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rusak2 [61]
3 years ago
13

A rubber ball is dropped from a height of 8m. After strikingthe floor, the ball bounces to a height of 5m. a. If the ball had bo

unced to a height of 8m, how would youdescribe the collision between the ball and the floor? b. If the ball had not bounced at all, how would you describethe collision between the ball and the floor? c. What happened to the energy lost by the ball during thecollision?
Physics
1 answer:
kifflom [539]3 years ago
5 0

Answer:

a) This means the collision between the ball and the floor is elastic.

b) This points to a perfectly inelastic collision between the ball and the floor as they stick together after collision

c) Check Explanation.

Explanation:

Collision of bodies are analysed according to whether both momentum and kinetic energy of the system is conserved, that is, if these two quantities before collision are equal to their values after collision.

In all types of collisions, momentum is usually conserved, but kinetic energy is conserved only in an elastic collision.

A ball dropped from a height of 8 m bounces up back to a height of 5 m.

a. If the ball had bounced to a height of 8m, how would you describe the collision between the ball and the floor?

The ball not bouncing back to a height of 8 m shows energy loss at some point in the total motion of the ball (most likely at the collision). If kinetic energy was conserved, the ball would bounce back up to the height at which it fell from (8 m) after the collision with the floor.

b. If the ball had not bounced at all, how would you describe the collision between the ball and the floor?

If the ball had not bounced at all, this means it lost all of its kinetic energy to the floor, and this points to a perfectly inelastic collision between the ball and the floor as they stick together after collision.

c. What happened to the energy lost by the ball during thecollision?

The energy lost during the collision is converted to another form, most likely responsible for some deformation on the ball & a minute deformation on the floor, converted to some form of heat as a result of the collision or into sound energy, usually, it's a combination of all This!

Hope this Helps!!!

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Glass absorbs ultraviolet (UV) rays from the Sun. Would a fraction of the incident UV light be reflected from the air/glass boun
fiasKO [112]

Answer:

Yes

Explanation:

Any transparent surface in practical is neither a perfect absorber of electromagnetic waves neither a perfect reflector. Generally all the transparent surfaces reflect some amount of irradiation and the other parts are absorbed and transmitted.

<u>That is given by as relation:</u>

\alpha+\rho+\tau=1

where:

\alpha= absorptivity which is defined as the ratio of the absorbed radiation to the total irradiation

\rho= reflectivity is defined as the ratio of reflected radiation to the total irradiation

\tau= transmittivity is defined as the ratio of total transmitted radiation to the total irradiation

6 0
3 years ago
A metal block has a mass of 5 grams and a volume of 1 cm3. What is the density of a piece of the same identical metal but with a
Fynjy0 [20]

Answer:

ρ = Mass / Volume     definition of density

ρ  = 5 g / 1cm^3 = 5 g / cm^3

Since the other object is made of the same metal its density is the same:

ρ = 5 g/cm^3

6 0
2 years ago
A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside
neonofarm [45]

Answer:

\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}, level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}

\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt}  = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}

By replacing all known variables:

\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0
3 years ago
Starting from rest, how far will a brick fall if it is going 15 m/s and accelerates at a rate of 9.8 m/s2?
masha68 [24]

Answer:

11.48 m

Explanation:

A brick starts from rest and gains a speed of 15 m/s and accelerates at 9.8 m/s^2

u = 0

v= 15

a= 9.8

s= ?

V^2 = U^2 + 2as

15^2 = 0^2 + 2 × 9.8×s

225= 19.6s

s= 225/19.6

s = 11.48m

Hence the brick will fall 11.48 m

7 0
3 years ago
Calculate the velocity of a 1650kg satellite that is in a circular orbit of 4.2 x 10^6m above the surface of a planet which has
Anastasy [175]
-- We're going to be talking about the satellite's speed. 
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.

-- The mass of the satellite makes no difference.

Since the planet's radius is  3.95 x 10⁵m  and the satellite is
orbiting  4.2 x 10⁶m  above the surface, the radius of the
orbital path itself is

                               (3.95 x 10⁵m) + (4.2 x 10⁶m)

                     =        (3.95 x 10⁵m) + (42 x 10⁵m)

                     =           45.95 x 10⁵ m

The circumference of the orbit is  (2 π R) =  91.9 π x 10⁵ m.

The bird completes a revolution every 2.0 hours,
so its speed in orbit is

                                     (91.9 π x 10⁵ m) / 2 hr

                        =        45.95 π x 10⁵  m/hr  x  (1 hr / 3,600 sec)

                        =           0.04 x 10⁵      m/sec

                        =              4 x 10³      m/sec  

                                     (4 kilometers per second)
6 0
3 years ago
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