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4 years ago

If a piece of space debris is too large to be a meteoroid and too small to be a planet, it could be

Physics

2 answers:

rewona [7]4 years ago

8 0

The answer is B. an asteroid

Vaselesa [24]4 years ago

3 0

an asteroid was the correct answer

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Green plants need light in order to survive. Structures in the leaves absorb light, which in turn, helps plants make their own f

Goryan [66]

The answer is green.

3 0

3 years ago

Read 2 more answers

20% Part (a) Use an "E Field Sensor" and move it along either equipotential. What can you say about the E field along an equipot

Alex

Answer:

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

Explanation:

As we know that the relation between electric field and electric potential is given as

$\Delta E = -\frac{dV}{dr}$

here if we say that potential is constant because electric field sensor is moving along equi-potential line.

Then we will say

V = constant

so we have

$\Delta E = 0$

so electric field will remain constant always in magnitude and always remains perpendicular to the surface

so we have

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

6 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

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

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

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>

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

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

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

Which type of circuit would be best to use for lights used for decorations? Question 1 options: Series circuit. One bulb could g

Masteriza [31]

Answer:

One bulb could go out and the strand will stay on.

Explanation:

In series circuit, there is only one path provided for the current to flow. So, all the lights are required to be in working condition, for the others to work. And if anyone light bulb goes out, the circuit will become incomplete and the rest of the strand will also go out. Because there is only one path for current flow which is now broken.

On the other hand, in parallel circuits, each light bulb has a separate connection with the source. Current path to each bulb is independent of the others. Therefore, if one bulb goes out, the rest of the strand will stay on.

So, the correct option is:

<u>One bulb could go out and the strand will stay on.</u>

6 0

3 years ago

What is the equivalent resistance of the following combination circuit? A 5.00 Ω resistance placed in series with a parallel com

Sunny_sXe [5.5K]

The equivalent resistance of the circuit is 8.6 Ω.

<h3>What is resistance?</h3>

Resistance can be defined as the opposition to current flow in an electric circuit. The S.I unit of resistance is ohms.

To calculate the equivalent resistance, we use the formula below.

Formula:

R' = R₁+(R₂//R₃)................ Equation 1

Where:

R' = Equivalent resistance of the circuit.

From the diagram,

Given:

R₁ = 5 .00 Ω
R₂ = 9.00 Ω
R₃ = 6.00 Ω

Substitute these values into equation 1

R' = 5+(9//6)
R' = 5+[(9×6)/(9+6)]
R' = 5+[54/15]
R' = 5+3.6
R' = 8.6 Ω

Hence, The equivalent resistance of the circuit is 8.6 Ω
Learn more about equivalent resistance here: brainly.com/question/12856032

6 0

2 years ago