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jek_recluse [69]
3 years ago
5

A train's velocity is slowly decreasing as it approaches a station. About 1 km before it reaches the station, several stowaways

jump off the train. What happens to the amount of acceleration if the force remains constant? Assume the train is traveling in the positive direction.
Physics
1 answer:
Mars2501 [29]3 years ago
6 0

Answer:

Increases

Explanation:

From Newton's second law:

F = ma

If F stays the same and m decreases, then a increases.

Therefore, the magnitude of the train's acceleration increases (the acceleration becomes more negative).

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A spacecraft is separated into two parts by detonating the explosive bolts that hold them together. The masses of the parts are
nordsb [41]

Answer:

  • 0.279 m/s

Explanation:

mass (m1) = 1150 kg

mass (m2) = 1900 kg

impulse (J) = 200 N.s

With what relative speed do the two parts separate because of the detonation?

since both parts separate, their relative speed = speed of m1 + speed of m2

  • speed of m1 = \frac{J}{m} = \frac{200}{1150} = 0.174 m/s
  • speed of m2 = \frac{J}{m} = \frac{200}{1900} = 0.105 m/s
  • relative speed = 0.174 + 0.105 = 0.279 m/s

6 0
3 years ago
SUBJECT: ASTRONOMY -- TOPIC: Time Dilation
inessss [21]

Answer:

You can determine if the ship is moving by lying down and measuring your height.

Explanation:

3 0
3 years ago
You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l
sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

h= v_it + \frac{1}{2}gt^2\\

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}

<u>vi =  9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

4 0
2 years ago
A 2.00-kg ball is moving at 2.20 m/s toward the right. It collides elastically with a 4.00-kg ball that is initially at rest. 1)
loris [4]
Elastic collision is when kinetic energy before = kinetic energy after

Ek= 1/2mv^2

total before
Ek=1/2(2)(2.2^2) = 4.84 J

total after
Ek= 1/2(2+4)(v^2) = 3v^2

Before = after
4.84=3v^2 | divide by 3
121/75 = v^2 | square root both sides
v=1.27 m/s
7 0
3 years ago
What force must be applied to move a 251 kg rock on a pavement like surface
Alchen [17]

If the rock is just sitting there and you want to SLIDE it, then you have to push it with a force of at least

(251 kg) x (9.8 m/s²) x (μ) =

(2,459 Newtons) x (the coefficient of static friction on that surface)


4 0
3 years ago
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