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Irina-Kira [14]
2 years ago
6

I could really use some help on this question guys! Will give brainliest!

Physics
2 answers:
aev [14]2 years ago
4 0

Answer:

I think it’s the third one

REY [17]2 years ago
3 0
I think it’s the Third one , hope this helps
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How do i calculate this?
Lesechka [4]

CAR 1

Momentum = Mass/Velocity

M = 2100/20

M = 105 m/s^2

CAR 2

Momentum = Mass/Velocity

M = 2100/30

M = 70 m/s^2

8 0
3 years ago
Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh
Alinara [238K]

Answer:

a) k_{avg}=6.22\times 10^{-21}

b) k_{avg}=8.61\times 10^{-21}

c)  k_{mol}=3.74\times 10^{3}J/mol

d)   k_{mol}=5.1\times 10^{3}J/mol

Explanation:

Average translation kinetic energy (k_{avg}) is given as

k_{avg}=\frac{3}{2}\times kT    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8  

k_{avg}=6.22\times 10^{-21}J

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416  

k_{avg}=8.61\times 10^{-21}J

c ) The translational kinetic energy per mole of an ideal gas is given as:

       k_{mol}=A_{v}\times k_{avg}

here   A_{v} = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}

          k_{mol}=3.74\times 10^{3}J/mol

d) now at T = 143° C

        k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}

          k_{mol}=5.1\times 10^{3}J/mol

8 0
3 years ago
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di
Vikki [24]

Answer:

Mass of the cart = 146 kg

Explanation:

A cart is pulled by a force of 250 N at an angle of 35° above the horizontal.

The cart accelerates at 1.4 m/s² horizontally.

Horizontal force = Fcosθ = 250 cos35° = 204.79N

We have F = ma

Substituting

        204.79 = m x 1.4  

              m = 146.28 kg = 146 kg

Mass of the cart = 146 kg

3 0
2 years ago
Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.
kotegsom [21]

Answer:

Explanation:

v= u + at

v is final velocity , u is initial velocity . a is acceleration and t is time

Initial velocity u = 0 . Putting the given values in the equation

v = 0 + g sin 18 x 3.5

= 10.6 m /s

3 0
3 years ago
Find the work performed when the given force f f is applied to an object, whose resulting motion is represented by the displacem
Ahat [919]

work = force x distance

fd foot pounds

3 0
2 years ago
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