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2 years ago

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compute the electric field experienced by a test a charge q = +0.08µC from a source charge q = +15µC in a vacuum when the test c

harge is placed 0.02 m away from the other charge.

1 answer:

Fudgin [204]2 years ago

3 0

<u>here</u><u> </u><u>is</u><u> </u><u>your</u><u> </u><u>answer</u><u> </u><u>hope</u><u> </u><u>it's</u><u> </u><u>helpful</u><u> </u><u>for</u><u> </u><u>you</u><u> </u><u>thankyou</u><u> </u><u>I</u><u> </u>

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Identify evidence that supports the theories of continental drift and plate tectonics. put responses in the correct input to ans

GenaCL600 [577]

The evidence that supports continental drift and plate tectonics includes different fossils, the same rocks and the shapes of continents that fit together.

<h3>What is continental drift?</h3>

Continental drift is a theory that states continents once were part of one big landmass known as Pangea.

Nowadays, the theory of continental drift proposed by Alfred Wegener has been replaced by plate tectonics.

In conclusion, the evidence that supports continental drift and plate tectonics includes fossils, the same rocks and the shapes of continents that fit together.

Learn more on the continental drift here:

brainly.com/question/394265

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2 years ago

Air escaping out from an air hose at a gas station always feels cool. Why?

kati45 [8]

Answer:

The air is contained at a high pressure in the tube. When it escapes from a small orifice, it suddenly expands. A large amount of its heat is absorbed in the process of expansion resulting in considerable fall in its temperature. This is why the escaping air feels cold.

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2 years ago

Why do the four outer planets have lower density than the four inner planets?

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4 years ago

A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th

ale4655 [162]

44.64m

Explanation:

Given parameters:

Mass of the car = 1500kg

Initial velocity = 25m/s

Frictional force = 10500N

Unknown:

Distance moved by the car after brake is applied = ?

Solution:

The frictional force is a force that opposes motion of a body.

To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.

-Frictional force = m x a

the negative sign is because the frictional force is in the opposite direction

m is the mass of the car

a is the acceleration of the car

a = $\frac{frictional force}{mass}$ = $\frac{10500}{1500}$ = -7m/s²

Now using;

V² = U² + 2as

V is the final velocity

U is the initial velocity

a is the acceleration

s is the distance moved

0² = 25² + 2 x 7 x s

0 = 625 - 14s

-625 = -14s

s = 44.64m

learn more:

Velocity problems brainly.com/question/10932946

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7 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago