A train travels 600 km in 4 hours. What is the speed of the train?

Physics

1 answer:

tekilochka [14]3 years ago

8 0

Answer:

150 km/h

Explanation:

I just took the test on edgenuity! Have a great day :)

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a gas has a volume of 8 liters at a temperature of 300 K the volume is then increased to 12 Liters what is the new temperature (

natali 33 [55]

300/8 = 37.5
37.5 x 12 = 450
New temp. = 450 K
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Answer:

self-contained, short-term light-source

Explanation:

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If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.503 X 104 T) at a distance of 15

dem82 [27]

Answer:

37.725 A

Explanation:

B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T

r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m

i = magnitude of current flowing through the wire = ?

Magnetic field by a long wire is given as

$B = \frac{\mu _{o}}{4\pi }\frac{2i}{r}$

Inserting the values

$0.503\times 10^{-4} = (10^{-7})\frac{2i}{0.15}$

i = 37.725 A

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Particles q1, 92, and q3 are in a straight line.

NNADVOKAT [17]

The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃

<h3 /><h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force,by the charge q₁ on the q₃;

$\rm F_{31}} = \frac{Kq_1q_3}{r^2} \\\\ \rm F_{31}} = \frac{9 \times 10^9 \times -28.1 \times 10^{-6}\times \times -47.9 \times 10^{-6}}{(0.600)^2} \\\\ F_{31}} =33.64 \ N$

The force,by the charge q₂ on the q₃;

$\rm F_{32}} = \frac{Kq_2q_3}{r^2} \\\\ \rm F_{32}} = \frac{9 \times 10^9 \times 25.5 \times 10^{-6}\times \times -74.9\times 10^{-6}}{(0.300)^2} \\\\ F_{32}} =-19.09 \ N$