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galben [10]
3 years ago
13

Hurryyyyyyyy

Physics
2 answers:
padilas [110]3 years ago
7 0
The answer would be C. Gamma Rays and High Frequency EM waves travel at the speed of light and are transverse waves.
Tcecarenko [31]3 years ago
7 0
They are transverse and travel at the speed of light. They are both parts of the electromagnetic spectrum so they travel at the speed of light. They are both types of transverse waves because the medium they travel through have particles that are perpendicular to the direction of the wave. Hope this helps!
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I need a short answer ?
mars1129 [50]

Answer:

Explanation:

7a) t = d/v = 100/45cos14.5 = 2.29533...= 2.30 s

7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m

  or

  h = (45sin14.5)² / (2(9.81)) = 6.47 m

which rounds to the same 6.5 m when limiting to the two significant digits of the initial velocity.

7 0
2 years ago
What is the greenhouse effect?
Zanzabum
A on that problem :)[email protected]
6 0
2 years ago
Read 2 more answers
Answer the following question: “Do the particles in a gas ever slow down and stop? Include information about the three states of
jarptica [38.1K]

Answer:

I hope it is no too late

Explanation:

hmmm,

In a gas, for example, the molecules are traveling in random directions at a variety of speeds - some are fast and some are slow. ... If more energy is put into the system, the average speed of the molecules will increase and more thermal energy or heat will be produced.

3 0
3 years ago
Find the magnitude and direction of an electric field that exerts a 4.80×10−17N westward force on an electron. (b) What magnitud
jok3333 [9.3K]

Explanation:

(a) E = F/q

E = 4.8×10^-17/1.6×10^-19

E = 300 N/C

(b) same magnitude of electric field is exerted on proton

4 0
2 years ago
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re
valentinak56 [21]

Answer:

a) 17.33 V/m

b) 6308 m/s

Explanation:

We start by using equation of motion

s = ut + 1/2at², where

s = 1.2 cm = 0.012 m

u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

E = electric field

m = mass of proton

a = acceleration

q = charge of proton

E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19

E = 2.77*10^-18 / 1.6*10^-19

E = 17.33 V/m

Final speed of the proton can be gotten by using

v = u + at

v = 0 + 1.66*10^9 * 3.8*10^-6

v = 6308 m/s

5 0
2 years ago
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