Which is NOT a characteristic of a solid?

Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way

VladimirAG [237]

Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E

The electric field between the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

3 0

3 years ago

At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is

MArishka [77]

Answer:

A) F_g = 8.9278 × 10^(-30)

B) F_e = 1.6 × 10^(-17) N upwards

C) F_m = 6.8 × 10^(-17) N downwards

Explanation:

A) Formula for gravitational force is;

F_g = m_e × g

m_e is mass of electron = 9.11 x 10^(-31) kg

F_g = 9.11 x 10^(-31) × 9.8

F_g = 8.9278 × 10^(-30) N downwards

B) Formula for Electric force is;

F_e = qe

q is charge on electron = 1.6 × 10^(-19) C

E is electric field = 100 N/C

F_e = 1.6 × 10^(-19) × 100

F_e = 1.6 × 10^(-17) N upwards

C) Magnetic force is given by the formula;

F_m = qVB

q is charge on electron = 1.6 × 10^(-19) C

V is velocity given as 8.50 × 10^(6) m/s

B is magnetic field = 50.0 μT = 50 × 10^(-6) T

F_m = 1.6 × 10^(-19) × 8.50 × 10^(6) × 50 × 10^(-6)

F_m = 6.8 × 10^(-17) N downwards

6 0

3 years ago

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.0 mm.

pantera1 [17]

Answer:

R_a/R_b=3

Explanation:

The resistance in terms of the area and the length of the wire is given by:

R=pL/A

if we have two wires, the first one is a solid wire with a diameter of d_A = 1 x 10^-3 m, and the second one is a hollow wire with inner diameter of d_B,i = 1 x 10^-3 m and outer diameter of d_B,σ= 2 x 10^-3 m, so the cross sectional area of the first wire is:

A_a=πr^2_a

A_a=πd^2_A/4

hence the resistance is:

R_a=(4*p*L_a)/π*d^2_A (1)

the area of the second wire is:

A_b=π*r^2_B,σ-π*r^2_B,i

A_b=π/4(d^2_B,σ-d^2_B,i)

hence the resistance is:

R_b=(4*p*L_b)/π(d^2_B,σ-d^2_B,i) (2)

To find the ratio between the resistances R_a/R_b, we divide (1) over (2) to get:

R_a/R_b=(d^2_B,σ-d^2_B,i)*L_a/(d^2_a*L_b)

but the wires have the same length, therefore:

R_a/R_b=(d^2_B,σ-d^2_B,i)/(d^2_a)

substitute with the given values to get:

R_a/R_b=3

6 0

3 years ago

Read 2 more answers

Consider two laboratory carts of different masses but identical kinetic energies and the three following statements. I. The one

kolbaska11 [484]

Answer:

d) I and III only.

Explanation:

Let be $m_{1}$ and $m_{2}$ the masses of the two laboratory carts and let suppose that $m_{1} > m_{2}$ . The expressions for each kinetic energy are, respectively:

$K = \frac{1}{2}\cdot m_{1}\cdot v_{1}^{2}$ and $K = \frac{1}{2}\cdot m_{2}\cdot v_{2}^{2}$ .

After some algebraic manipulation, the following relation is constructed:

$\frac{m_{1}}{m_{2}} = \left(\frac{v_{2}}{v_{1}}\right)^{2}$

Since $\frac{m_{1}}{m_{2}} > 1$ , then $\frac{v_{2}}{v_{1}} > 1$ . That is to say, $v_{1} < v_{2}$ .

The expressions for each linear momentum are, respectively:

$p_{1} = \frac{2\cdot K}{v_{1}} = m_{1}\cdot v_{1}$ and $p_{2} = \frac{2\cdot K}{v_{2}} = m_{2}\cdot v_{2}$

Since $v_{1} < v_{2}$ , then $p_{1} > p_{2}$ . Which proves that statement I is true.

According to the Impulse Theorem, the impulse needed by cart I is greater than impulse needed by cart II, which proves that statement II is false.

According to the Work-Energy Theorem, both carts need the same amount of work to stop them. Which proves that statement III is true.

6 0

3 years ago

What mathematical formula correctly expresses the relationship between force, mass and acceleration? A. m = F + a B. F = m / a C

IRINA_888 [86]

D. F=ma

F is for force, and that equals two things, M for mass and A for acceleration. When mass is accelerated, it gives you force. Force equals multiplying mass and its acceleration.

3 0

3 years ago