The dances created and performed collectively by the ordinary people.

Two blocks are placed at the ends of a horizontal massless board, as in the drawing. The board is kept from rotating and rests o

Andrew [12]

Answer:

The magnitude of the angular acceleration ∝ = $\frac{rxF}{2.8[tex]r^{2}$ }[/tex]

Explanation:

The angular acceleration ∝ is equal to the torque (radius multiplied by force) divided by the mass times the square of the radius. The magnitude of angular acceleration ∝ will have the equation above but we have to replace the mass in the equation by 2.8kg as stated.

7 0

3 years ago

In what setting do all frequencies of electromagnetic waves travel at the same speed?

Irina-Kira [14]

All electromagnetic waves travel in vacuum (space) at the speed of light (3 * 10^8 m/s). Radio waves is just a member of the electromagnetic spectrum. All electromagnetic waves follow the wave equation: speed = frequency * wavelength. With all electromagnetic waves, the speed in space is the same.

6 0

3 years ago

In building a particle accelerator, you manage to produce a uniform electric field of magnitude 6.03 × 10 5 N/C in one 35.5 cm s

mamaluj [8]

Answer:

V = 2.14×10⁵ V.

W = 3.424×10⁻¹⁴ J.

Explanation:

Electric Potential: This can be defined as the work done in bringing a unit positive charge from infinity to that point, against the action of a field.

The S.I unit is V.

The expression containing electric potential, distance and electric field is given as,

V = E×r .............. Equation 1

Where V = Electric potential difference across the length of the accelerator's section, E = Electric Field, r = Length of the section.

Given: E = 6.03×10⁵ N/C, r = 35.5 cm = 0.355 m.

Substitute into equation 1

V = 6.03×10⁵×0.355

V = 2.14065×10⁵ V.

V ≈ 2.14×10⁵ V.

amount of Work required to move a proton through the section is given as,

W = qV ............... Equation 2

Where W = work required to move a proton through the section, q = charge on a proton V = Electric potential.

Given: V = 2.14×10⁵ V, q = 1.60 x 10⁻¹⁹ C.

Substitute into equation 2

W = (2.14×10⁵)(1.60 x 10⁻¹⁹)

W = 3.424×10⁻¹⁴ J.

6 0

3 years ago

5.

Kamila [148]

Answer: (x - 4)² + (y - 7)² = 9

<u>Explanation:</u>

The equation of a circle is: (x - h)² + (y - k)² = r² where

(h, k) is the center
r is the radius

Given: (h, k) = (4, 7)

Find the intersection of the given equation and the perpendicular passing through (4, 7).

3x - 4y = -1

-4y = -3x - 1

$y=\dfrac{3}{4}x-1$

$m=\dfrac{3}{4}$ --> $m_{\perp}=-\dfrac{4}{3}$

$y-y_1=m_{\perp}(x-x_1)\\\\y-7=-\dfrac{4}{3}(x-4)\\\\\\y=-\dfrac{4}{3}x+\dfrac{16}{3}+7\\\\\\y=-\dfrac{4}{3}x+\dfrac{37}{3}$

Use substitution to find the point of intersection:

$x=\dfrac{29}{5}=5.8,\qquad y=\dfrac{23}{5}=4.6$

Use the distance formula to find the distance from (4, 7) to (5.8, 4.6) = radius

$r=\sqrt{(5.8-4)^2+(4.6-7)^2}\\\\r=\sqrt{3.24+5.76}\\\\r=\sqrt9\\\\r=3$

Input h = 4, k = 7, and r = 3 into the circle equation:

(x - 4)² + (y - 7)² = 3²

(x - 4)² + (y - 7)² = 9

4 0

3 years ago

The VLBA (Very Long Baseline Array) uses a number of individual radio telescopes to make one unit having an equivalent diameter

Alex787 [66]

Answer:

d₂=232.8 m

Explanation:

We know that formula for resolution given as

θres = 1.22 λ/d

Where

λ=wavelength

d=diameter

Given that resolution should be same.So we can say that

1.22 λ₁/d₁= 1.22 λ₂/d₂

λ₁/d₁= λ₂/d₂

$d_2=\dfrac{\lambda _2}{\lambda _1}d_1$

Given that

λ₂=582 nm

λ₁=2 cm

d₁=8000 km

$d_2=\dfrac{582\times 10^{-9}\times 8000\times10^3}{0.02}\ m$

d₂=232.8 m

8 0

3 years ago