Sports car goes from a velocity of zero to a velocity of 12 m/s East in two seconds. What is the cars acceleration?

Two identical circular, wire loops 35.0 cm in diameter each carry a current of 2.80 A in the same direction. These loops are par

defon

Answer:

The answer is " $4659.2 \times 10^{-24} \ N$ "

Explanation:

The magnetic field at ehe mid point of the coils is,

$\to B=\frac{\mu_0 i R^2}{(R^2+x^2)^{\frac{3}{2}}}\\\\$

Here, i is the current through the loop, R is the radius of the loop and x is the distance of the midpoint from the loop.

$\to B=\frac{(4\pi\times 10^{-7})(2.80\ A) (\frac{0.35}{2})^2}{( (\frac{0.35}{2})^2+ (\frac{0.24}{2})^2)^{\frac{3}{2}}}\\\\$

$=\frac{(12.56 \times 10^{-7})(2.80\ A) \times 0.030625}{( 0.030625+ 0.0144)^{\frac{3}{2}}}\\\\=\frac{ 1.07702 \times 10^{-7} }{0.0095538976}\\\\=112.730955 \times 10^{-7}\\\\=1.12\times 10^{-5}\ \ T\\$

Calculating the force experienced through the protons:

$F=qvB=(1.6 \times 10^{-19}) (2600)(1.12 \times 10^{-5})= 4659.2 \times 10^{-24}\ N$

7 0

3 years ago

A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed

Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

$\int EdS=\frac{Q_{in}}{\epsilon_o}$ (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

$\int EdS=ES=E(4\pi r^2)$ (2)

Qin is calculate by using the charge density:

$Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho$ (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

$\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}$

Next, you use the results of (3), (2) and (1):

$E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})$

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

$E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}$

hence, the electric field is 1580594.95 N/C

7 0

3 years ago

Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.55 μC charge and flies due west at a speed of

Illusion [34]

Answer:

3.014 x 10⁻⁸ N

Explanation:

q = magnitude of charge on the supersonic jet = 0.55 μC = 0.55 x 10⁻⁶ C

v = speed of the jet = 685 m/s

B = magnitude of magnetic field in the region = 8 x 10⁻⁵ T

θ = angle between the magnetic field and direction of motion = 90

magnitude of the magnetic force is given as

F = q v B Sinθ

F = (0.55 x 10⁻⁶) (685) (8 x 10⁻⁵) Sin90

F = 3.014 x 10⁻⁸ N

6 0

3 years ago

A bird is standing on an electric transmission line carrying 3000 A of current. A wire like this has about 3.0 x 10-5 22 of resi

Darya [45]

Answer:

13.5 x 10^-9 A

Explanation:

Yes

6 0

3 years ago

Find :

Pepsi [2]

Answer:

Explanation:

b) Gravity reduces the initial upward velocity to zero in a time of

t = v/g = 40/10 = 4 s

a) h = v₀t + ½gt² = 40(4) + ½(-10)4² = 80 m

or

v² = u² + 2as

h = (0² - 40²) / 2(-10) = 80 m

8 0

2 years ago

Sports car goes from a velocity of zero to a velocity of 12 m/s East in two seconds. What is the cars acceleration￼?

Sports car goes from a velocity of zero to a velocity of 12 m/s East in two seconds. What is the cars acceleration?