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Andru [333]
2 years ago
8

A 6.0-v battery maintains the electrical potential difference between two parallel metal plates separated by 1.0 mm. what is the

electric field between the plates?
Physics
1 answer:
Gnoma [55]2 years ago
5 0
The voltage<span> difference between the two plates can be expressed in terms of the </span>work<span> done on a positive test charge q when it moves from the positive to the negative plate.</span><span>
E=V/d
where V is the voltage and d is the distance between the plates.
 
So,

E=6.0V/1mm= 6000 V/m. The electric field between the plates is 6000 V/m.</span>
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All stars in a stellar cluster have roughly the same:___.
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All stars in a stellar cluster have roughly the same distance.

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1 year ago
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xxTIMURxx [149]
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3 0
2 years ago
A cylinder of mass 250 kg and radius 2.60 m is rotating at 4.00 rad/s on a frictionless.
aleksandrvk [35]

Answer:

The angular momentum of a cylinder, when it is rotating with constant angular velocity is Lini =Iωi

. When two cylinders are added to the rotating cylinder, which are identical in their dimensions, the moment of inertia of the entire system increases (since mass increases). The final moment of inertia will be 3I

Since friction exist, all the cylinders start rotating with same angular velocity, the new angular velocity can be calculated using conservation of angular momentum

Thus, Iωi =3Iωf ⟹ωf =ωi/3 = 0.33ωi

8 0
3 years ago
true or false/ force field that exists in the space around every massive body is called electric field​
musickatia [10]

Answer:

True

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3 0
3 years ago
Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the
tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement on Earth = 1.2 m

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a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m

The displacement of the ball on the moon is 7.04 m

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3 years ago
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