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Kitty [74]
3 years ago
7

What type of material can charges move through easily?

Physics
1 answer:
Katyanochek1 [597]3 years ago
4 0
Pass through it easily is called a Metals such as iron, steel, copper, and aluminum are good conductors.
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A 12000 kg train engine moving at 2.2 m/s hits and locks into 3 boxcars with a total mass of 25000 kg sitting still. If the coll
PilotLPTM [1.2K]

Answer:

V = 0.714m/s

Explanation:

Full solution calculation can be found in the attachment below.

From the principle of conservation of linear momentum, the sum of momentum before collision equals the sum of momentum after collision.

Before collision only the train had momentum. After the collision the train and the boxcars stick together and move as one body. The initial momentum of the train is now shared with the boxcars as they move together as one body. The both move with a common velocity v.

See the attachment below for the solution calculation.

5 0
3 years ago
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A
Pani-rosa [81]
The answer would be center of mass, B
8 0
3 years ago
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regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac
solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum p_{i} before the collision must be equal to the final momentum p_{f} after the collision:

p_{i}=p_{f} (1)

Being:

p_{i}=MV_{i}+mU_{i}

p_{f}=(M+m) V

Where:

M=480 g \frac{1 kg}{1000 g}=0.48 kg the mas of the peregrine falcon

V_{i}=45 m/s the initial speed of the falcon

m=240 g \frac{1 kg}{1000 g}=0.24 kg is the mass of the pigeon

U_{i}=0 m/s the initial speed of the pigeon (at rest)

V the final speed of the system falcon-pigeon

Then:

MV_{i}+mU_{i}=(M+m) V (2)

Finding V:

V=\frac{MV_{i}}{M+m} (3)

V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg} (4)

V=30 m/s (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

F=\frac{\Delta p}{\Delta t} (6)

Where:

F is the force exerted on the pigeon

\Delta t=0.015 s is the time

\Delta p is the pigeon's change in momentum

Then:

\Delta p=p_{f}-p_{i}=mV-mU_{i} (7)

\Delta p=mV (8) Since U_{i}=0

Substituting (8) in (6):

F=\frac{mV}{\Delta t} (9)

F=\frac{(0.24 kg)(30 m/s)}{0.015 s} (10)

Finally:

F=480 N

7 0
3 years ago
An air-track glider attached to a spring oscillates between the 14.0 cm mark and the 71.0 cm mark on the track. The glider compl
horsena [70]

Answer:

     A = 2,8333  s

Explanation:

El periodo es definido como el tiene que toma de dar una oscilación.

En este caso realiza varias osicilacion por lo cual debemos encontrar el promedio del perdono.

              T = t/n

calculemos

              A = 34,0/ 12,0

              A = 2,8333  s

6 0
3 years ago
What happens to the temperature of a gas when it is compressed? a the temperature does not change. b the temperature increases.
nasty-shy [4]
I think the temperature increases
5 0
4 years ago
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