A 4.2 kg parachutist is moving straight downward with a speed of 3.85 m/s

Mashcka [7]

Answer:

3. When the number of turns, N is doubled, the strength of the electromagnet is also doubled

4. Doubling the voltage, doubles the strength of the electromagnet

5. The number of paper clips a 7.5 V battery would pick is approximately 28 paper clips

The number of paper clips a 7.5 V battery would pick is 59 paperclips

6. For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is approximately 7 paperclips

For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is 16 paperclips

Explanation:

3. The Magnetomotive Force, MMF = The Number of Turns on the Coil, N × The Current I Flowing in the Coil, I

∴ MMF = N × I

When the number of turns, N is doubled, the magnetomotive force, MMF is also doubled, and the strength of the electromagnet is doubled

4. Given that the voltage, V applied to the coil = The current, I flowing × The resistance, R of the coil, we have

V = I × R

Therefore, for a given constant resistance in the coil, doubling the voltage, doubles the current and therefore doubles the strength of the electromagnet

5. The average slope for the 25-coil electromagnet = (23 - 12)/(6 - 3) = 3. $\bar 6$

The number of paper clips a 7.5 V battery would pick = 12 + (7.5 - 3) × 11/3 = 28.5 paperclips ≈ 28 paper clips

The average slope for the 50-coil electromagnet = (48 - 26)/(6 - 3) = 7. $\bar 3$

The number of paper clips a 7.5 V battery would pick = 26 + (7.5 - 3) × 22/3 = 59 paperclips

6. The slope calculated from a start point of approximately 0.4 V, is given as follows;

The slope for the 25-coil electromagnet = (12 - 6)/(3 - 0.4) = 30/13

Therefore, for the 25-coil electromagnet, the average number of paper clips a 1 V battery would pick = 6 + (1 - 0.4) × 30/13) = 96/13 ≈ 7 paperclips

The slope for the 50-coil electromagnet = (26 - 13)/(3 - 0.4) = 5

Therefore, for the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick = 13 + (1 - 0.4) × 5 = 16 paperclips

8 0

3 years ago

Read 2 more answers

A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it

Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface $\Phi$ it is possible to calculate the magnitude $\frac{\Phi}{A}=B\ cos \theta$ .

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux $\Phi$ is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is ( $m^{2}$ ). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with $\theta$ being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

$\Phi=B\ A\ cos\theta$

We are told the values of $\Phi$ and B, then we can calculate the magnitude

$\frac{\Phi}{A}=B\ cos\theta$

3 0

3 years ago

What are some possible materials you could use to make your battery? DON’T FORGET TO include information about why it is importa

Dima020 [189]

Explanation:

Commercially available batteries use a variety of metals and electrolytes. Anodes can be made of zinc, aluminum, lithium, cadmium, iron, metallic lead, lanthanide, or graphite. Cathodes can be made of manganese dioxide, mercuric oxide, nickel oxyhydroxide, lead dioxide or lithium oxide. Potassium hydroxide is the electrolyte used in most battery types, but some batteries use ammonium or zinc chloride, thionyl chloride, sulfuric acid or lithiated metal oxides. The exact combination varies by battery type. For example, common single-use alkaline batteries use a zinc anode, a manganese dioxide cathode, and potassium hydroxide as the electrolyt

4 0

3 years ago

An object is moving north with an initial velocity of 14 m/s accelerates 5m/s for 20 seconds. What is the final velocity of the

olga2289 [7]

Use the kinematic equation: Vf=Vi+at

Then plug;

Vi=14 m/s

a=5 m/s²

t=20 s. Therefore;

Vf=14+(5*20)

Vf=114 m/s.

6 0

3 years ago

a 5.00 kg plane is speeding up on the ground with an applied force of 706 N. If the net force is 450 N, what would be the force

loris [4]

Answer:

The force due to air resistance is 256 N.

Explanation:

Given;

mass of the plane, m = 5 kg

applied force on the plane, Fa = 706 N

the net force on the plane, ∑F= 450 N

Let the force due to air resistance = Fr

The net force on the plane is given as;

Net force = applied force - force due to air resistance

∑F = Fa - Fr

Fr = Fa - ∑F

Fr = 706 - 450

Fr = 256 N.

Therefore, the force due to air resistance is 256 N.

5 0

3 years ago