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Pavel [41]
2 years ago
7

Astronomers determine that a certain square region in interstellar space has an area of approximately 2.4 \times 10^72.4×10 ​7 ​

​ (light-years)2, where a light-year is a unit of astronomical distance. Assuming that this region is a square, then a single side of the square would be approximately how many light-years long?
Physics
1 answer:
mafiozo [28]2 years ago
5 0

Answer:

1.5 × 10³⁶ light-years

Explanation:

A certain square region in interstellar space has an area of approximately 2.4 × 10⁷² ​​ (light-years)². The area of a square can be calculated using the following expression.

A = l²

where,

A is the area of the square

l is the side of the square

l = √A = √2.4 × 10⁷² ​​ (light-years)² = 1.5 × 10³⁶ light-years

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A dolphin is able to tell in the dark that the ultrasound echoes received from two sharks come from two different objects only i
Vlad1618 [11]

Answer:

a) Wavelength of the ultrasound wave = 0.0143 m <<< 3.5m, hence its ability is not limited by the ultrasound's wavelength.

b) Minimum time difference between the oscillations = Period of oscillation = 0.00952 ms

Explanation:

The frequency of the ultrasound wave = 105 KHz = 105000 Hz. The speed of ultrasound waves in water ≈ 1500 m/s. Wavelength = ?

v = fλ

λ = v/f = 1500/105000 = 0.0143 m <<< 3.5m

This value, 0.0143m is way less than the 3.5m presented in the question, hence, this ability is not limited by the ultrasound's wavelength.

b) Minimum time difference between the oscillations = The period of oscillation = 1/f = 1/105000 = 0.00000952s = 0.00952 ms

Hope this helps!

6 0
2 years ago
A floating leaf oscillates up and down two complete cycles in one second as a water wave passes by. The wave's wavelength is 10
postnew [5]

Answer:

C) 20 m/s

Explanation:

Wave: A wave is a disturbance that travels through a medium and transfers energy from one point to another, without causing any permanent displacement of the medium itself. Examples of wave are, water wave, sound wave, light rays, radio waves. etc.

The velocity of a moving wave is

v = λf ............................ Equation 1

Where v = speed of the wave, λ = wave length, f = frequency of the wave.

Given: f = 2 Hz (two complete cycles in one seconds), λ = 10 meters

Substituting these values into equation 1

v = 2×10

v = 20 m/s.

Thus the speed of the wave = 20 m/s

The right option is C) 20 m/s

7 0
2 years ago
A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle ϕ1 with
strojnjashka [21]

Answer:

x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}

Explanation:

Let 'F₁'  and 'F₂' be the forces applied by left and right wires on the bar as shown in the diagram below.

Now, the horizontal and vertical components of these forces are:

F_{1x} = -F_1cos(\phi_1)\\F_{1y}=F_1sin(\phi_1)\\\\F_{2x}=F_2cos(\phi_2)\\F_{2y}=F_2sin(\phi_2)

As the system is in equilibrium, the net force in x and y directions is 0 and net torque about any point is also 0. Therefore,

\sum F_x=0\\F_{1x}=F_{2x}\\F_1cos(\phi_1)=F_2cos(\phi_2)\\\frac{F_1}{F_2}=\frac{cos(\phi_2)}{cos(\phi_1)}-------1

Now, let us find the net torque about a point 'P' that is just above the center of mass at the upper edge of the bar.

At point 'P', there are no torques exerted by the F₁x and F₂x nor the weight of the bar as they all lie along the axis of rotation.

Therefore, the net torque by the forces F_{1y}\ and\ F_{2y} will be zero. This gives,

-F_{1y}\times x + F_{2y}(L-x) = 0\\F_{1y}\times x=F_{2y}(L-x)\\x=\frac{F_{2y}(L-x)}{F_{1y}}

But, F_{1y}=F_1sin(\phi_1)\ and\ F_{2y}=F_2sin(\phi_2)

Therefore,

x=\frac{F_2sin(\phi_2)(L-x)}{F_1sin(\phi_1)}\\\textrm{From equation (1),}\frac{F_2}{F_1}=\frac{cos(\phi_1)}{cos(\phi_2)}\\\therefore x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times (L-x)\\x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times L-\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times x\\\\

x(1+\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L\\x(1+\frac{cos(\phi_1)}{sin(\phi_1)}\times \frac{sin(\phi_2}{cos(\phi_2)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L

We know,

tan(\phi)=\frac{sin(\phi)}{cos(\phi)}\\\\cot(\phi)=\frac{cos(\phi)}{sin(\phi)}

∴x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}

6 0
1 year ago
A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision
TEA [102]

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

0 = m(3.5 sin30) - mv_y

v_y = 1.75 m/s

Now similarly in X direction we will have

m(4.5) = m(3.5 cos30 ) + mv_x

v_x = 1.47 m/s

now final kinetic energy of both puck after collision is given as

KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)

KE_f = 8.73 m

initial kinetic energy of both pucks is given as

KE_i = \frac{1}{2}m(4.5^2) + 0

KE_i = 10.125 m

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

5 0
2 years ago
According to Newton’s law of universal gravitation, which statements are true?
Arte-miy333 [17]

Answer: The statement first and the fourth statement are true.

Explanation:

According to Newton's gravitational law, every particle in the universe attracts every other particle with the force of attraction between the masses is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

As we move to higher altitude, the force of gravity on use decreases because the force of gravity is inversely proportional to the distance.

If the masses of the two objects are more then there will be greater force of gravity between them.

Therefore, the statement first and the fourth statement are true.

7 0
2 years ago
Read 2 more answers
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