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Yuki888 [10]
3 years ago
14

In part B of the lab, when the current flows through the orange part of the wire from right to left, the wire deflects (or moves

) ____. This is in accordance with the right-hand-rule.
Physics
1 answer:
s2008m [1.1K]3 years ago
7 0

Answer:

This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.

We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:

F = L*(IxB)

if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:

I = i*(1, 0, 0)

And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.

To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.

Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.

For example, if the magnetic field is in the positive z-axis, we will point upwards.

Now the palm of our hand tells us in which direction the force is applied.

This is the right-hand rule.

For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.

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Mark and David are loading identical cement blocks onto David’s pickup truck. Mark lifts his block straight up from the ground t
Pepsi [2]

Answer:

b) true. The jobs are equal

Explanation:

The work on a body is the scalar product of the force applied by the distance traveled.

    W = F. d

Work is a scalar, the work equation can be developed

    W = F d cos θ

Where θ is the angle between force and displacement

Let's apply these conditions to the exercise

a) False, if we see the expression d cosT is the projection of the displacement in the direction of the force, so there may be several displacement, but its projection is always the same

b) true. The jobs are equal dx = d cosθ

c) False, because the force is equal and the projection of displacement is the same

d) False, knowledge of T is not necessary because the projection of displacement is always the same

e) False mass is not in the definition of work

5 0
3 years ago
A small metal sphere has a mass of 0.19 gg and a charge of -23.0 nCnC. It is 10.0 cmcm directly above an identical sphere that h
RoseWind [281]

Answer:

  a = -7.29 m / s²

Explanation:

For this exercise we must use Newton's second law,

          F -W = m a

Force is electrical force

         F = k q₁ q₂ / r²

         k q₁ q₂ / r² -mg = m a

indicate that the charge of the two spheres is equal

         q₁ = q₂ = q

         a = (k q² / r² - m g) / m

         a = k q² / m r² - g

Let's reduce the magnitudes to the SI system

        m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg

        q1 = q2 = q = -23.0 nC (1C / 10⁹ nC) = -23.0 10⁻⁹ C

        r = 10.0 cm (1m / 100cm) = 0.1000 m

let's calculate

        a = 9 10⁹ (23.0 10⁻⁹)² / (0.1000² 1.9 10⁻⁴) - 9.8

        a = -7.29 m / s²

The negative sign indicates that the direction of this acceleration is downward

6 0
3 years ago
The pilot of an aircraft wishes to fly due west in a 33.9 km/h wind blowing toward the south. The speed of the aircraft in the a
diamong [38]

Answer:\theta =9.96^{\circ} North of west

Explanation:

Given

Plane wishes to fly in west

but wind with speed 33.9 km/h towards south obstructing its path

so plane must fly at an angle of \theta w.r.t west such that it final velocity is towards west

Plane absolute speed=195 km/h

To fly towards west velocity in Y direction should be zero

thus 195sin\theta =33.9

\theta =9.96^{\circ}

so Plane should head towards 9.96^{\circ} North of west in order to fly in west.

So plane

actual velocity is

v=-195cos9.96\hat{i}+195sin9.96\hat{j}

5 0
3 years ago
A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

5 0
3 years ago
un litro de un gas es calentado a presión constante desde 20°C hasta 60°C que volumen final ocupará dicho gas? ​
WINSTONCH [101]

Answer:

Final volume, V2 = 3 Litres

Explanation:

Given the following data;

Initial volume, V1 = 1 litre

Initial temperature, T1 = 20°C

Final temperature, T2 = 60°C

To find the final volume, we would use Charles' law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

V1/T1 = V2/T2

Making V2 as the subject formula, we have;

V1T2 = V2T1

V2 = (V1T2)/T1

Substituting into the formula, we have;

V2 = (1 * 60)/20

V2 = 60/20

Final volume, V2 = 3 Litres

4 0
3 years ago
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