8 m

now we will have

g = \frac{(6.67 \times 10^{-11})(1.98 \times 10^30)}{(6.95 \times 10^8)^2}

g = 273.4 m/s^2

now we need to find the ratio of weight on surface of sun and on surface of Earth

R = \frac{mg_{sun}}{mg_{earth}}

R = \frac{273.4}{9.8} = 28 times

so weight will increase by 28 times

6 0
3 years ago
Look at this picture of a frog.
fredd [130]

Answer:

Biosphere

Explanation:

The biosphere consist of all living organisms

6 0
3 years ago
Read 2 more answers
I was driving along at 20 m/s , trying to change a CD and not watching where I was going. When I looked up, I found myself 46 m
vampirchik [111]

Answer:

6.46393559312 m/s²

Explanation:

Time taken to cover 56 m

t=\dfrac{56}{29}\ s

Distance covered in 0.42 seconds

0.42\times 20=8.4\ m

From equation of linear motion

s=ut+\frac{1}{2}at^2

8.4+20(\dfrac{56}{29}-0.42)+\dfrac{1}{2}a(\dfrac{56}{29}-0.42)^2=46\\\Rightarrow a=(46-8.4-20(\dfrac{56}{29}-0.42))\times\dfrac{2}{(\dfrac{56}{29}-0.42)^2}\\\Rightarrow a=6.46393559312\ m/s^2

The minimum acceleration is 6.46393559312 m/s²

5 0
3 years ago
if an object is moving with a velocity of 24m/s and has an acceleration of -4m/s how long will it take to stop
ololo11 [35]
The negative sign on the acceleration is only a vector quantity that means the object is accelerating to the left. Hence, we can only focus on it magnitude which is 4 m/s^2. Acceleration is the change in velocity over time. The change in velocity must be 24 m/s - 0 m/s, if you want the object to stop. Therefore,

a = (v2 - v1)/t
4 = (24 - 0)t
t = 6 seconds

The object will stop after 6 seconds.
7 0
3 years ago
It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you
Debora [2.8K]
You'd get an extra 40/60 of the energy, or 2/3. Multiply 5/3 by the required energy to get the actual consumption.
7 0
3 years ago
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