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Paraphin [41]
3 years ago
7

Find the displacement in meters a runner would travel in 5 hours at an average velocity of 12km/h to the southwest

Physics
1 answer:
vova2212 [387]3 years ago
8 0

Answer:

60,000m

Explanation:

Convert km/h to m/s by multiplying with 1000/3600.

Convert hours to seconds by multiplying with 3600.

Because displacement is a vector quantity and deals with the shortest distance between points, simply plug it into the equation s=vt.

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How much kinetic energy does a system containing 3 moles of an ideal gas at 300 K possess? What is the heat capacity at constant
ElenaW [278]

Answer:

1. K.E = 11.2239 kJ ≈ 11.224 kJ

2. C_{V} = 37.413 JK^{- 1}

3. Q = 10.7749 kJ

Solution:

Now, the kinetic energy of an ideal gas per mole is given by:

K.E = \frac{3}{2}mRT

where

m = no. of moles = 3

R = Rydberg's constant = 8.314 J/mol.K

Temperature, T = 300 K

Therefore,

K.E = \frac{3}{2}\times 3\times 8.314\times 300

K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ

Now,

The heat capacity at constant volume is:

C_{V} = \frac{3}{2}mR

C_{V} = \frac{3}{2}\times 3\times 8.314 = 37.413 JK^{- 1}

Now,

Required heat transfer to raise the temperature by 15^{\circ} is:

Q = C_{V}\Delta T

\Delta T = 15^{\circ} = 273 + 15 = 288 K

Q = 37.413\times 288 = 10774.9 J = 10.7749 kJ

8 0
3 years ago
Any one watch SerieS here
Illusion [34]

Answer:

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Explanation:

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8 0
3 years ago
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Which of the following scenerios fits all of the criteria for the two-source interference equations to be valid?
vekshin1

Answer:

answer the correct  is B

Explanation:

For the interference phenomenon to occur, some conditions must be met.

* You must have a light in phase and coherent, for this you can for a light from an incandescent source through a single slit, the light that comes out is coherent

* This light must strike two slits

* the light that passes through the slits must hit a distant screen and be able to see the phenomenon of constructive interference

when examining the different answer the correct one is B

8 0
3 years ago
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Vikki [24]

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8 0
3 years ago
a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi
zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

                     v = u + at

                     a = (v - u) / t

Substituting the given values in the above equation

                    a = (10 - 0) / 3

                       = 3.33 m/s²

Using the second equations of motion

                      s = ut + 1/2 at²

                         = 0 + 0.5 x 3.33 x 3²

                         = 15 m

The force exerted by the car

                         F = m x a

                            = 1500 Kg x 3.33 m/s²

                            = 5000 N

The work done by the car,

                          W = F x S

                               = 5000 N x 15 m

                               = 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

                            P = W / t

                                = 75,000 J / 3 s

                                 = 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0
4 years ago
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