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slega [8]
3 years ago
11

An out-of-control train is racing toward the Metropolis terminal train station - only Superman can help. The train has a mass of

75000 kg, and Superman has a mass of 115 kg. If the train has a velocity of 35 m/s, how fast does Superman have to fly in the opposite direction to stop it in a totally inelastic steel-Man-of-Steel collision?
Physics
1 answer:
Finger [1]3 years ago
8 0

Answer:

22826.09 m/s

Explanation:

From the law of conservation of momentum,

Sum of momentum before collision = sum of momentum after collision.

For an inelastic collision, the train and the superman have a common velocity

Note: For the superman to stop the train in an opposite direction, the common velocity after collision is zero, and such the total momentum after collision is zero

Therefore,

MU + mv = 0

MU = - mu............................................ Equation 1

Making u the subject of the equation

u = -MU/m......................................... Equation 2

Where M = mass of the train, U = initial velocity of the train, m = mass of the super man, u = initial velocity of the superman.

Given: M = 75000 kg, U = 35 m/s, m = 115 kg.

u = -(75000×35/115)

u = -22826.09 m/s

Note: The velocity is negative because the direction of the superman is opposite the direction of the train.

Hence the superman have to fly 22826.09 m/s in the opposite direction

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The total resistance of a parallel circuit is 25 ohms. If the total current is 100mA, how much current is through a 220 ohm resi
gulaghasi [49]

Answer:

The current across the resistance is 0.011 A.

Explanation:

Total resistance, R = 25 ohms

Total current, I = 100 mA = 0.1 A

Let the voltage is V.

By the Ohm's law

V = I R

V = 0.1 x 25 = 2.5 V

Now the resistance is R' = 220 ohm

As they are in parallel so the voltage is same. Let the current is I'.

V = I' x R'

2.5 = I' x 220

I' = 0.011 A

7 0
3 years ago
A student examines the effect of the number of D batteries in a closed circuit on the brightness of a light bulb. In the experim
xxTIMURxx [149]

Answer:

Placing cells in series increases the

voltage in the circuit by 1.5 V for each

cell. Increasing the voltage increases the

brightness of the bulb

Explanation:

I. MAKING THE CONNECTION

How many terminals are located on the battery? 2

How many terminals are located on the bulb? 2

II. PLACING CELLS IN SERIES

What is the effect on the brightness of the bulb by increasing the number of cells?

The bulbs become brighter when increasing the number of cells.

What changes occur in the current in the circuit when increasing the number of cells?

Increasing the number of cells increases the current in the circuit.

What changes occur in the voltage in the circuit when increasing the number of cells?

Increasing the number of cells increases the voltage (for cells in series the voltage is additive).

What changes occur in the resistance in the circuit when increasing the number of cells?

The resistance is determined by the number of bulbs. The resistance in the circuit remains unchanged.

III. PLACING BULBS IN SERIES

What is the effect on the brightness of the bulbs by increasing the number of bulbs?

Increasing the number of bulbs decreases the brightness of the bulbs.

What changes occur in the resistance in the circuit as more bulbs are added?

The resistance increases. In a series circuit, adding bulbs increases the resistance in the circuit.

What changes occur in the current in the circuit as more bulbs are added?

Increasing the resistance decreases the current.

Observations on unscrewing one bulb. Explain your observations.

A complete circuit requires the electrons to move from the negative terminal of the battery to the

positive terminal. When one bulb is unscrewed it opens the circuit preventing a complete circuit and

the electrons cannot return to the battery.

IV. PLACING BULBS IN PARALLEL

Compare the brightness with two bulbs in parallel with two bulbs in series.

Two bulbs in parallel are brighter than two bulbs in series.

How does increasing the number of circuits (bulbs) change the current and resistance?

In a parallel circuit each bulb is in its own circuit. As bulbs are added the resistance in the circuit

decreases since each circuit is another pathway for electrons to move from one end of the circuit

4 0
4 years ago
What happens to the intensity of solar energy as latitude increases?it stays the same. it increases. it doubles. it decreases?
vekshin1
The intensity of solar energy decreases as latitude increases
3 0
3 years ago
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A 21.3 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.7 cm from
Bezzdna [24]

Answer:

The magnetic field strength due to current flowing in the wire is9.322 x 10⁻⁶ T.

Explanation:

Given;

electric current, I = 21.3 A

distance of the magnetic field from the wire, R = 45.7 cm = 0.457 m

The strength of the resulting magnetic field at the given distance is calculated as;

B = \frac{\mu_o I}{2\pi R}

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A

B = \frac{\mu_o I}{2\pi R}\\\\B = \frac{4\pi*10^{-7} *21.3}{2\pi(0.457)}\\\\B = 9.322 *10^{-6} \ T

Therefore, the magnetic field strength due to current flowing in the wire is 9.322 x 10⁻⁶ T.

7 0
4 years ago
A protostar's radius decreases by a factor of 100 and its surface temperature increases by a factor of two before it becomes a m
Sliva [168]

Answer:

L_f = K (\frac{r}{100})^2 * (2T)^4

L_f = K \frac{r^2}{10000} * 16 T^4

L_f = \frac{16}{10000} k r^2 T^4 = \frac{1}{625} k r^2 T^4

L_f = \frac{1}{625} L_i

So then we see that the final luminosity decrease by a factor of 625 so then the correct answer for this case would be:

B. Decreases by a factor of 625

Explanation:

For this case we can use the formula of luminosity in terms of the radius and the temperature given by:

L_i = K r^2 T^4

Where L_i = initial luminosity, r= radius and T = temperature.

We know that we decrease the radius by a factor of 100 and the temperature increases by a factor of 2 so then the new luminosity would be:

L_f = K (\frac{r}{100})^2 * (2T)^4

L_f = K \frac{r^2}{10000} * 16 T^4

L_f = \frac{16}{10000} k r^2 T^4 = \frac{1}{625} k r^2 T^4

L_f = \frac{1}{625} L_i

So then we see that the final luminosity decrease by a factor of 625 so then the correct answer for this case would be:

B. Decreases by a factor of 625

6 0
3 years ago
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