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Aliun [14]
1 year ago
12

Do you know what NBA basketball player is this. And what is he doing to stay healthy.

Physics
1 answer:
Ede4ka [16]1 year ago
4 0

Answer:

That's Chris Paul.

Explanation:

He focuses on his body maintenance, he's vegan. He insures he's getting the right nutritional food in his diet. He stretches a lot and does foam rolls.

HOPE THIS HELPS! GOOD LUCK!

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What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?​
givi [52]

Answer:

1.2 x 10^-12

Explanation:

3/2.5 x 10^-6/10^6

1.2 x 10^-6 x 10^-6

1.2 x 10^-12

6 0
2 years ago
A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails
harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

y=h+ut+\frac{1}{2}gt^2

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

v^2-u^2 = 2ad

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

a=\frac{v-u}{t}

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

8 0
2 years ago
A spring-loaded gun can fire a projectile to a height h if it is fired straight up. if the same gun is pointed at an angle of 45
Naily [24]
When firing straight up:
v^2 = u^2 - 2gh, where v = final velocity = 0, u = initial velocity, g = gravitational acceleration, h = maximum height attained.

Then,
0 = u^2 - 2gh
u = Sqrt (2gh) ---- (1)

When firing at 45°,
Initial velocity, U = u Sin 45 = Sqrt (2gh)·Sin 45

Maximum height, H = U^2*(Sin Ф)^2/2g

substituting;
H = [Sqrt (2gh)·Sin 45]^2*(Sin 45)^2]/2g
H = [2gh*(Sin 45)^2*(Sin 45)^2]/2g
H = [h*(Sin 45)^4] = h/4

Therefore, maximum height when the gun fires at 45° is a quarter of maximum height when the gun fires vertically up.
6 0
3 years ago
Fill in the blank
ohaa [14]
That's the accuracy of the measurement.
6 0
3 years ago
Read 2 more answers
What happens when a sound wave passes from water to air?
s344n2d4d5 [400]
It slows down. The molecules/particles aren't as close together, resulting in slower movement.
4 0
2 years ago
Read 2 more answers
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