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slega [8]
3 years ago
11

An out-of-control train is racing toward the Metropolis terminal train station - only Superman can help. The train has a mass of

75000 kg, and Superman has a mass of 115 kg. If the train has a velocity of 35 m/s, how fast does Superman have to fly in the opposite direction to stop it in a totally inelastic steel-Man-of-Steel collision?
Physics
1 answer:
Finger [1]3 years ago
8 0

Answer:

22826.09 m/s

Explanation:

From the law of conservation of momentum,

Sum of momentum before collision = sum of momentum after collision.

For an inelastic collision, the train and the superman have a common velocity

Note: For the superman to stop the train in an opposite direction, the common velocity after collision is zero, and such the total momentum after collision is zero

Therefore,

MU + mv = 0

MU = - mu............................................ Equation 1

Making u the subject of the equation

u = -MU/m......................................... Equation 2

Where M = mass of the train, U = initial velocity of the train, m = mass of the super man, u = initial velocity of the superman.

Given: M = 75000 kg, U = 35 m/s, m = 115 kg.

u = -(75000×35/115)

u = -22826.09 m/s

Note: The velocity is negative because the direction of the superman is opposite the direction of the train.

Hence the superman have to fly 22826.09 m/s in the opposite direction

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Convert 68852 millijoules into Calories. (Write your answer in the decimal form. Do not include units in your answer).
alex41 [277]

Answer: 68852 millijoules = 16.46 calories

Explanation:

Given;

Convert 68852 millijoules to calories.

1 calorie = 4.184J = 4184millijoules

Therefore,

1 millijoule = 1/4184 calories

68852 millijoule = 68852 × 1/4184 calories

= 16.46 calories

6 0
3 years ago
Two identical conducting spheres, A and B, carry equal charge. They are stationary and are separated by a distance much larger t
podryga [215]

Answer:

8F_i = 3F_f

Explanation:

When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.

Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.

Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.

The electrostatic force, Fi, in the initial configuration can be calculated as follows.

F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f

7 0
3 years ago
A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Bess [88]

Answer:

Explanation:

given,

initial velocity of the ball = 20 m/s

angle of ramp = 22°

ball travel at a distance  = 5 m

a) for friction less  

  \dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh

             v^2 = u^2 - 2gh

             v = \sqrt{u^2- 2 g h cos 22^0}

             v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }

                    v = 17.58 m/s

b) considering the friction

  \dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl

             v^2 = u^2 - 2gh-2\mu_kmgl

             v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}

             v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }

                   v = 17.16 m/s

7 0
3 years ago
2. Interpret Graphs How does temperature change with depth in Earth's<br> mantle?
fomenos
The Earth gets hotter as one travels towards the core, known as the geothermal gradient. The geothermal gradient is the amount that the Earth's temperature increases with depth. ... On average, the temperature increases by about 25°C for every kilometer of depth.
3 0
2 years ago
WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of
nekit [7.7K]

Answer:

82.1 km

Explanation:

We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:

A_x = (72) sin 30^{\circ} =36.0 km\\A_y = (72) cos 30^{\circ} = 62.4 km

Vector B is 48 km south, so:

B_x = 0\\B_y = -48

Finally, vector C:

C_x = -(100) cos 30^{\circ} =-86.6 km\\C_y = (100) sin 30^{\circ} = 50.0 km

Now we add the components along each direction:

R_x = A_x + B_x + C_x = 36.0 + 0 +(-86.6)=-50.6 km\\R_y = A_y+B_y+C_y = 62.4+(-48)+50.0=64.6 km

So, the resultant (which is the distance in a straight line between the starting point and the final point of the motion) is

R=\sqrt{R_x^2+R_y^2}=\sqrt{(-50.6)^2+(64.6)^2}=82.1 km

4 0
3 years ago
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