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taurus [48]
3 years ago
10

A 0.5 kg ball moving at a speed of 3 m/s rolls up a hill. How does the ball roll before it stops

Physics
1 answer:
sergij07 [2.7K]3 years ago
7 0
It will move 1.5 minutes before it comes to a stopping point.
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A physics student swings a pail of water in a vertical circle 1.0 m in radius at a constant speed. If the water is NOT to spill
love history [14]

Answer:

(A) 3.1 m/s

(B) 2.0 s

Explanation:

At the minimum speed, the force of gravity equals the centripetal force.

mg = m v² / r

v = √(gr)

v = √(9.8 m/s² × 1.0 m)

v = 3.1 m/s

The time is the circumference divided by the speed.

t = (2π × 1.0 m) / (3.1 m/s)

t = 2.0 s

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3 years ago
Is the voltage of two identical lamps the same?​
Dahasolnce [82]

Answer:

It depends if they have the same lightbulb in them.

Explanation:

8 0
2 years ago
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The graph below shows the velocity of a car over time.
jekas [21]

Answer:

D. Calculate the area under the graph.

Explanation:

The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)

You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.

Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.

3 0
3 years ago
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Explain what happens with light causing you to see colored opaque objects
algol13

answer:

an opaque object is one that doesn't let light pass through it. instead, it reflects or absorbs the light that strikes it.

explanation:

in this case, we see the material because of the transmitted light. therefore, the wavelength of the transmitted light determines the color that the object appears.

good luck :)

i hope this helps

have a nice day !!

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2 years ago
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A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp
Jobisdone [24]

Answer:

The tension is  T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by hinge   Fx= \frac{11}{4\sqrt{3} } Mg

Explanation:

   From the question we are told that

          The mass of the beam  is   m_b =M

          The length of the beam is  l = L

           The hanging mass is  m_h = 3M

            The length of the hannging mass is l_h = \frac{3}{4} l

            The angle the cable makes with the wall is \theta = 60^o

The free body diagram of this setup is shown on the first uploaded image

The force F_x \ \ and \ \ F_y are the forces experienced by the beam due to the hinges

      Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

     So

           \sum F =0

Now about the x-axis the moment is

              F_x -T cos \theta  = 0

     =>     F_x = Tcos \theta

Substituting values

            F_x =T cos (60)

                 F_x= \frac{T}{2} ---(1)

Now about the y-axis the moment is  

           F_y  + Tsin \theta  = M *g + 3M *g ----(2)

Now the torque on the system is zero because their is no rotation  

   So  the torque above point 0 is

          M* g * \frac{L}{2}  + 3M * g \frac{3L}{2} - T sin(60) * L = 0

            \frac{Mg}{2} + \frac{9 Mg}{4} -  T * \frac{\sqrt{3} }{2}    = 0

               \frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}

               T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }

                   T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by the hinge is

             F_x= \frac{T}{2} ---(1)

Now substituting for T

              F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}

                  Fx= \frac{11}{4\sqrt{3} } Mg

4 0
3 years ago
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