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2 years ago

Can a apple fall from a tree and not break open?

2 answers:

6 0

Well as you've seen before most likely, an apple can fall right off a tree, and stay perfectly fine. its all dependent on the variables. What angle is the apple falling at? Whats the speed, height, even the size of the apple? We must take into aware anything can happen, but its all behind the variables.

Brainliest if helpful

P.S Fact Check with the second person who just answered

Mrac [35]2 years ago

3 0

Answer:

yes

Explanation:

i had an apple tree and it happens all the time most often they don't break open.

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Which of the following is an example of a population?

svlad2 [7]

D the gray wolves in the forest.is the correct answer.

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3 years ago

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Once you start pulling your object with less force than friction, what should you expect your object to do? What about when your

forsale [732]

#Case -1

If Pulling force is less than frictional force the object won't move .

#Case-2

If Pulling force is greater than frictional force then object will be .

In order to calculate friction force you need Limiting friction first .

$\\ \sf\longmapsto F_L=\mu sN$

u s is coefficient of static friction and N is normal reaction

$\\ \sf\longmapsto F_L=\mu smg$

As N=mg

8 0

2 years ago

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A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, $v_{o} = 20.0 m/s$

Angle made by the ramp, $\theta = 22.0^{\circ}$

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH$

where

H = $dsin22^{\circ} = 5sin22^{\circ}$

g = $9.8 m/s^{2}$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ}}$ = 19.06 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m$

(b) now, considering the coefficient of friction bhetween ramp and the ball, $\mu = 0.150$ :

velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH - \mu gd$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5}$ = 18.7 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m$

6 0

2 years ago

What do echolocation and ultrasounds have in common?

Natali5045456 [20]

They both are mechanical waves.

Explanation:

Echolocation and ultrasounds are both mechanical waves that require a medium to trace through. They both process and transfer information through waves; the difference is that we humans cannot hear ultrasound waves.

8 0

1 year ago

A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f

zalisa [80]

Explanation:

Charge=5.00 E-5

Force=0.751N

F=qE

0.751=5.00 E-5*E

E=1.502*10⁴

option a

8 0

3 years ago

Read 2 more answers