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hodyreva [135]
3 years ago
6

In which direction does the sun appear to move across the sky

Physics
2 answers:
blsea [12.9K]3 years ago
6 0

Answer:east

Explanation:Earth rotates or spins toward the east, and that's why the Sun, Moon, planets, and stars all rise in the east and make their way westward across the sky.

bulgar [2K]3 years ago
6 0

Answer:

it rises east and moves west through out the day and sets west

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Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support
alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

r_{M\perp}(F_M)-r_{W\perp}(W)=0

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

r_{M\perp}(F_M)=r_{W\perp}(W)

Now, we divide by the distance of the muscle:

(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}

Now, we substitute the values:

F_M=\frac{(2.4cm)(50N)}{5.1cm}

Now, we solve the operations:

F_M=23.53N

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

\Sigma F_v=F_j-F_M-W

Since there is no vertical movement the sum of vertical forces is zero:

F_j-F_M-W=0

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

F_j=F_M+W

Now, we plug in the values:

F_j=23.53N+50N

Solving the operations:

F_j=73.53N

Therefore, the force is 73.53 Newtons.

8 0
1 year ago
A Transformer has a primary coil with 650 turns if the output voltage is 500 volts and an input voltage of 200 volts how many te
maxonik [38]

Answer:

1625

Explanation:

Ns/Np = Vs/Vp

Ns / 650 = 500/200

Ns = 2.5 x 650

Ns = 1625

6 0
3 years ago
The electric field of a sinusoidal electromagnetic wave obeys the equation E=(375V/m)cos[(1.99×107rad/m)x+(5.97×1015rad/s)t].a.
andrey2020 [161]

Answer:

A) max = Eo = 375 V/m, B)  B = 125 10⁻⁸ T , C) f = 9.50 101⁴ Hz, D) λ = 3.158 10⁻⁷ m, E)  T = 1.05 10⁻¹⁵ s , F) invisible for humans   G)  v = c = 3 10⁸ m/s

Explanation:

The expression given for the electric field is

       E = 375 cos (1.99 107x + 5.97 105t)

The general formula for the electric field of a transverse traveler is

     E = Eo cos (kx-wt)

Where k is the wave number and w the angular velocity

A) The amplitudes as electric is

     Emax = Eo = 375 V / m

B) the electric and magnetic field are related

     E / B = c

     B = E / c

     B = 375/3 108

     B = 125 10⁻⁸ T

C) angular velocity and frequency is related

    .w = 2π f

     f = w / 2π

     f = 5.97 10¹⁵ / 2π

     f = 9.50 101⁴ Hz

D) the speed of light has the formula

      c = λ f

      λ = c / f

      λ = 3 10⁸ / 9.50 10¹⁴

      λ = 3.158 10⁻⁷ m

E) The period

     T = 1 / f

     T = 1 / 9.5 10¹⁴

     T = 1.05 10⁻¹⁵ s

F) let's reduce the wavelength nm

   λ = 3.158 10⁻⁷ m (10⁹nm / 1m)

   λ = 3.158 10²nm = 315.3 nm

The visible radiation range is between 400nm and 700nm. This radiation is ultraviolet and is invisible humans

G) All electromagnetic radiation has a speed at the speed of light (c)

      v = c = 3 10⁸ m/s

5 0
3 years ago
The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le
lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = \sqrt{2gh}

 V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

  = 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

     = 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

    = - 0.329 J

Hence, kinetic energy is not conserved.

8 0
3 years ago
the planet neptune is approximately 4.5*10^9 kilometers from the sun. The planet Venus is approximately 1.1*10^8 kilometers from
Firlakuza [10]

Answer:

Neptune is approximately 41 times as far from the sun as Venus

Explanation:

Estimate = distance of Neptune from the sun ÷ distance of Venus from the sun = 4.5×10^9 ÷ 1.18×10^8 = 40.9 (approximately 41)

7 0
4 years ago
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