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3 years ago

Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speed up

Physics

1 answer:

Amiraneli [1.4K]3 years ago

6 0

Answer: 5 m/s^2

Explanation: You have to use a kinematics equation that Includes what you are given. So in this case you have Initial velocity, final velocity, and time. You’re looking for acceleration so you have everything you need to use an equation, the one without distance.

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An upright spring with a 96g mass on it is compressed 2 cm. When

Alexeev081 [22]

Answer:

I only know answer A and it's 2825.28 N/m, with rounding it's 2825.5

Explanation:

Use the m*g*h=1/2*k*x^2 equation

96*9.81*60=1/2*k*2^2

5650.56=2k

5650.56/2=2825.28N/m

8 0

3 years ago

If molecules in a substance move FASTER will the TEMPERATURE increase or decrease?

statuscvo [17]

The temperature will increase

4 0

3 years ago

He user of a machine applies force to the machine over the _____ distance.

gtnhenbr [62]

The blank distance is your answer

5 0

2 years ago

A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The

VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

$\dfrac{1}{2}mu^2 = m g d sin \theta$

$u^2 = 2 g d sin30^0$

$u= \sqrt{2\times 9.8 \times 10 sin30^0}$

u = 9.89 m/s

Using second law of motion

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

Using third equation of motion ,

v² - u² = 2 a s

0² - 9.89² = 2 x 2.45 x s

s = 20 m

the distance moved before stopping is 20 m

3 0

3 years ago

a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its

lilavasa [31]

Answer:

D/H =15

Explanation:

We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
At this point, we can find the value of H, applying the following kinematic equation:

$v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)$

If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

$H = \frac{v_{0} ^{2} }{2*g} (2)$

We can use the same equation, to find the value of D, as follows:

$v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)$

In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
When we replace these values in (1), we find that v₁ = -v₀.
Replacing in (3), we have:

$(4*v_{0})^{2} - (-v_{0}) ^{2} = 2* g* D\\ \\ 15*v_{0}^{2} = 2*g*D$

Solving for D:

$D = \frac{15*v_{0} ^{2} }{2*g}$

From (2) we know that H can be expressed as follows:

$H = \frac{v_{0} ^{2} }{2*g}$

⇒ D = 15 * H

$\frac{D}{H} = 15$

3 0

2 years ago