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VikaD [51]
3 years ago
7

If you see a rainbow near the time of sunset, where in the sky will the rainbow be?

Physics
1 answer:
Igoryamba3 years ago
8 0
Closest to the sun f thats a choice or closest to the atmoshpere
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A single-turn square loop carries a current of 16 A . The loop is 15 cm on a side and has a mass of 3.8×10^−2kg . Initially the
DiKsa [7]

Answer:

The minimum magnetic field is 0.078 T.

Explanation:

Given that,

Current = 16 A

Side = 15 cm

Mass m= 3.8\times10^{-2}\ kg

Mass each segment in given square loop is

m=\dfrac{3.8\times10^{-2}}{4}

We need to calculate the torque due to gravity

Using formula of torque

\tau_{g}=2mg(\dfrac{L}{2})+mgL

\tau_{g}=2mgL

The torque due to magnetic field

\tau_{B}=FL

\tau_{B}=BIL^2

The equilibrium condition

\tau_{B}=\tau_{g}

Put the value into the formula

BIL^2=2mgL

B=\dfrac{2mgL}{IL^2}

B=\dfrac{2mg}{IL}

Put the value into the formula

B=\dfrac{2\times\dfrac{3.8\times10^{-2}}{4}\times9.8}{16\times15\times10^{-2}}

B=0.078\ T

Hence, The minimum magnetic field is 0.078 T.

7 0
3 years ago
Graphing Motion
KIM [24]
The answer will be D
3 0
3 years ago
80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting te
vodka [1.7K]

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C

Therefore, the resulting temperature is 23.37 ⁰C

6 0
3 years ago
Which statement is true about a planet’s orbital motion?
lana66690 [7]

Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

                                    V = \sqrt{\frac{GM}{R + h} }

Where

                   V - velocity of the orbit at a height h from the surface

                    R - Radius of the second object

                    G - Gravitational constant

                    h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

                         1/2 mV^{2} = GMm/R

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

3 0
3 years ago
A ceiling fan has three blades. The moment of inertia of a blade is 0.2kgm^2. The net torque exerted on fan blades is 8Nm. Find
olchik [2.2K]

Answer:

(A) the angular acceleration of the blades is 13.33 m/s.

Explanation:

Given;

moment of inertia of a blade, I = 0.2 kgm²

net torque exerted on fan blades, ∑τ =  8Nm

Torque is given as product of moment of inertia and angular acceleration;

τ = Iα

where;

α is the angular acceleration

Since there are three blades of the ceiling fan, the net torque is given as;

∑τ = (3I)α

∑τ = 3Iα

α = ∑τ / 3I

α = (8) / (3 x 0.2)

α = 13.33 m/s

Therefore, the angular acceleration of the blades is 13.33 m/s.

8 0
2 years ago
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