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3 years ago

What are some examples and non-examples of volume

Physics

2 answers:

Jlenok [28]3 years ago

4 0

Answer:

Explanation:

Some correct non-examples are: A glass half-empty; Anything in two dimensions; The amount that covers something.

gregori [183]3 years ago

4 0

Answer:The glass is half full is an example of volume.

The glass is half empty is an example of non-volume.

A blank page is a non-example of everything including volume.

Explanation:

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a light wave is traveling straight up out of the ocean then moves out into the air how do you expect the movement of the light w

Mariulka [41]

If the light is traveling straight up, then it hits the interface (surface
or boundary) between water and air perpendicularly (90° to the surface).

This direction is the direction of the 'normal' to the surface. So the
angle of incidence is zero, and that means the angle of refraction is
also zero. The light just keeps going in the same direction when it
emerges into the air, and is not bent.

However, its speed increases in air, and that means its wavelength
also becomes longer than it was in the water.

6 0

4 years ago

It is well known that runners run more slowly around a curved track than a straight one. One hypothesis to explain this is that

barxatty [35]

Answer:

$percentage = 6.27 %$

Explanation:

As we know that when runner is moving on straight track then the net force on his feet is given as

$F_s = mg$

while when runner is moving on circular track then we have

$F_c = \sqrt{(mg)^2 + (\frac{mv^2}{R})^2}$

$F_c = m\sqrt{g^2 + \frac{v^4}{R^2}}$

$F_c = m\sqrt{9.8^2 + \frac{8^4}{18^2}}$

$F_c = m(10.42)$

now percentage change is given as

$percentage = \frac{F_c - F_s}{F_s} \times 100$

$percentage = \frac{m(10.42) - m(9.81)}{m(9.81)}\times 100$

$percentage = 6.27 %$

6 0

3 years ago

Which change will always result in an increase in the gravitational force between two objects? increasing the masses of the obje

abruzzese [7]

By definition, we have that the gravitational force is given by:

$F = \frac{G * m1 * m2}{r ^ 2}$

Where,

G: gravitational constant

m1: mass of object number 1

m2: mass of object number 2

r: distance between both objects.

Therefore, for the gravitational force to increase, the following conditions must be met:

1) Increase the mass of the objects so that the numerator of the equation is greater.

2) Decrease the distance between the objects so that the denominator of the equation is smaller.

Answer:

A change that will always result in an increase in the gravitational force between two objects is:

increasing the masses of the objects and decreasing the distance between the objects

4 0

3 years ago

Read 2 more answers

Using a diagram explain how acceleration can be obtained from a velocity time graph

Oksi-84 [34.3K]

Answer:

For any v-t graph the acceleration is the slope of the graph. For average acceleration in a time period ‘t’ consider the change in velocity in time t and divide it by the time t. For instantaneous acceleration you need to go into the realm of differential calculus.

Explanation:

I hope this helps! I would really appreciate it if you would please mark me brainliest! Have a blessed day!

3 0

3 years ago

A 23 g bullet is accelerated in a rifle barrel 62 cm long to a speed of 593 m/s. Use the work-energy theorem to find the average

Anuta_ua [19.1K]

Answer:

The average force exerted on the bullet while it is being accelerated is 6,522.52 N.

Explanation:

Given;

mass of the bullet, m = 23 g = 0.023 kg

length of the barrel, L = 62 cm = 0.62 m

speed of the bullet, v = 593 m/s

Applying work-energy theorem;

the work done in accelerating the bullet in the riffle = kinetic energy acquired by the bullet.

W = K.E

F x d = ¹/₂mv²

where;

d is the is the distance traveled by the bullet in the riffle = L

F(0.62) = ¹/₂ x 0.023 x (593²)

F(0.62) = 4043.964

F = (4043.964) / (0.62)

F = 6,522.52 N

Therefore, the average force exerted on the bullet while it is being accelerated is 6,522.52 N.

6 0

3 years ago