All categories

2 years ago

Mark all the units for speed

Physics

1 answer:

mr Goodwill [35]2 years ago

6 0

Answer:

Speed. I am speed.

Explanation:

The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.

You might be interested in

If you walk a distance of 8 blocks and then 3 blocks south from home, what is your position compared to home? What distance did

Harman [31]

The position compared to that of home is a reference to displacement, I believe.

Displacement = x total - x initial

So I believe the answer is 5 blocks due north (if you’re walking linearly from your home), unless the questions is referring to relative displacement, in which then you’d need to use the Pythagorean theorem to find the hypotenuse between both positions. And then you’d have to find theta for the degrees between the south direction and the other unmentioned direction. But I don’t think that’s the case.

Distance refers to x total and doesn’t care for direction, as this refers to a scalar quantity opposed to a vector. Thus the equation is just

d = x

So 8 blocks + 3 blocks = a distance of eleven blocks walked total

7 0

3 years ago

Two objects have a force of gravity of 100 N. If the distance between both objects decreases by a factor of 7, while the masses

Shkiper50 [21]

The gravitational force F between two masses M and m a distance r apart is

F = G M m / r ²

Decrease the distance by a factor of 7 by replacing r with r / 7, and decrease both masses by a factor of 8 by replacing M and m with M / 8 and m / 8, respectively. Then the new force F* is

F* = G (M / 8) (m / 8) / (r / 7)²

F* = (1/64 × G M m) / (1/49 × r ²)

F* = 49/64 × G M m / r ²

In other words, the new force is scaled down by a factor of 49/64 ≈ 0.7656, so the new force has magnitude approx. 76.56 N.

8 0

3 years ago

The pointer of an analog meter is connected to a

dangina [55]

C. coil suspended by bearings.
but im not 100% sure

8 0

3 years ago

Read 2 more answers

A temporary license permits the holder to drive for up to _ days while the application is reviewed

Ugo [173]

60 days, tell me if I'm correct please.

7 0

3 years ago

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0

3 years ago