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Ede4ka [16]
3 years ago
5

Please help question #70

Physics
1 answer:
Keith_Richards [23]3 years ago
8 0

can i get the question so that i can answer your question

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What is the order of the electromagnetic spectrum from highest to lowest energy?
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<span> Radio waves, microwaves, infra red, visible light, ultra violet, x-rays, gamma rays. </span>

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3 years ago
Mr. Phillips' car iso parked on a steep hill
Wewaii [24]
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7 0
4 years ago
Regardless of their frequency, wavelength, or energy, all electromagnetic waves: A. travel only through the vacuum of space. B.
agasfer [191]
All electromagnetic waves travel at the same speed in a vacuum: 3.0 x 10^5 (300,000) kilometres per second. some electromagnetic waves are part of the visible light spectrum and some do emit harmful radiation, but certainly not all. they travel fine on earth without the vacuum of space too. 
6 0
3 years ago
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
liq [111]

Answer:

A) total time = 55.5 seconds

B) average velocity = 25.27 m/s

Explanation:

It starts from rest, so initial velocity, u = 0 m/s

We are given;

acceleration; a = 2 m/s²

Final velocity; v = 31 m/s

From Newton's first law of motion,

v = u + at

So, 31 = 0 + 2t

t = 31/2

t = 15.5 sec

We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;

(a) Total time in which car is in motion = 15.5 + 35 +5 = 55.5 seconds

b)Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;

S = ut + ½at²

S1 = 0 + ½(2 * 15.5²)

S1 = 240.25 m

Distance traveled in 35 sec with with velocity of 31 m/sec is;

S2 = velocity x time

S2 = 35 × 31 = 1085 m

Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity(u) will be 31 m/s.

From the first equation of motion,

a = (v - u)/t

a = (0 - 31)/5

a = -6.2 m/s²

So, distance travelled is;

S3 = ut + ½at²

S3 = (31 × 5) + ½(-6.2 × 5²)

S3 = 155 - 77.5

S3 = 77.5 m

So overall total distance = S1 + S2 + S3

Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m

Average velocity = total distance/total time

Average velocity = 1402.75/55.5 = 25.27 m/s

6 0
3 years ago
Review 1: A plane is located x = 40 miles (horizontally) away from an airport at an altitude of h miles. Radar at the airport de
-Dominant- [34]

Explanation:

Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,

s(t)^2={h^2+x^2}...........(1)

Given, h = 4, x = 40 and s(t) = -20 mph

Differentiate equation (1) wrt t

2s(t)s'(t)=2x(t)x'(t)

x'(t)=\dfrac{s(t)s'(t)}{x(t)}

When x = 40, s(t)=\sqrt{40^2+4^2}=40.19\ m

x'(t)=\dfrac{-240s(t)}{x(t)}

x'(t)=\dfrac{-240\times 40.19}{40}

x'(t)=-241.14\ m/s

So, the speed of the airplane is 241.14  m/s. Hence, this is the required solution.

8 0
3 years ago
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