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mylen [45]
4 years ago
13

what are the 3 properties of components of the universe that can be determined using electromagnetic radiation?

Physics
1 answer:
stepan [7]4 years ago
4 0
Your answer is electricity, light and magnetism.  They can be determined usinf elecromagnetic radioation. 
<span>
Even the energy  can't be detected by our eyes, there are a lot of measurement  instruments that can measure infrared (IR), gamma rays, radio or X-rays or ultraviolet (UV)</span>
You might be interested in
Question 18
Sedaia [141]

Answer:

5.5 kg

Explanation:

The gravitational potential energy of an object is given by

U=mgh

where

m is its mass

g is the acceleration of gravity

h is the height of the object above the ground

For the rock in the problem, we know:

h = 12 m is the height above the ground

U = 650 J

g = 9.8 m/s^2

Therefore, we can solve the equation for m, the mass of the rock:

m=\frac{U}{gh}=\frac{650}{(9.8)(12)}=5.5 kg

3 0
3 years ago
Suppose we could shrink the earth without changing its mass..?At what fraction of its current radius would the free-fall acceler
Drupady [299]

Answer:

at R/\sqrt{3}

Explanation:

The free-fall acceleration at the surface of Earth is given by

g=\frac{GM}{R^2}

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

The formula can be rewritten as

R=\sqrt{\frac{GM}{g}} (1)

We want to shrink the Earth at a radius R' such that the acceleration of gravity becomes 3 times the present value, so

g' = 3g

Keeping the mass constant, M, and substituting into the equation, we have

3g=\frac{GM}{R'^2}

R'=\sqrt{\frac{GM}{3g}}=\frac{1}{\sqrt{3}}\sqrt{\frac{GM}{g}}=\frac{R}{\sqrt{3}}

5 0
3 years ago
Read 2 more answers
a train engineer drives a train 235 km north along a straight track . then he drives the train back south 126 km . finally , he
yuradex [85]

let here North direction is along +Y and South direction is along -Y

similarly East direction is towards +X and west direction is towards -X

now it is given that

first it moves 235 km north

d_1 = 235 \hat j

then it moves back 126 km south

d_2 = - 126 \hat j

then again he moves north

d_3 = 45 \hat j

now the total displacement is given as

d = d_1 + d_2 + d_3

d = 235 - 126 + 45 = 154 km

so final displacement of train is 154 km North

5 0
4 years ago
A hockey puck oscillates on a frictionless, horizontal track while attached to a horizontal spring. The puck has mass 0.160 kg a
marshall27 [118]

Explanation:

The given data is as follows.

     mass (m) = 0.160 kg,            spring constant (k) = 8 n/m,

     Maximum speed (v_{m}) = 0.350 m/s

Formula for angular frequency is as follows.

          \omega = \sqrt{\frac{{k}{m}}

    \omega = \sqrt{\frac{{8}{0.160}}

    \omega = 7.07 rad/sec

(a) Formula to calculate the amplitude is as follows.

            \nu_{max} = A \omega

                  A = \frac{\nu}{\omega}

                      = \frac{0.35}{7.07}

                      = 0.05 m

Hence, value of amplitude is 0.05 m.

(b)   Displacement = 0.030 m

Formula for mechanical energy is as follows.

            M.E = \frac{1}{2}kA^{2}

Putting the values into the above formula as follows.

            M.E = \frac{1}{2}kA^{2}

                   = \frac{1}{2} \times 8 \times (0.05)^{2}

                   = 9.8 \times 10^{-3} Joule

For x = 0.03,

As,     P.E = \frac{1}{2} \times kx^{2}

                = \frac{1}{2} \times 8 \times (0.03)

                = 3.6 \times 10^{-3}

Hence, calculate the kinetic energy as follows.

            K.E = M.E - P.E

                  = (9.8 \times 10^{-3} - 3.6 \times 10^{-3}) J

                  = 6.2 \times 10^{-3} J

Thus, we can conclude that kinetic energy of the puck when the displacement of the glider is 0.0300 m is 6.2 \times 10^{-3} J.

7 0
3 years ago
HELP!!
tresset_1 [31]

Answer:

So do 2400 divided by 70. I got 34.285714 and the numbers behind the decimal are repeating. If you round it you get 34.3

3 0
3 years ago
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