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2 years ago

14

The student toses a ball into the air and then watches the ball fall back down to earth. When does the ball have the most potent

ial energy

1 answer:

pishuonlain [190]2 years ago

6 0

Answer:

when the ball is at its highest in the air.

Explanation:

I don't know for sure, but when the ball is in the air it has potential energy to fall(or something like that).

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.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c

ohaa [14]

Answer:

a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

F t = 2 (m v)

therefore the average force is

F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

Em_f = K = ½ m v²

energy is conserved in this system

Em₀ = Em_f

m g h = ½ m v²

v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

F = 2m √(2gh) /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

F = m a

a = F / m

a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

a = 2 √ (2 9.8 1.5) / 10⁻³

a = 1.1 10⁵ m / s²

6 0

2 years ago

Two different liquids are poured into a jar until it is half full. The jar is then sealed shut and shaken. The liquids undergo a

Lilit [14]

Answer:

A closed system.

Explanation:

The three major types of system are: open, closed and isolated. Open system interacts with its surroundings with respect to its particles and energy. A closed system interacts with its surroundings with respect to energy but not its particles. While an isolated system does not interact with its surroundings in any way.

Therefore, after the jar is sealed, it is an example of a closed system. This is because the emitted gas could not escape into the surroundings, but thermal energy was emitted into its surroundings after the chemical reaction has taken place.

7 0

3 years ago

When a 70 kg man sits on the stool, by what percent does the length of the legs decrease? Assume, for simplicity, that the stool

allochka39001 [22]

The diameter of one leg of the stool is missing and it's 2cm.

Answer:

(ΔL/L) = 0.00729%

Explanation:

If the Weight of the man is W, the weight will be distributed equally on the 3 legs and so the reactions for each leg will be W/3 or F/3.

Now, Youngs modulus(Y) of douglas fir wood is about 1.3 x 10^(10) N/m^2. Gotten from youngs modulus of common materials.

Now, weight of man is 70kg.

Now diameter of one leg is 2cm.so radius of one leg = 2/2 = 1cm = 1 x 10^(-2)m

Area for one leg is; π( 1 x 10^(-2)m)^2 = 3.14 x 10^(-4)m

Now as stated earlier, the force on one leg is; F/3.

Now F = mg = 70 x 9.81 = 686.7N

So, force on one leg = 686.7/3 = 228. 9N

Now we know youngs modulus(Y) = Stress/Strain.

Stress = F/A while Strain = ΔL/L

Therefore Y = (F/A) / (ΔL/L)

And therefore, (ΔL/L) = F/(AY)

So (ΔL/L) = 228.9/(3.14 x 10^(-4))x(1.3 x 10^(10)) = 7. 29 x 10^(-5)

When expressed in percentage, it becomes 0.00729%

7 0

3 years ago

A rural mail carrier logged the mileage on the odometer at the start of the mail route as 31,500 miles. After 6.5 hours, the car

dusya [7]

Answer: $32.77\ mi/hr$

Explanation:

Given

Initially Reading on the odometer is $31,500\ miles$

Final reading on the odometer is $31,713\ miles$

Time taken is $t=6.5\ hr$

average velocity $=\dfrac{\text{Displacement}}{\text{time}}$

$v_{avg}=\dfrac{31713-31500}{6.5}$

$v_{avg}=\dfrac{213}{6.5}$

$v_{avg}=32.77\ mi/hr$

Thus the average velocity of mail truck is $32.77\ mi/hr$

8 0

3 years ago

Principal of simple machine?

sergij07 [2.7K]

Answer:

if there is no friction in a simple machine, work output and work input are found equal in that machine

Explanation:

3 0

3 years ago