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RideAnS [48]
3 years ago
5

A rock slides down from A to B along the inside surface of a frictionless hemispherical bowl. As the rock slides, mechanical ene

rgy is conserved (ignoring air resistance) because: (Mark all that apply.) Group of answer choices the bowl is hemispherical in shape. the normal force on the rock is balanced by the centrifugal force. the normal force on the rock is balanced by the centripetal force. the normal force on the rock acts perpendicular to the bowl's surface. the rock's acceleration is perpendicular to the bowl's surface. NONE of the other choices.
Physics
1 answer:
DIA [1.3K]3 years ago
6 0

Answer:

the normal force on the rock acts perpendicular to the bowl's surface.

Explanation:

As we know that Normal force is the reaction force of two contact surfaces which always act perpendicular to the contact surfaces

Here we know that the rock is moving inside the bowl

So Normal force on the rock must perpendicular to the surface of the bowl which always passes through the center of the bowl.

Since the rock is moving in vertical plane so it must have two acceleration

1) Tangential acceleration which will increase the magnitude of the speed along the tangential path

2) Centripetal acceleration which will change the direction of the rock

So here only correct option will be

the normal force on the rock acts perpendicular to the bowl's surface.

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Which is NOT a property of magnetic field lines?
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Why is electric current scalar quantity with having direction ?​
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A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
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Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
3 years ago
Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper
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Explanation:

Expression to calculate thermal resistance for iron (R_{I}) is as follows.

             R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}  

where,   L_{I} = length of the iron bar

             k_{I} = thermal conductivity of iron

             A_{I} = Area of cross-section for the iron bar

Thermal resistance for copper (R_{c}) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where,  L_{c} = length of copper bar

             k_{c} = thermal conductivity of copper

            A_{c} = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

           Q = \frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

 Putting the given values into the above formula as follows.

       Q = \frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

  = \frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}

           = 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

                 P = \frac{Q}{T}

Here, T is 1 second so, power conducted is equal to heat transferred.

So,           P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

7 0
3 years ago
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