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Flura [38]
3 years ago
9

Answer and I will give you brainiliest Please heeeelp​

Physics
1 answer:
Lostsunrise [7]3 years ago
5 0

Answer:

T = 4.905[N]

Explanation:

In order to solve this problem we must perform a sum of forces on the vertical axis.

∑Fy = 0

We have two forces acting only, the weight of the body down and the tension force T up, as the body does not move we can say that it is system is in static equilibrium, therefore the sum of forces is equal to zero.

T-m*g=0\\T=0.5*9.81\\T=4.905[N]

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Kinetic and
beks73 [17]

Answer:

raise the board to a higher angle

Explanation:

Static friction is the force opposite to the applied force.

Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.

So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.

Hence, the correct option is "raise the board to a higher angle".

5 0
3 years ago
Scientists described light as a wave because the results of many experiments with light demonstrated that light behaves as a wav
Ksju [112]

The answer is Data collected from many experiments could not be explained.


Explanation:

With pretty much every science theory, the theory can not change without valid experimentation by multiple professionals.

4 0
3 years ago
Read 2 more answers
Two fully charged cylindrical capacitors are connected to two identical batteries. The capacitors are identical except that the
Leni [432]

Answer:

Part(a):  The relative capacitance is \dfrac{C_{A}}{C_{B}} = 0.33

Part(b): The relative energy stored is \dfrac{U_{A}}{U_{B}} = 0.33

Part(c): The relative charge stored is \dfrac{Q_{A}}{Q_{B}} = 0.33

Explanation:

We know the capacitance (C) of a capacitor having charge (Q) and subjected to a potential difference of (V) is given by

C = \dfrac{Q}{V}

Also, the energy (U) stored by a capacitor can be written as

U = \dfrac{1}{2}C~V^{2}

Let us assume that the inner radius of the Capacitor B, as shown in the figure, be \textbf{r_{i}^{B}}\bf{r_{i}^{B}}, the outer radius be \bf{r_{o}^{B}}, the inner radius of Capacitor A be \bf{r_{i}^{A}} and the outer radius be \bf{r_{o}^{B}}.

Given in the problem,

&& r_{o}^{B} = 2~r_{B}^{i}\\&& r_{o}^{A} = 4~r_{B}^{i}\\&& and~r_{i}^{B} = 4~r_{o}^{B} = 8~r_{B}^{i}

Now, the capacitance (C) of a cylindrical capacitor is given by,

\bf{C = \dfrac{2~\pi~\epsilon_{0}~L}{ln(\dfrac{r_{o}}{r_{i}})}}

where \epsilon_{o} is the permittivity of the free space, L is the length of the cylindrical capacitor.

Part(a):

The capacitance of capacitor A,

C_{A} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{A}}{r_{i}^{A}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{8~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(8)}

and the capacitance of capacitor B,

C_{B} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{2~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(2)}

giving the relative capacitance of each capacitor to be

\dfrac{C_{A}}{C_{B}} = \dfrac{ln(2)}{ln(8)} = \dfrac{ln(2)}{3~\ln(2)} = \dfrac{1}{3} = 0.33

Part(b):

Energy stored by capacitor A,

U_{A} = \dfrac{1}{2}~C_{A}~V^{2}

Energy stored by capacitor B,

U_{B} = \dfrac{1}{2}~C_{B}~V^{2}

giving the relative energy stored by each capacitor to be

\dfrac{U_{A}}{U_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33

Part(c):

The charge stored by capacitor A,

Q_{A} = C_{A}~V

The charge stored by capacitor B,

Q_{B} = C_{B}~V

giving the relative charge stored by each capacitor to be

\dfrac{Q_{A}}{Q_{B}} =  \dfrac{C_{A}}{C_{B}} = 0.33

8 0
3 years ago
HURRY WILL GIVE BRAINLIESTHow much heat is needed to melt 45.00 g of ice at 0°C if the latent heat of fusion of water is 333.7 J
Allisa [31]

D. 15,020 J

heat = 45 x 333.7 j = 15016.5j

7 0
3 years ago
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Why are stars given an absolute magnitude?
mafiozo [28]

Answer:

The solution was to implement an absolute magnitude scale to provide a reference between stars. To do so, astronomers calculate the brightness of stars as they would appear if it were 32.6 light-years, or 10 parsecs from Earth.

Explanation:

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