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svlad2 [7]
2 years ago
15

Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Physics
1 answer:
Hitman42 [59]2 years ago
7 0

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

h = \frac{v^{2}}{2g}

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

\frac{mv^{2}}{2} = mgh

h = \frac{v^{2}}{2g}

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

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Which of the following provides evidence that geologic activity occurred in the past on Earth's Moon?
Bingel [31]

Answer: d) the presence of solidified lava flows on the Moon

Explanation:

A geological activity means an occurrence of event such as volcanic eruption, earthquake, sedimentation, erosion etc. The revolution of the Moon around the Earth , the axial tilt of the Moon  or the phases of the Moon are not surface features. hence, these events cannot provide the evidence of geological activity in the past of Moon.

The surface features of moon such as Mares, Craters, mountains, Rays and rills are the proof of some geological activity on the Moon. Mares are the dark patches on the moon's surface formed of solidified lava. Due to negligible atmosphere on the moon, the meteors strike its surface and cause craters to form. Thus, the correct answer is d.

6 0
3 years ago
Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceed
photoshop1234 [79]

Answer:

The railroad tracks are 13 m above the windshield (12 m without intermediate rounding).

Explanation:

First, let´s calculate the time it took the driver to travel the 27 m to the point of impact.

The equation for the position of the car is:

x = v · t

Where

x = position at time t

v = velocity

t = time

x = v · t

27 m = 17 m/s · t

27 m / 17 m/s = t

t = 1.6 s

Now let´s calculate the distance traveled by the bolt in that time. Let´s place the origin of the frame of reference at the height of the windshield:

The position of the bolt will be:

y = y0 + 1/2 · g · t²

Where

y = height of the bolt at time t

y0 = initial height of the bolt

g = acceleration due to gravity

t = time

Since the origin of the frame of reference is located at the windshield, at time 1.6 s the height of the bolt will be 0 m (impact on the windshield). Then, we can calculate the initial height of the bolt which is the height of the railroad tracks above the windshield:

y = y0 + 1/2 · g · t²

0 = y0 -1/2 · 9.8 m/s² · (1.6 s)²

y0 = 13 m

8 0
3 years ago
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc
julia-pushkina [17]

Answer:

E = 2.84 * 10^5 N/C

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

v^2 = u^2 + 2as

(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C

8 0
3 years ago
A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled
Ksivusya [100]

Answer:

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring  (T) = ma

Tension force in the spring  (T) = 9 × 2.10

Tension force in the spring  (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N  × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

8 0
3 years ago
The thrust from a car is 2000 N. The air resistance and friction together total 680 N. What happens to the speed of the car?
zheka24 [161]
The car will speed up because if the air resistance and friction forces are reacting in the direction opposite to the Applied force, The Net force will still be in the direction of the car's motion and since F=ma a will cause your velocity to increase
6 0
4 years ago
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