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2 years ago

Anything that has volume or mass

Physics

1 answer:

irinina [24]2 years ago

8 0

Any object, except antimatter, :)

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Select the correct answer from each drop-down menu.

-Dominant- [34]

Answer:

And the force of ( Attraction or repulsion) between the poles A and D ( maximum or minimum)

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2 years ago

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A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds

vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

$\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega = (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s$

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s x 29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

8 0

2 years ago

How many different types of atoms are present in one molecule of aluminum hydroxide, Al(OH)3?

tatyana61 [14]

There are approximately 3 different types of atoms that are present in one molecule of aluminum hydroxide, AI(OH)3.

8 0

2 years ago

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If an equation is dimensionally correct is thar equation a right equation<br>

Nady [450]

If an equation is dimensionally correct, it does not mean that the equation must be true. On the other hand, when the equation is dimensionally correct, the equation cannot be true. Dimensional analysis is a technique used to check whether a relationship is correct

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2 years ago

An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2

Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

$M=-\frac{q}{p}$

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

$M=-\frac{200 cm}{50 cm}=-4.0$

7 0

2 years ago