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IgorLugansk [536]
3 years ago
10

Gamma rays have a higher frequency than visible light waves. What can you conclude from this?

Physics
2 answers:
Finger [1]3 years ago
7 0

Answer:

D

Explanation:

The higher the frequency, the longer the wavelength, so if gamma rays have a higher frequency, than gamma rays have a longer wavelength than visible light waves

NeTakaya3 years ago
3 0

Answer:

i think it  d

Explanation:

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Distance v. Time
Allushta [10]

Answer:

4

(m)

2 ( s )

Explanation:

ok...........

3 0
3 years ago
A bowler who always left the same three pins standing could be considered a(n) ____ bowler.
RideAnS [48]
A bowler who always left the same 3 pins standing could be considered a C. Precise bowler as from bowling countless number of times he has observed the same amount of pins knocked down each time.
4 0
3 years ago
Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

8 0
3 years ago
An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X
Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

u'=0.625 c

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

5 0
3 years ago
What is the overall charge of 1.5 x 10^10 electrons?
Troyanec [42]

Answer: Charge = -2.4x10^-9 Coulombs

Explanation:

The charge of one electron is e = -1.6x10^-19 C

Then, the charge of 1.5 x 10^10 electrons is equal to 1.5 x 10^10 times the charge of one electron:

Here i will use the relation (a^b)*(a^c) = a^(b + c)

Charge = ( 1.5 x 10^10)*( -1.6x10^-19 C) = -2.4x10^(10 - 19) C  

Charge = -2.4x10^-9 C

7 0
2 years ago
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