Fill in the blanks please.

A 5.0 kg cannonball is fired horizontally at 68 m/s from a 15-m-high cliff. A strong tailwind exerts a constant 12 N horizontal

jeka57 [31]

Answer:

3.7 m

Explanation:

ASSUMING this means extra distance beyond where the cannonball would land WITHOUT the wind assistance but in general ignoring air resistance. Hmmmmmm...tricky

The ball drops from vertical rest to ASSUMED horizontal ground 15 m below in a time of

t = √ (2h/g) = √(2(15)/9.8) = 1.75 s

Without the tail wind, the ball travels horizontally

d = vt = 68(1.75) = 119 m

The tailwind exerts a constant acceleration on the ball of

a = F/m = 12/5.0 = 2.4 m/s²

The average horizontal velocity during the flight is

v(avg) = (68 + (68 + 2.4(1.75)) / 2 = 70.1 m/s

so the distance with tailwind is

d = v(avg)t = 70.1(1.75) = 122.675 m

The extra distance is 122.675 - 119 = 3.675 = 3.7 m

8 0

3 years ago

object x and y fall from a same height and object x is heavier than y which object would fall faster qnd y

hjlf

Answer: They’d fall at the same speed, because air resistance is the only thing that makes an object fall faster than another. There’s a video somewhere on the internet of a bowling ball and a feather falling at the same speed in a vacuum, if you look for it. Hope this helps!

4 0

3 years ago

Where do most metamorphic processes take place?

inysia [295]

Answer:

Most metamorphic processes take place deep underground, inside the earth's crust.

Explanation:

During metamorphism, protolith chemistry is mildly changed by increased temperature (heat), a type of pressure called confining pressure, and/or chemically reactive fluids. hope this helps you :)

8 0

4 years ago

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

Liula [17]

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

v = v₀ - a t

a = (v₀- v) / T

On planet X

v = v₀ - a' t’

a ’= (v₀-v) / 2T

Let's substitute the land values in plot X

a’= a / 2

Now let's use Newton's second law

W = ma

m g = m a

a = g

We substitute

a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

4 0

3 years ago

if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav

allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, $R_{max}$ = 3 m

The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth

The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, $R_{max}$ , of a projectile motion is found by using the following formula

$R_{max} = \dfrac{u^2}{g}$

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

$R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }$

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, $g_{Moon}$ = 1/16 × g

∴ $g_{Moon}$ = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

$R_{max \ Moon} = \dfrac{u^2}{g_{Moon}} = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m$

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is $R_{max \ Moon}$ ≈ 48 m

8 0

3 years ago