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Ede4ka [16]
3 years ago
12

The height of a flagpole is 8.5 meters tall. If the length of the meter stick shadow is 1.5 meters long, what would be the lengt

h of the flagpole’s shadow?
Physics
1 answer:
marin [14]3 years ago
3 0
Well, you could easily sow that for every full meter of height of the object, the length of the corresponding shadow would be 50 centimeters longer. therefore, you can multiply 8.50 meters by 1.5 and you would get an answer of 12.75 meters
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What does special relativity reveal about the speed of light relative to its source?
Vladimir [108]

Regardless of the source's mobility, light travels at the same speed.

<h3>What makes special relativity so crucial?</h3>

In the calculating and interpretation of high-velocity phenomena, as well as on our methods of thinking, Einstein's special relativity has had a significant influence on the area of physics. Today, we have a considerably better knowledge of space and time than we did at the start of the century.

<h3>Why is special relativity thus named?</h3>

Because it exclusively uses inertial frames to apply the concept of relativity, the theory is known as "special". General relativity, which Einstein created, applies the principle broadly, that is, to any frame, and this theory takes the gravitational forces into account.

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5 0
2 years ago
An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f
konstantin123 [22]
A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force  F=m \frac{v^2}{r}, directed toward the center of the circle (so, horizontally)
- the weight of the plane: W=mg, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to a=12 m/s^2, we can find the magnitude of this other force F by using Newton's second law:
F+mg=ma
F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
R= \sqrt{(F_c^2+(W+F)^2}=
= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2}  =2.23 \cdot 10^6 N

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
\tan \theta =  \frac{-(W+F)}{F_c}= -0.5
And so, the angle is
\theta = \arctan (-0.5)=-26.8 ^{\circ}
 
7 0
3 years ago
What is the sound intensity level in a car when the sound intensity is 0.525 μW/m2 ? Use I0 = 1.00×10−12 W/m2 for the reference
never [62]

Answer:

The sound intensity level in the car is 57.2 dB.

Explanation:

Sound intensity level in decibels, β = 10 log (I/I₀); where I = 0.525 × 10⁻⁶ W/m², I₀ = 1.0 × 10⁻¹² W/m²

β (dB) = 10 log ((0.525 × 10⁻⁶)/(1.0 × 10⁻¹²)) = 10 × 5.72 = 57.2 dB

Hope this Helps!!!

8 0
3 years ago
If the faster stone takes 12.0 s to return to the ground, how long will it take the slower stone to return?
iren [92.7K]
Same speed, because mass is neglected. The things that affect the speed are the distance and speed of the rock.
5 0
3 years ago
Which contributions did Johannes Kepler make? Check all that apply.
gayaneshka [121]

Answer:

Its 4,5,

Explanation:

7 0
3 years ago
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