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AnnyKZ [126]
3 years ago
9

Which statement about how light travels is true? A Light waves need a medium to travel, and they travel at the same speed even i

f the light source is moving. B Light waves can travel in a vacuum and travel at a constant speed even if the light source is moving. C Light waves can travel in a vacuum and will travel faster if the light source is moving forward. D Light waves need a medium to travel, and they travel faster if the light source is moving forward.
Physics
2 answers:
creativ13 [48]3 years ago
8 0
The answer is a B Light wave can travel in a vacuum and travel at a contest speed even if the light source is moving
Pie3 years ago
7 0

Answer:

Light wave can travel in a vacuum and travel at a contest speed even if the light source is moving

Explanation:

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The object moves with BLANK from A to B. it BLANK from B to C. it moves with BLANK from C to D.
Ganezh [65]

Note that this is a position vs time graph.

From A to B, the graph is a straight line with a nonzero slope. This indicates a constant velocity.

From B to C, the graph is a straight line with 0 slope. This indicates a constant position, i.e. the object remains stationary.

From C to D, the graph is a straight line with a nonzero slope. This indicates a constant velocity.

4 0
3 years ago
Two charges of +4x10-SC and -4x10-5C are placed 1 meter apart. Using Coulomb's law, the force between them is 14.4 N.
Snowcat [4.5K]

Answer:

d. the force is reduced by one quarter to 3.6 n

Explanation:

6 0
2 years ago
The circumference of a sphere was measured to be
professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

\phi = 76cm

Error (dr) = 0.5cm

The radius then would be

\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm

And

\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr

PART A ) For the Surface Area we have that,

A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)

Maximum error:

dA = 4(\frac{38}{\pi})(0.5)

dA = \frac{76}{\pi}cm^2

The relative error is that between the value of the Area and the maximum error, therefore:

\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}

\frac{dA}{A} = 0.01315 = 1.31\%

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

V = \frac{4}{3} \pi r^3

V = \frac{4}{3} \pi (\frac{38}{\pi})^3

V = \frac{219488}{3\pi^2}

Therefore the Maximum Error would be,

\frac{dV}{dr} = \frac{4}{3} 3\pi r^2

dV = 2r^2 (2\pi dr)

dV = 4r^2 (\pi dr)

Replacing the value for the radius

dV = 4(\frac{38}{\pi})^2(0.5)

dV = \frac{2888}{\pi^2} cm^3

And the relative Error

\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }

\frac{dV}{V} = 0.03947

\frac{dV}{V} = 3.947\%

3 0
3 years ago
Find the sum of the vectors:11 km N ,11km E
vladimir1956 [14]

The resultant vector is 11√2 km due north east.

<h3><u>Explanation:</u></h3>

The vector is a type of quantity which has both magnitude and direction. This quantities when expressed needs to specify both magnitude and direction.

We need to calculate the magnitude and direction separately.

Here firstly for the magnitude,

The magnitudes are both 11 km and they are at right angles to each other.

So, the resultant magnitude = √(11² +11²) km

=11√2 km

Now for the direction, one vector is due north and the other is due east.

So the resultant vector is due north east.

So the final vector is 11√2 km due North-East.

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3 years ago
Llammmmaaaa or chickennnss?
Semenov [28]

Answer:

chickennnss

Explanation:chickennnss

8 0
3 years ago
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