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tigry1 [53]
3 years ago
10

What is the CGS unit of current?

Physics
2 answers:
tangare [24]3 years ago
6 0

The answer you are looking for is biot.

melisa1 [442]3 years ago
3 0

biot

 

i think it will help you very much

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Answer for a thanks
mariarad [96]
Well, if your question is how light affects plants,

then you would want to design an experiment that plays aruond with the amount of light a plant gets

thus the thing changing (or variable) would be amount of light
7 0
3 years ago
A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction
zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, F_c=\frac{mv^2}{R}

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

f=\mu N

As there is no vertical motion, therefore, N=mg. So,

f=\mu mg

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816

Therefore, option (c) is correct.

7 0
3 years ago
1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x
allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }

With (\epsilon_{0}) been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    }

Explanation:

(A) Considering a uniform linear density λ_{0} on the ring, then:

dQ=\lambda dl (1)⇒Q=\lambda_{0} 2\pi R(2)⇒\lambda_{0}=\frac{Q}{2\pi R}(3)

Applying the technique of charge integration for finite charges:

V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r'  }} \, dQ(4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

r'=\sqrt{R^{2} +Z^{2}}(5)

Using the expressions (1),(4) and (5) you obtain:

V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}}  }} \, d\phi

Integrating results:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }   (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

|E| =-\nabla V (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

|E| =-\frac{dV(z)}{dz} (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    } (S_b)

4 0
3 years ago
Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A
boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

(c) The average force on the right sled is 922.625 N

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴  22.7 × v₁'  = -3.63 × 3.05

v₁' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law,  m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴  22.7 × v₂'  = -3.63 × 3.05

v₂' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = F_{average} × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}

∴ F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s}  = 922.625 \ kg\cdot m/s^2 = 922.625 \ N  

The average force on the right sled applied by the cat while landing, \mathbf{F_{average}} = 922.625 N

Learn more about conservation of momentum here:

brainly.com/question/7538238

brainly.com/question/20568685

brainly.com/question/22257327

8 0
2 years ago
How will you define motion?
Naddik [55]
Motion by definition is the movement of things above or below the surface of gravity
3 0
3 years ago
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