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RUDIKE
19 days ago
14

In this lab, you will use a dynamics track to generate collisions between two carts. If momentum is conserved, what variable cha

nge would result in a velocity change after a collision? In the space below, write a scientific question that you will answer by doing this experiment.
Physics
2 answers:
Keith_Richards [2.2K]19 days ago
8 0

In collision-related situations, momentum remains constant.

Thus, we can deduce:

m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}

To resolve these problems we also need another equation concerning the restitution coefficient:

v_{2f} - v_{1f} = e(v_{1i} - v_{2i})

With these two equations, we can calculate the velocities after a collision.

Here, the contact force brings about changes in velocity during their interaction.

This represents the impulse that occurs when the two objects are briefly in contact during the collision, which modifies their velocities.

Maru [2.3K]19 days ago
5 0

How does altering the mass impact the interactions between colliding objects?

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The amount of electric energy consumed by a 60.0-watt lightbulb for 1.00 minute could lift a
Sav [2226]

Answer:

Explanation:

For a 60W light bulb used for 1 minute:

P = 60 W

t = 1 minute = 60 seconds

This energy is capable of lifting an object weighing 10N.

W = 10N

This indicates conversion of electrical energy into potential energy.

Let's calculate the electrical energy:

Power describes the rate of work done.

Power = Work / time

Thus, work = power × time

Work = 60 × 60

Work = 3600 J

Potential energy calculation:

P.E = mgh

Where the weight is given by:

W = mg

Therefore, P.E = W·h

P.E = 10·h

Thus, we equate:

Potential energy = Electrical energy

P.E = Work

10·h = 3600

Dividing both sides by 10 gives:

h = 3600 / 10

h = 360m

The object can be lifted to a height of 360m.

6 0
27 days ago
What upward gravitational force does a 5600kg elephant exert on the earth?
Keith_Richards [2256]
The force is calculated by multiplying mass and gravitational acceleration (F= mg). To find the solution, the mass of the elephant (5600 kg) is multiplied by gravity (9.8 m/s²). The result is 55,880 N, representing the upward gravitational force the elephant exerts on the Earth.
8 0
9 days ago
Read 2 more answers
liquid methane and liquid water are different, but they do have some things in common. what are some things that all liquids hav
Keith_Richards [2256]

Answer: In contrast, methane consists of one carbon atom and 4 hydrogen atoms. Similar to water, the bonds present are covalent.... Water has strong hydrogen bonding with other molecules, resulting in the need for a significant amount of energy to break these bonds, and its boiling point is around 100 degrees Celsius.

Explanation:

5 0
8 days ago
The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (
Keith_Richards [2256]

Answer:

Momentum is expressed as p = 7.2 g-m/s

Explanation:

It is given that,

The momentum of the object is p=7.2\times 10^{-3}\ kg-m/s

We need to represent momentum in alternative equivalent units. There are various solutions to this issue. For instance, mass units can be represented in grams, milligrams, etc., and length can be in meters, millimeters, etc.

Since, 1 kg = 1000 grams

Thus, p=7.2\times 10^{-3}\times 10^3\ g-m/s

Therefore, the momentum of the object is 7.2 g-m/s. This is the solution needed.

6 0
1 month ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
Maru [2355]

1) 9.18 s

During the initial phase, the rocket accelerates at

a_1=13.5 m/s^2

over a time interval of

t_1=3.50 s

The final velocity after this period is found using the SUVAT formula:

v_1=u+a_1t_1

with initial velocity u = 0. Substituting a1 and t1 gives:

v_1=(13.5)(3.50)=47.3 m/s

Afterward, the rocket decelerates uniformly at

a_2 = -5.15 m/s^2

until it stops, meaning the final velocity is

v_2 = 0

Again using the SUVAT formula,

v_2 = v_1 + a_2 t_2

and knowing v_1 = 47.3 m/s, solve for t2, the time until the rocket halts:

t_2 = -\frac{v_1}{a_2}=-\frac{47.3}{5.15}=9.18 s

2) 299.9 m

Calculate distances covered in both phases.

Distance in the first phase:

d_1 = ut_1 + \frac{1}{2}a_1 t_1^2

Substituting values from part 1,

d_1 = 0 + \frac{1}{2}(13.5) (3.50)^2=82.7 m

Distance in the deceleration phase:

d_2= v_1 t_2 + \frac{1}{2}a_2 t_2^2

Substituting known values,

d_2 = (47.3)(9.18) + \frac{1}{2}(-5.15) (9.18)^2=217.2 m

Total distance traveled is

d = 82.7 m + 217.2 m = 299.9 m

6 0
1 month ago
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