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Dahasolnce [82]
3 years ago
10

A train is traveling north. it goes 200 km /hr for 2.5 hours how far did the train go.

Physics
1 answer:
DerKrebs [107]3 years ago
5 0

Answer:

500Km

Explanation:

This question is an example of speed, distance, time.

The formula for working out the distance is: Speed*Time

To work this out you would need to multiply the speed of 200Km/h by the time of 2.5 hours, this gives you 500 Km. This is because the formula to work out the distance is speed * time.

In adittion, if the train is traveling at 200km per hour then that means that over the course of an hour the distance travelled would be 200Km. So, by multiplying the speed by the time you are working out how far it travelled in that given time.

1) Multiply 200 by 2.5.

200*2.5=500Km

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alexandr402 [8]

Answer:

Can't see anything, please share clearly

4 0
3 years ago
When an object is balanced about a pivot, the total clockwise moment must be equal to the total __________ __________. What two
Keith_Richards [23]

Answer:

When an object is balanced, about a pivot, the total clockwise moment must be equal to the total anticlockwise moment about that pivot.

Hope that helps.

5 0
2 years ago
At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle wh
Afina-wow [57]

Answer:

W = 1.22 \times 10^9 J

Explanation:

Initial potential energy of the given spacecraft is given as

U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}

so we have

U = - \frac{Gm}{r}(M_e + M_m)

so we have

M_e = 5.98 \times 10^{24} kg

M_m = 7.35 \times 10^{22} kg

m = 1160 kg

r = 3.84 \times 10^8 m

U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})

U = -1.22 \times 10^9 J

now total work done to move it to infinite is given

W = 0 - U

W = 1.22 \times 10^9 J

6 0
3 years ago
a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec
MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.  

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

\frac{1}{2}  mu_1^2+0=\frac{1}{2}  mv_1^2+\frac{1}{2}  mv_2^2

∴  \frac{1}{2}  m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2

∴ \frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2}  mv_1^2+\frac{1}{2} mv_2^2

∴ mv₁v₂ = 0

  1. It is impossible for the mass to be zero.
  2. Because the second ball moves, velocity v2 cannot be zero.
  3. As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

#SPJ4

3 0
1 year ago
Can some help me plz
kakasveta [241]

Answer:

D. I'm guessing

The gold-foil experiment showed that the atom consists of a small, massive, positively charged nucleus with the negatively charged electrons being at a great distance from the centre. 

5 0
3 years ago
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