All categories

1 year ago

What is the acceleration of a 600,000 kg freight train, if each of itsthree engines can provide 100,000 n of force?

1 answer:

6 0

The acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

<h3>How to calculate acceleration?</h3>

The acceleration of a freight train can be calculated using the following formula:

Force = mass × acceleration

According to this question, a 600,000kg freight train can produce 100,000N of force. The acceleration is as follows:

100,000 = 600,000 × a

100,000 = 600,000a

a = 0.167m/s²

Therefore, the acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

Learn more about acceleration at: brainly.com/question/12550364

#SPJ1

You might be interested in

A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2

stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

$s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2$

For the bottom of window (position B)

$s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2$

$\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673$

We also have

$t_B-t_A=0.27$

Solving

$t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s$

So after 1.11 seconds ball reaches at top of window,

We have equation of motion v = u + at

$v_A=0+9.81\times 1.11=10.89m/s$

Speed at which the ball passes the window’s top = 10.89 m/s

7 0

3 years ago

A magic medallion is suspended from a string inside a compartment of Hogwarts Express which is running straight westwards on hor

Serggg [28]

Answer:

a = 1 m/s² and

Explanation:

The first two parts can be seen in attachment

We use Newton's second law on each axis

Y axis

Ty - W = 0

Ty = w

X axis

Tx = m a

With trigonometry we find the components of tension

Sin θ = Ty / T

Ty = T sin θ

Cos θ = Tx / T

Tx = T cos θ

We calculate the acceleration with kinematics

Vf = Vo + a t

a = (Vf -Vo) / t

a = (20 -10) / 10

a = 1 m/s²

We substitute in Newton's equations

T Sin θ = mg

T cos θ = ma

We divide the two equations

Tan θ = g / a

θ = tan⁻¹ (g / a)

θ = tan⁻¹ (9.8 / 1)

θ = 84º

We see that in the expression of the angle the mass does not appear therefore you should not change the angle

4 0

2 years ago

A book has been lifted 1.5 meters into the air giving it 30J of potential energy. How much force was used to lift it

joja [24]

Answer:

Force on the object is 20 N

Explanation:

As we know that work done to raise the book from initial position to final position is known as potential energy stored in it

So here we know that

$U = F.s$

here we know that

U = 30 J

s = displacement = 1.5 m

so we have

$30 = F(1.5)$

$F = 20 N$

6 0

3 years ago

What is the process called when the moon begins to fade from a full moon to the new moon?

nataly862011 [7]

Waning is what it’s called

8 0

2 years ago

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

juin [17]

Answer:

Explanation:

Given

For first case

launch angle $\theta =45^{\circ}C$

at highest point $h=150 m/s$

$150=u\cos 45$

$u=\frac{150}{\cos 45}=212.132 m/s$

For second case

$\theta _2=37^{\circ}C$

at highest Point velocity is $u\cos \theta _2$

$=212.132\times \cos 37$

$=169.41 m/s$

as there is no acceleration in x direction therefore horizontal velocity is same

7 0

2 years ago

Read 2 more answers