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Viktor [21]
3 years ago
14

A train slows down from 34 m/s to 4 m/s in 60 seconds. what is it's acceleration?

Physics
1 answer:
OverLord2011 [107]3 years ago
3 0
Acceleration = change in velocity / time

change in velocity = final velocity - initial velocity

because the object is slowing down, the acceleration will be negative.

acceleration = (4-34) / 60
acceleration = -30 / 60
acceleration = -0.5 m/s^2

answer : the train is accelerating at -0.5 m/s^2
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the correct answer is 0.286 s

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Clara made a chart to summarize some of the evidence that supports the big bang theory.
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The temperature of the CMB is cooler, not hotter, than at the time of the big bang.

Explanation:

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6 0
3 years ago
Read 2 more answers
A 30-kg child sits at the top of a 3-meter slide. After sliding down, the child is traveling 4 m/s. How much PE does he start wi
Jobisdone [24]
At the top:

         Potential Energy = (mass) x (gravity) x (height)

                                       = (30 kg) x (9.8 m/s²) x (3 meters)

                                       =      882 joules

At the bottom:

           Kinetic Energy  =  (1/2) x (mass) x (speed)²

                                       = (1/2) x (30 kg) x (3 m/s)²

                                       =        (15 kg)  x  (9 m²/s²)

                                       =              135 joules .

He had  882 joules of potential energy at the top,
but only  135 joules of kinetic energy at the bottom.

Friction stole  (882 - 135) = 747 joules of his energy while he slid down.
The seat of his jeans must be pretty warm.
6 0
3 years ago
Can someone help me figure out how to do a function formula for fx and fy
masha68 [24]

Answer: How to solve for FX and FY?

to find fx(x, y): keeping y constant, take x derivative; • to find fy(x, y): keeping x constant, take y derivative. f(x1,...,xi−1,xi + h, xi+1,...,xn) − f(x) h . ∂y2 (x, y) ≡ ∂ ∂y ( ∂f ∂y ) ≡ (fy)y ≡ f22. similar notation for functions with > 2 variables.

Explanation:

4 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
3 years ago
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