What is the velocity of the 100 kg cart at point b?

A round flat metal coil has 180 turns and negligible resistance. It is connected in a series circuit with a 17-Ω resistor, with

evablogger [386]

Answer: C) 0.25 m

Explanation: In order to explain this problem we have to consider the Faraday law, we have:

ε=-dФ/dt where ε the emf induced by the change of the magnetic field given by dФ/dt.

then ε=I*R=17*6=102 V

We have a coil then we have the magnetic flux as follow:

Ф=N*A*B then we have

dФ/dt= N*A*dB/dt where A and N is the area and number of turn of the coil.

A=π*R^2 where R is the radius of teh coil.

Finally we have;

dФ/dt= N*π*R^2*dB/dt then

R= [dФ/dt/(N*π*dB/dt)]^1/2= [102/(180*π*3)]1/2=245.2*10^-3=≅0.25m

6 0

4 years ago

There are 100 cm in every meter. If a measurement is 1.67 m, what is the measurement expressed in centimeters?

timofeeve [1]

100 cm is in every meter means that the conversion from meter to centimeters is multiplying by 100. So you would multiply 1.67 by 100, which is 167cm.

8 0

3 years ago

Earths gravitational pull just got 3 times stronger what happens to your weight?

djverab [1.8K]

You get 3 times heavier

6 0

4 years ago

Read 2 more answers

tom does not reallt want to give away blue marbles and would like to change the probability that he chooses a blue marble to one

Llana [10]

We don't know how many of ANY color are in the bag right now, so there's no way to calculate an answer.

What Tom has to do is make sure that the number of marbles that are NOT blue is NINE TIMES the number of blue ones in the bag.

4 0

3 years ago

There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

4 0

3 years ago