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evablogger [386]
3 years ago
14

An k incline plane has 2 = 40.0º and : = 0.15. Starting from rest, how long will it take a 4.0 kg block to reach a speed of 12 m

/s?

Physics
2 answers:
ahrayia [7]3 years ago
4 0
Here's how I did it.  Let me know if you have a question

Alekssandra [29.7K]3 years ago
4 0

Answer: 2.31s

Explanation:

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What is the difference between circumstances and circumference??<br><br>pls ans asap​
dimulka [17.4K]

Answer:

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Explanation:

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6 0
3 years ago
Two fans are watching a baseball game from different positions. One fan is located directly behind home plate, 18.3 m from the b
EleoNora [17]

Answer:

  • 0.317 s

Explanation:

distance of fan A = 18.3 m

distance of fan B = 127 m

speed of sound (s) = 343 m/s

What is the time difference between hearing the sound at the two locations?

time (T) = distance / speed

  • time for sound to reach fan A = 18.3 / 343 = 0.053 s
  • time it takes for sound to reach fan B = 127 / 343 = 0.370 s
  • time difference = 0.370 - 0.053 = 0.317 s
5 0
3 years ago
1. What layer of the Earth is<br> flowing, solid and plastic?
melisa1 [442]

Answer:

The mantle

Explanation:

The mantle is a plastic solid of varying densities which allow convection currents to flow molten rock towards the earth's surface resulting in volcanic activity, tectonic plate movement, earthquakes, and movement of continents.

5 0
3 years ago
Read 2 more answers
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0
3 years ago
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
3 years ago
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