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stepan [7]
3 years ago
8

3. Why is a train so hard to stop even when it is going slowly?

Physics
2 answers:
Nesterboy [21]3 years ago
6 0

Answer:

For trains the wheels and the rail are both steel, and the steel-steel friction coefficient is around 0.25. So the stopping time and distance will, at best, be three to four times greater than a car.

Explanation:

Nitella [24]3 years ago
5 0

Answer: them things weigh tons and tons. like hundreds of tons. whenever the train is even slightly moving, the force is still so great that it makes it very difficult to stop.

Explanation: my brain knows

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Papessa [141]

A. 2

B. F - f

Hope this helps!

6 0
3 years ago
Read 2 more answers
Selena, who moves with the grace and fluid motions of a cat, has been described as
MaRussiya [10]
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6 0
3 years ago
calculate the force exerted on a rocket given that the propelling gases are expelled at a rate of 1500 kg/s with a speed of 4.0x
Tanya [424]

we know that momentum is given as

P = mv

where P = momentum , m = mass , v = velocity

taking derivative both side relative to "t"

dP/dt = v (dm/dt)

we know that :  dP/dt  = F = force

hence

F = v (dm/dt)

given that :

dm/dt = rate of mass expelled = 1500 kg/s

v = velocity = 4 x 10⁴ m/s

hence  , inserting the values in the above formula

F = (4 x 10⁴) (1500)

F = 6 x 10⁷ N



3 0
4 years ago
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th
Ghella [55]

Answer:

<em><u>Assuming that the vertical speed of the ball is 14 m/s</u></em> we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

 

Explanation:

a) <u>Assuming that the vertical speed is 14 m/s</u> (founded in the book) the initial speed of the ball can be calculated as follows:  

V_{f}^{2} = V_{0}^{2} - 2gh

<u>Where:</u>

V_{f}: is the final speed = 14 m/s

V_{0}: is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s  

b) The maximum height is:

V_{f}^{2} = V_{0}^{2} - 2gh

h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m

c) The time can be found using the following equation:

V_{f} = V_{0} - gt

t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s

d) The flight time is given by:

t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s

         

I hope it helps you!    

3 0
4 years ago
What is the weight of a 15 kg object
s344n2d4d5 [400]

to get weight multiply Mass by 9.8

weight =9.8x15=147N

8 0
4 years ago
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