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Basile [38]
3 years ago
5

An object with charge q=−6.00×10−9C is placed in a region of uniform electric field and is released from rest at point A. After

the charge has moved to point B, 0.500 m to the right, it has kinetic energy 3.00×10−7J.
(a) If the electric potential at point A is +30.0 V, what is the electric potential at point B?
(b) What are the magnitude and direction of the electric field?
Physics
1 answer:
shtirl [24]3 years ago
8 0

Answer:

its a

Explanation:

used google

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Make the following conversion. 120 mL = _____ cm³ 12.0 1200 120 1.20
omeli [17]
120 is your answer. 120 mL = 120 cm^3
7 0
3 years ago
Read 2 more answers
Can the resistors in an "unbalanced" Wheatstone bridge circuit be treated as a combination of series and/or parallel resistors?
OlgaM077 [116]

Answer:

Explanation:

The resistors in a unbalanced wheat stone bridge cannot be treated as a combination of series and parallel combination of resistors.

In case of balanced wheat stone bridge, the resistors can be treated as the combination of series and parallel combination.

Here, In the balanced wheat stone bridge

R1 and R2 be in series and Ra and Rx is series and then their combination is in parallel combination.

4 0
3 years ago
Newton’s first law of motion is often called the law of?
jeka57 [31]

Answer:

Newton's first law is often called the law of inertia.

searched it up to verify

Also welcome to brainly

6 0
2 years ago
If the magnitude of the initial velocity of the ball v0 = 7.94 ± 0.03, and the "gun" is tilted 31 ± 0.4º upwards, what is the un
ycow [4]

Answer:

0.05

Explanation:

Given a variable C which is the product of two variables A, B:

C=A\cdot B

Then the absolute error on C is given by:

\frac{\sigma_C}{C}=\frac{\sigma_A}{A}+\frac{\sigma_B}{B}

where \sigma_A, \sigma_B, \sigma_C are the uncertanties on the measure of A, B and C, respectively.

In this problem, the horizontal component of the velocity v_x is given by

v_x = v_0 cos \theta

Therefore, the uncertainty on vx is given by:

\frac{\sigma_{v_x}}{v_x}=\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta} (1)

where we have:

v_0 = 7.94

\sigma_{v_0}=0.03

\theta=31^{\circ}, so

cos \theta=cos 31^{\circ}=0.857

The uncertainty on cos \theta is given by:

\sigma_{cos \theta}=|sin \theta|\sigma_\theta

where:

|sin \theta| = |sin 31^{\circ}|=0.515

and

\sigma_\theta=0.4^{\circ}=0.007 rad

So

\sigma_{cos \theta}=|0.515|\cdot 0.007 = 0.0036

Also,

v_x = v_0 cos \theta = (7.94)(cos 31^{\circ})=6.81 m/s

So, combininb everything into (1), we find:

\sigma_{v_x}=(\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta})v_x=(\frac{0.03}{7.94}+\frac{0.0036}{0.857})(6.80)=0.05

4 0
3 years ago
Gravitational force on the moon is only 1/6 that of the gravitational force on earth. what would be the weight of a 10-kg object
MA_775_DIABLO [31]
According to Newton's Second Law of Motion, the force is equal to the mass of an object multiplied by its acceleration. When you talk about gravitational force, the acceleration referred to here is the acceleration due to gravity. This is very familiar to us in physics. The acceleration due to gravity on Earth is equal to 9.81 m/s². It actually depends on the location. According to the Universal Law of Gravitation:

F = Gm₁m₂/d²

The force is a factor of the product of two masses and their distance from each other. The G is a constant called the universal gravitational constants. So, gravitational force is actually a relative force exerted by one body to another. 

Going back the Second Law of Motion, we can modify the equation to:

F = mg

Since it is mentioned that the gravity on the moon is only 1/6 of the Earth, then the gravity for moon is:
g,moon = 1/6(9.81) = 1.635 m/s²
So, let's compare the weight of the object with a mass of 10 kg. The weight is actually the force due to gravity pulling you towards the center of the body.

Weight on Earth = (10 kg)(9.81 m/s²) = 98.1 N
Weight on Moon = (10 kg)(1.635 m/s²) = 16.35 N

The mass, on the other hand, is not affected by gravity. It is always constant. Therefore, the mass of the object on the moon is the same with its mass on the Earth.
6 0
3 years ago
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