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Vitek1552 [10]
3 years ago
14

A railroad car with mass m is moving with an initial velocity v when it collides and connects with a second railroad car with a

mass of 3m, which is initially at rest. How do the speed and momentum of the connected car system compare with those of the car with mass m before the collision
Physics
1 answer:
inessss [21]3 years ago
7 0

Answer:

Same momentum but speed is reduced 4 times.

Explanation:

According to the law of momentum conservation, the total momentum of the system before and after the collision is the same

Before the collision, the bigger car is at rest, only the 1st car of mass m is moving at speed v

p = mv

After the collision, both cars are connected and moving at speed V

P = (m + 3m)V = 4mV

These 2 momentum are equal

p = P

mv = 4mV

V = v/4

So after the collision, they have the same momentum but the speed decreased 4 times

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A company has been in business for 40 years. It has tens of thousands of customer addresses in three different computer systems
alexira [117]

The person in charge of information management is the database administrator.

<h3>What is a database?</h3>

The term database refers to the arranging of the data that concerns clients in relevant entry point such as computers and file cabinets. This is the way by which companies are bale to keep track of their customers.

The person that is most helpful in the management of the tens of thousands of customer addresses in three different computer systems and written on paper in filing cabinets is the database administrator.

Learn more about database:brainly.com/question/6447559

#SPJ1

7 0
1 year ago
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her
Marysya12 [62]

Answer:

 μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

          fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ  = 0

          fr sin θ  - cos θ  (W / 2 + 0,3 W_painter) = 0

          fr = cotan θ  (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

     

X axis

        F1 - fr = 0

        F1 = fr

the friction force has the expression

       fr = μ N

Y Axis

       N - W - W_painter = 0

       N = W + W_painter

       

we substitute

      fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

     μ (W + W_painter) = cotan θ  (W / 2 + 0,3 W_painter)

      μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values ​​they give

      μ = cotan θ  (12/2 + 0.3 55) / (12 + 55)

      μ = cotan θ  (22.5 / 67)

      μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

       cotan 45 = 1 / tan 45 = 1

the result is

    μ = 0.336

5 0
3 years ago
A soccer ball was kicked off the roof of a building going
WINSTONCH [101]

Answer:

Final vertical velocity = -29m/s

Horizontal distance = 100m

Height = 20.41m

Explanation:

1. The vertical final velocity can be calculated thus: vy = vyo - gt

Where;

vy = vertical velocity (m/s)

vyo = vertical initial velocity (20m/s)

g = acceleration due to gravity (9.8m/s²)

t = time (5s)

Hence, vy = vyo - gt

vy = 20 - (9.8 × 5)

vy = 20 - 49

vy = -29m/s

2. x = V0 x t

Where;

x = horizontal distance (m)

Vo = initial velocity

t = time (s)

x = 20 × 5

x = 100m

3. Maximum height = (voy)²/2g

= 20²/ 2 × 9.8

= 400/19.6

= 20.41m

4 0
3 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
3 years ago
Which of the following provides evidence that there must be at least two types of electrical charge, but that there is only one
polet [3.4K]

Answer:

Option D (On the...............dominate) would be the right approach.

Explanation:

The Gravitational constant (G) will be:

= 6.67\times 10^{-11}

The Coulomb's law constant (K) will be:

= 9\times 10^9

  • Throughout particular, these have been determined that among 2 substances with almost the similar form of charge, the combination of electromagnetic as well as the force does seem to be usually the following:

⇒ \frac{f_e}{f_g}\sim 10^{42}

  • By that same argument, the electrostatic force including its planet's atmosphere would have strongly influenced the effect, as well as maybe the planet's atmosphere, would have crashed, or perhaps the earth would have shifted at a much longer exposure from one another and.
  • Throughout particular, astronomical distance statutory framework that gravity seems to be predominant, whereas electrostatic forces have been generally ignored. It is quite since there are so many categories of allegations throughout the planet's atmosphere that balance out someone else's effects, there's only yet another form of momentum, because although the forces are still cumulative, as well as therefore offering to help everything hold to the universe, encouraging the universe just to rotate across the sun.

The latter three choices aren't connected to either the situation mentioned in the clarification segment elsewhere here.

5 0
3 years ago
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