8 miles per hour
(extra space)
Answer:
The magnitude of the resultant decreases from A+B to A-B
Explanation:
The magnitude of the resultant of two vectors is given by

where
A is the magnitude of the first vector
B is the magnitude of the second vector
is the angle between the directions of the two vectors
In the formula, A and B are constant, so the behaviour depends only on the function
. The value of
are:
- 1 (maximum) when the angle is 0, so the magnitude of the resultant in this case is

- then it decreases, until it becomes 0 when the angle is 90 degrees, where the magnitude of the resultant is

- then it becomes negative, and continues to decrease, until it reaches a value of -1 when the angle is 180 degrees, and the magnitude of the resultant is

Answer:
The value of resistance when power is 1100 watts =
= 50 ohms
Explanation:
Power
= 2200 Watts
Resistance
= 25 ohms
Power
= 1100 Watts
Resistance
= we have to calculate
Given that the power in an electric circuit varies inversely with the resistance
⇒ P ∝ 
⇒
= 
⇒
= 
⇒
= 50 ohms
This is the value of resistance when power is 1100 watts.
Answer:
<span>C) Both have rocky composition
Explanation
Earth has rocks. Many many rocks.
It is made of many type of rocks such as </span>sedimentary<span>, </span>metamorphic<span>, and </span><span>igneous rocks. and guess what a asteroid is? That is right ! A ROCK.
</span>A asteroid is normally made of <span>stony silicate rocks , but not always , sometimes I can be diamonds!! The point is , they both are rocks floating around space.</span><span>
</span>
Answer:
when the object goes from the focal length to twice the focal length the image goes from infinity to twice the focal length, this image is real and inverted
Explanation:
Let's use the constructor equation to describe the image of a concave mirror
1 / f = 1 / p + 1q
where f is the focal length, p and q the distance to the object and the image, respectively
1 /q = 1/f - 1/p
tell us that the image is between the focal and twice the focal, let's calculate the position of the image
for both ends
case 1, distance to the object p = f
1 / q = 1 / f -1 / f
1 / q = 0
q = ∞
the image is in infinity
case2, distance to object p = 2f
1 / q = 1 / f - 1 / 2f
1 / q = 1 / 2f
q = 2f
the image is twice the focal length, the object and the image are at the same point
therefore the image when the object goes from the focal length to twice the focal length the image goes from infinity to twice the focal length, this image is real and inverted