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Artemon [7]
4 years ago
6

Helpppppp please anybody who knows

Physics
1 answer:
levacccp [35]4 years ago
3 0
We know that the intensity of light decreases as the square of the distance
from the source.

Lamp-1 is twice (2x) as far from the detector as Lamp-2 is.
If they appear equal to the detector, then Lamp-1 must be (2)² = 4 times
as bright as Lamp-2.

4 x 15 candelas  =  60 candelas  


Question #15).
A). is true, but has nothing to do with particle behavior
B). Diffraction is a purely wave thing, so it doesn't help.
C). is true, but doesn't say anything about particle behavior
D). is true, but isn't an indication of particle behavior.

Looks to me like NONE of the choices is any help
to suggest particle behavior.
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The observation deck of a skyscraper is 420 m above
Reil [10]
2e min :)) pls park braliest
5 0
3 years ago
A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}

t=1.597s

Substituting (6) in (1):

67.1 =V_{o} cos(0\°) 1.597-------------------(4)

Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

V_0 = \frac{6.71}{1.597}

V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

5 0
3 years ago
How much force is needed to accelerate a vehicle with a mass of 1000 kg at a rate<br> of 5 m/s2?
Mashcka [7]
The force needed to accelerate a vehicle with a mass of 1000kg at a rate of 5m/s2 would be 5000
8 0
2 years ago
Consider a 4-m-long, 4-m-wide, and 1.5-m-high above-the-ground swimming pool that is filled with water to the rim. (a) Determine
pashok25 [27]

Answer:

44100 N

Explanation:

Each wall will have dimension of 4 m x 1.5 m

Whole force will act on central point of wall situated at a depth of 1.5 /2 = .75m

pressure at CM = h d g , h = .75 , d ( density of water = 10³ )

pressure at CM = .75 x 10³ x 9.8

= 7350 N / m²

Total force on each wall

= pressure x area

= 7350 x 4 x 1.5

= 44100 N Ans

b ) If h = 1.5 x 2 = 3

Pressure = hdg

1.5 x 10³ x 9.8

= 14700 N / m²

Force

= pressure x area

14700 x 3 x 4

= 176400 N

Which is 4 times 44100 N

So force will quadruple.

It is so because both area and height have become twice.

3 0
3 years ago
A toy car is placed at 0 on a number line. It moves 9 cm to the left, then 4 cm to the right, and then 6 cm to the len
MA_775_DIABLO [31]

Answer:19 cm

Explanation:

6 0
3 years ago
Read 2 more answers
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