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3 years ago

14

4. What are three tips parents can use to improve communication with their children? Discuss one that you think is most importan

t

1 answer:

Mumz [18]3 years ago

4 0

Have them give up there phone for 15 minutes
talk to them when there in a good mood
watch a movie then have a disscution about it

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Sam's job at the amusement park is to slow down and bring to a stop the boats in the log ride. you may want to review ( pages 29

meriva

The work should be at least equivalent to the kinetic energy possessed by the boat and its riders. The formula for kinetic energy is written below:

W = KE
W = 1/2*mv²
W = 1/2*(1,000 kg)(1.3 m/s)²
<em>W = 845 J</em>

7 0

3 years ago

Read 2 more answers

What is the electric flux Φ3 through the annular ring, surface 3? Express your answer in terms of C , r1, r2, and any constants.

mr Goodwill [35]

Answer:

Pls see attached file

Explanation:

4 0

3 years ago

A 6,000 kg train car is moving to the right at 10 m/s and connects to a 4,000-kg train car that wasn't moving. What is the veloc

vredina [299]

1) The final velocity of the two trains is 6 m/s to the right

2) It is an inelastic collision

Explanation:

1)

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

Therefore we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 6,000 kg$ is the mass of the first train

$u_1 = 10 m/s$ is the initial velocity of the first train

$m_2 = 4,000 kg$ is the mass of the second train

$u_2 = 0$ is the initial velocity of the second train (initially at rest)

$v$ is the final combined velocity of the two trains

Solving the equation for v, we find the final velocity of the two trains:

$v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(6000)(10)+0}{6000+4000}=6 m/s$

2)

There are two types of collisions:

Elastic collision: in an elastic collision, both the total momentum and the total kinetic energy of the system are conserved
Inelastic collision: in an inelastic collision, only the total momentum is conserved, while the total kinetic energy is not (part of it is converted into thermal energy due to internal frictions)

To verify what type of collision is this, we can compare the total kinetic energy before and after the collision:

Before:

$K=\frac{1}{2}m_1 u_1^2 = \frac{1}{2}(6000)(10)^2=300,000 J$

After:

$K=\frac{1}{2}(m_1 +m_2)v^2 = \frac{1}{2}(6000+4000)(6)^2=180,000 J$

As we can see, the kinetic energy is not conserved, so this is an inelastic collision.

Learn more about momentum here:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

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#LearnwithBrainly

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3 years ago

What is happening if energy input remains constant and voltage remains the same in a circuit, but the current decreases?

atroni [7]

The resistance of the circuit will definitely increase. This can be proved by the Ohm's Law where it is said that the potential difference across a conductor is directly proportional to the current and the proportionality constant is known as the resistance. The law is expressed as V = IR. As we can see, when the voltage remains constant while the current decreases then the resistance will increase.<span />

3 0

3 years ago

A 30-m-long rocket train car is traveling from Los Angeles to New York at 0.5c when a light at the center of the car flashes. Wh

Novosadov [1.4K]

Given that,

Distance =30 m

speed = 0.5c

(A). We need to find the bell and siren simultaneous events for a passenger seated in the car

According to given data

The distance travelled by the light to reach either side of the rocket train car is same.

So, The two events are simultaneous and the bell and siren are the simultaneous events for a passenger seated in the car.

(B). We need to calculate time interval between the events

Using formula of time dilation

$\Delta t=\dfrac{\Delta t'}{\sqrt{1-\dfrac{v^2}{c^2}}}$ .....(I)

Where, $\delta t'$ = proper time

$\delta t$ = time interval between the events

The time interval between the events measured in a reference frame

The proper time in this case is

$\Delta t'=\Delta t_{1}-\dfrac{v\Delta x}{c^2}$

For the second interval,

Put the value of $\Delta t'$ in the equation (I)

$\Delta t_{2}=\dfrac{\Delta t_{1}-\dfrac{v\Delta x}{c^2}}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value in the equation

$\Delta t_{2} = \dfrac{0-\dfrac{0.5c\times30}{c^2}}{\sqrt{1-\dfrac{0.5^2c^2}{c^2}}}$

$\Delta t_{2}=\dfrac{-15}{3\times10^{8}\sqrt{1-0.25}}$

$\Delta t_{2}=-5.77\times10^{8}\ s$

Negative sign shows the siren rings before the bell ring.

Hence, (A). Yes, the bell and siren are simultaneous events.

(B). The siren sounds before the bell rings.

8 0

3 years ago