A train travelling at 20 m/s has 2 000 000 J of kinetic energy. What is the mass of the train?

Physics

2 answers:

Slav-nsk [51]2 years ago

8 0

Answer:

10,000 kg

Explanation:

Svetlanka [38]2 years ago

3 0

Answer:

Explanation:

The mass of the train can be found by using the formula

$m = \frac{2k}{ {v}^{2} } \\$

k is the kinetic energy

v is the velocity

From the question we have

$m \frac{2(2000000)}{ {20}^{2} } = \frac{4000000}{400} = \frac{40000}{4} \\$

We have the final answer as

Hope this helps you

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It takes 6 sec for a stone to fall the the bottom of a mineshaft. calculate the

Stels [109]

Ok so this is simple projectile motion problem.

if we have an object falling in free fall it is subject to gravity of -9.80m/s^2

so it says it takes 6 sec to fall and we know initial velocity was zero so we know that h=vt+1/2gt^2 so we get h=0+1/2*9.80*6^2 = 176.4m

so solving for final speed we get KE=PE = 1/2mv^2=mgh = 1/2v^2=gh so
v=sqrt(2*g*h) = sqrt(2*9.8*176.4m) = 58.8m/s final speed when it hits the ground

hope this helps you! Thanks!!

8 0

2 years ago

What could be the possible answer to the question ?<br><br>thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$