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omeli [17]
3 years ago
5

The slope of a displacement vs. time graph gives

Physics
1 answer:
Sauron [17]3 years ago
3 0
The answer is b. Velocity
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Water has physical properties that can be<br> measured and
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B. A 20kg wagon is pulled up a 30° incline at an acceleration of 2.5ms?. The force pulling the wagon is parallel to the incline
yawa3891 [41]

The value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.

<h3>Coefficient of the kinetic friction</h3>

The value of coefficient of kinetic friction is calculated as follows;

F - Ff = ma

F - μmgcosθ = ma

where;

  • F is applied force
  • μ is coefficient of kinetic friction
  • m is mass of the wagon
  • a is acceleration of the wagon

182 - μ(20 x 9.8 x cos30) = 20(2.5)

182 - 169.74μ = 50

182 - 50 = 169.74μ

132 = 169.74μ

μ = 132/169.74

μ = 0.78

Thus, the value of the coefficient of kinetic friction between the wagon and inclined surface is 0.78.

Learn more about coefficient of friction here: brainly.com/question/20241845

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3 years ago
As food passes through the alimentary canal, the presence of chemical secretions will occur in which order?
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The first one....................
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What are the two groups of planets based on composition?
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OK so yeah the mercury Venus Earth and Mars a rocky planets and Jupiter Saturn Uranus and Neptune our gaseous planets so I think that’s your answer correct me if I’m wrong:D
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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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