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2 years ago

Apparent brightness depends on which two properties?i’ll give brainlist?:)

Physics

2 answers:

Norma-Jean [14]2 years ago

8 0

Answer:

Explanation:

I just did it on a p e x

Eddi Din [679]2 years ago

6 0

Answer:

Explanation:

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Please I need your help

frez [133]

1.14 km = Distance
2.30 m/s = Speed
5.12 cm/s2 = Speed
6.150 mph = Distance
8.3.2 sec = Speed
9.25 ft = Distance

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2 years ago

a tire has a tread pattern with a crevice every 4.00 cm. each crevice makes a single vibration as the tire moves. what is the fr

koban [17]

the frequency (in hz) of these vibrations if the car moves at 24.2 m/s is 605 HZ .

Calculation :

frequency = $\frac{number of contacts}{second}$

frequency = $\frac{24.2 m/s}{1}*100cm*\frac{1}{4cm}$

= 605 HZ

Frequency describes the number of waves passing through a particular location in a particular time. So if the wave takes 1/2 second to travel, the frequency is 2 per second. If it takes 1/100th of an hour, the frequency is 100 per hour.

Frequency is the number of occurrences of a repeating event per unit time. ... sometimes called time-frequency for clarity,

Learn more about frequency here : brainly.com/question/254161

#SPJ4

5 0

1 year ago

What creates the changes in step 1,2,3 of the nitrogen cycle

Naya [18.7K]

1-fixation ( Bacteria Converts nitrogen to ammonium so plants can use it )
2-nitrification ( bacteria changes ammonium to nitrates and plants )
3 - Assimilation (plants absorb nitrates it is then used for Chlorophyll..)

6 0

3 years ago

A man pushes his lawnmower with a velocity of +0.75 m/s relative to the ground. A girl rides by on her bike with a velocity of +

SashulF [63]

Answer:

B. +5.75 m/s

Explanation:

When there are two bodies, a and b, whose velocities measured by a third observer (in this case, the ground) are $V_a$ and $V_b$ respectively, the relative velocity of B with respect to A is given by:

$V_{ba}=V_b-V_a$

Thus, the velocity of the girl relative to the lawnmower is:

$V_{ba}=6.5\frac{m}{s}-0.75\frac{m}{s}\\V_{ba}=5.75\frac{m}{s}$

8 0

3 years ago

Read 2 more answers

A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o

Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m) ⇒ h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv² ⇒ d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0

2 years ago