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frozen [14]
3 years ago
5

A train can travel at 300km/h. How long will it take to to travel 12.000km?

Physics
2 answers:
Leokris [45]3 years ago
8 0

Answer:

Train will take 156 sec to travel 12km.

Amanda [17]3 years ago
8 0
156 seconds or 2 minutes & 36 seconds
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A bus that was moving at a high speed stopped suddenly. One of
fiasKO [112]

Answer:

this doesn't make sinces

7 0
3 years ago
A 2-kW electric heater takes 15 min to boil a quantity of water. If this is done once a day and power costs 10 cents per kWh, wh
frozen [14]

Answer:

<h2> $1.50</h2>

Explanation:

Given data

power P= 2 kW

time t= 15 min to hours = 15/60= 1/4 h

cost of power consumption per kWh= 10 cent = $0.1

We are expected to compute the cost of operating the heater for 30 days

but let us computer the energy consumption for one day

Energy of heater  for one day= 2* 1/4 = 0.5 kWh

the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50

<u><em>Hence it will cost  $1.50 for 30 days operation</em></u>

4 0
3 years ago
A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two
AfilCa [17]

Answer:

please the answer below

Explanation:

(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\

k=8.89*10^9

For both cases we have

k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}

(b) by replacing this values of r in the expression for V we obtain

k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}

hope this helps!!

3 0
3 years ago
Read 2 more answers
V=xf-xi/t solve for t
irina [24]

Answer:

t=\frac{x_f-x_i}{v}

Explanation:

Starting from the equation:

v=\frac{x_f-x_i}{t}

First of all, let's multiply by t on both sides:

v\cdot t = \frac{x_f-x_i}{t}\cdot t \\vt = x_f - x_i

And then, let's divide by v on both sides:

\frac{vt}{v}=\frac{x_f-x_i}{v}\\t=\frac{x_f-x_i}{v}

So, finally

t=\frac{x_f-x_i}{v}

7 0
3 years ago
A surveyor is using a magnetic compass 5.6 m below a power line in which there is a steady current of 140 A. (a) What is the mag
ArbitrLikvidat [17]

Answer:

(a) B = 5.6 micro Tesla

Explanation:

Current in the wire, i = 140 A

distance, r = 5 m

The formula for the magnetic field at a distance r due to the current carrying wire

B=\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}

B=10^{-7}\times \frac{2\times140}{5}

B = 5.6 x 10^-6 Tesla

B = 5.6 micro Tesla

(b) As the magnetic field of earth at this site is 20 micro tesla so the magnetic field due to current carrying wire interfere the magnetic compass.

4 0
3 years ago
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