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NNADVOKAT [17]
3 years ago
13

Which statement is not true about radiation

Physics
1 answer:
Mekhanik [1.2K]3 years ago
4 0
I don’t know i’m sorry
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A car has a speed of 12m/s. The mass of the car and its passengers is 1250 kg. What is the total momentum of the car/passengers?
Temka [501]

Answer:

15000kgm/s

Explanation:

Given parameters:

Speed of the car  = 12m/s

Mass of car and passengers = 1250Kg

Unknown:

Momentum of the car  = ?

Solution:

Momentum is the quantity of motion a body posses;

  Momentum  = mass x velocity

Now insert the given parameters and solve;

  Momentum  = 12 x 1250  = 15000kgm/s

5 0
3 years ago
The Solar System extends to a distance of at least 100,000 A.U., which is the same as saying...
Mariulka [41]

Answer:

a. It extends to about 100,000 times the distance between the Earth and the Sun.

d. Stars are born from clouds of gas and dust, and when they die they return gas and dust to the galaxy from which new stars can be born.

b. Kind of interesting

Explanation:

  • The average distance between the sun and the earth is called one astronomical unit.
  • It is the unit of length to measure the distance in our solar system.
  • It is equal to 1.49 x 10¹¹ m
  • It is analogous to the light-year to measure the distance of distant stars.
  • So according to this unit, our solar system extends about 100,000 times the distance between the Earth and the Sun.
5 0
4 years ago
The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the
Evgen [1.6K]

Question Continuation

Derive an expression for x in terms of m, M, and D. b) If the net force is zero a distance ⅔D from the planet, what is the ratio R of the mass of the planet to the mass of the moon, M/m?

Answer:

a. x = (D√M/m)/(√M/m + 1)

b. The ratio R of the mass of the planet to the mass of the moon=4:1

Explanation:

Given

m = Mass of moon

M = Mass of the planet

D = Distance between the centre of the planet and the moon

Net force = 0

Let Y be a point at distance x from the planet

Let mo = mass at point Y

a.

Derive an expression for x in terms of m, M and D.

Formula for Gravitational Force is

F = Gm1m2/r²

Y = D - x

Force on rest mass due to mass M (FM) =Force applied on rest mass due to m (Fm)

FM = G * mo * M/x²

Fm = G * mo * m/Y²

Fm = G * mo * m/(D - x)²

FM = Fm = 0 ------ from the question

So,

G * mo * M/x² = G * mo * m/(D - x)² ----- divide both sides by G * mo

M/x² = m/(D - x)² --- Cross Multiply

M * (D - x)² = m * x²

M/m = x²/(D - x)² ---_ Find square roots of both sides

√(M/m) = x/(D - x) ----- Multiply both sides by (D - x)

(D - x)√(M/m) = x

D√(M/m) - x√(M/m) = x

D√(M/m) = x√(M/m) + x

D√(M/m) = x(√(M/m) + 1) ------- Divide both sides by √M/m + 1

x = (D√M/m)/(√M/m + 1)

b. Here x = ⅔D

FM = G * mo * M/x²

Fm = G * mo * m/(D - x)²

FM = Fm

G * mo * M/x² = G * mo * m/(D - x)² ----- divide both sides by G * mo

M/x² = m/(D - x)² --- (Substitute ⅔D for x)

M/(⅔D)² = m/(D - ⅔D)²

M/(4D/9) = m/(⅓D)²

9M/4D = m/(D/9)

9M/4D = 9m/D ---- Divide both side by 9/D

M/4 = m

M = 4m

M/m = 4

M:m = 4:1

So, the ratio R of the mass of the planet to the mass of the moon=4:1

3 0
3 years ago
Group 7A of the periodic table contains the?
olga2289 [7]

The correct answer is A. Most reactive non metals.

Firstly if some one knows how to read the periodic table he would have no confusion in deciding whether group 7 has metals or non metals. Group 7 contains non metals so basically we can easily cancel out two options of metals.

Secondly group 7 non metals are the most reactive non metals as they need only one electron in order to complete their valence shell and become stable.


4 0
3 years ago
Read 2 more answers
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
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