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3 years ago

Better climate prediction is important for which if the following reasons

Physics

1 answer:

brilliants [131]3 years ago

3 0

Where the answer choice then i can answer it

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Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.110 m, calcula

velikii [3]

Answer:

6.1328 kg

60.16284 N

Explanation:

r = Radius of ball = 0.11 m

$\rho$ = Density of fluid = $1.1\times 10^3\ kg/m^3$ (Assumed)

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of ball

V = Volume of ball = $\frac{4}{3}\pi r^3$

The weight of the bowling ball will balance the buouyant force

$W=F_b\\\Rightarrow mg=V\rho g\\\Rightarrow m=\frac{V\rho g}{g}\\\Rightarrow m=V\rho\\\Rightarrow m=\frac{4}{3}\pi 0.11^3\times 1.1\times 10^3\\\Rightarrow m=6.1328\ kg$

The mass of the bowling ball will be 6.1328 kg

Weight will be $6.1328\times 9.81=60.16284\ N$

5 0

3 years ago

During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t

tia_tia [17]

1.A) 4.9 m

AL2006 Ace

The instant it was dropped, the ball had zero speed.

After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered 4.9 meters.

ANYTHING you drop does that, if air resistance doesn't hold it back.

Read more on Brainly.com - brainly.com/question/11776597#readmore

2 idk sorry

5 0

3 years ago

Read 2 more answers

At what height h above the ground does the projectile have a speed of 0.5v?

maw [93]

Answer:

$h=\dfrac{3v^2}{8g}$

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

$v=u+at$

$v=u-gt$

Initial speed of the projectile is v and final speed is 0.5 v.

$0.5v=v-gt$

$t=\dfrac{v}{2g}$

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

$h=vt-\dfrac{1}{2}gt^2$

$h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2$

$h=\dfrac{3v^2}{8g}$

So, the height of the projectile above the ground is $\dfrac{3v^2}{8g}$ . Hence, this is the required solution.

6 0

3 years ago

Please help! This is due in 10 minutes

Elenna [48]

Answer:

Atom - the basic particle of matter

Density - calculated from measurements of mass and volume

Motion - calculated from measurements of distance and time

Energy - can change form and move matter

Matter - the scientific word for <em>stuff</em>

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Hope this helps! Please mark brainliest if correct :D

8 0

2 years ago

I’ll literlly give you points <br> and make you brainiest

sesenic [268]

Answer:

Explanation:

4 0

3 years ago