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Solnce55 [7]
3 years ago
12

URGENT!!!!!!!!!: If an electromagnetic wave has a frequency of 6 x 10^5 Hz, what is its

Physics
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

A

Explanation:

because the speed divide by the frequency is equal to the wavelength(in meters)

5×10² m

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In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward
Vladimir79 [104]

Answer:

The answer is "1.01 \times 10^{-13}"

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

KE=U=\frac{kqq'}{r}\\\\

Calculating the closest distance:

\to r=\frac{kqq'}{KE}\\\\

=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\

=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}

5 0
3 years ago
Why forces are balanced and unbalanced? need help with this the lesson is tommorow
Naya [18.7K]
"Balanced" means that if there's something pulling one way, then there's also
something else pulling the other way. 

-- If there's a kid sitting on one end of a see-saw, and another one with the
same weight sitting on the other end, then the see-saw is balanced, and
neither end goes up or down.  It's just as if there's nobody sitting on it.

-- If there's a tug-of-war going on, and there are 300 freshmen pulling on one
end of a rope, and another 300 freshmen pulling in the opposite direction on
the other end of the rope, then the hanky hanging from the middle of the rope
doesn't move.  The pulls on the rope are balanced, and it's just as if nobody
is pulling on it at all.

-- If a lady in the supermarket is pushing her shopping cart up the aisle, and her
two little kids are in front of the cart pushing it in the other direction, backwards,
toward her.  If the kids are strong enough, then the forces on the cart can be
balanced. Then the cart doesn't move at all, and it's just as if nobody is pushing
on it at all.

From these examples, you can see a few things:

-- There's no such thing as "a balanced force" or "an unbalanced force".
It's a <em><u>group</u> of forces</em> that is either balanced or unbalanced.

-- The group of forces is balanced if their strengths and directions are
just right so that each force is canceled out by one or more of the others.

-- When the group of forces on an object is balanced, then the effect on the
object is just as if there were no force on it at all.
4 0
3 years ago
Read 2 more answers
Gold in its pure form is too soft to be used for most jewelry. Therefore, the gold is mixed with other metals to produce an allo
Novay_Z [31]

Answer:

Explanation:18kt alloy contains

i) 75% of gold

rhogold=19.3g/cm^3

=75/100×19.3

=14.475g/cm^3

ii) 16% of silver

rhosilver=10.5g/cm^3

=16/100×10.5

=1.68g/cm^3

iii) 9% of copper

rhocopper =8.90g/cm^3

=9/100×8.9

=0.801g/cm^3

Overall density of 18kt gold

=(0.801+1.68+14.475)g/cm^3

=16.956g/cm^3

=17g/cm^3 to 3s.f

6 0
3 years ago
A 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?
Marianna [84]

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

<u>Given the following data;</u>

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

<em>P.E = 3430J</em>

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

<em>Workdone = 3430Nm</em>

6 0
3 years ago
A transmission diffraction grating with 420 lines/mm is used to study the light intensity of di event orders (n). A screen is lo
Goshia [24]

Answer:

Explanation:

Diffraction grating is used to form interference pattern of dark and bright band.

Distance between adjacent slits (a ) = 1 / 420 mm

= 2.38 x 10⁻³ mm

2.38 x 10⁻⁶ m

wave length of red light

= 680 x 10⁻⁹ m

For bright red band

position x on the screen

= n λD / a ,         n = 0,1,2,3 etc

D = distance of screen

putting n = 1 , 2 and 3 , we can get three locations of bright red band.

x₁ = λD / a

=  680 x 10⁻⁹ x 2.8 / 2.38 x 10⁻⁶

= .8 m

= 80 cm

Position of second bright band

= 2 λD / a

= 2 x 80

= 160 cm

Position of third bright band

= 3 λD / a

= 3 x 80

= 240 cm

5 0
3 years ago
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