The symbol for the charge on a proton is:

The magnetic field at point P due to a 2.0-A current flowing in a long, straight, thin wire is 8.0 μT. How far is point P from t

Tanzania [10]

Answer:

r = 0.05 m = 5 cm

Explanation:

Applying ampere's law to the wire, we get:

$B = \frac{\mu_oI}{2\pi r}\\\\r = \frac{\mu_oI}{2\pi B}$

where,

r = distance of point P from wire = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = 2 A

B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T

Therefore,

$r = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(2\ A)}{2\pi(8\ x\ 10^{-6}\ T)}\\\\$

r = 0.05 m = 5 cm

8 0

2 years ago

A car slows down from 27.7 m/s to 10.9 m/s in 2.37 s. What is its acceleration?

Sunny_sXe [5.5K]

Answer:

- 7.088 m/s²

Explanation:

As we know that,

★ Acceleration = Change in velocity/Time

→ a = (v - u)/t

Here,

Initial velocity (u) = 27.7 m/s
Final velocity (v) = 10.9 m/s

→ a = (10.9 m/s - 27.7 m/s)/2.37 s

→ a = -16.8/2.37 m/s²

→ a = -7.088 m/s² [Answer]

Negative sign denotes that the velocity is decreasing.

4 0

2 years ago

Read 2 more answers

A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se

Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

$v = \frac{Bqr}{m}$

$v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}$

v = 661098.9 = 661099 m/s

(b)

$\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}$

e / m = 1.76 x 10^14 C / kg

(c) Let K be the kinetic energy

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

(d) if the acceleration voltage is doubled

V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

$r=\frac{mv}{Bq}$

$r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}$

r = 0.0113 m = 1.13 cm

7 0

3 years ago

Add the following displacement vectors: 12 m south and 15 m 55° E of N??

daser333 [38]

12m S=0m E, -12m N
15m 55d E of N = 15 sin 55, 15 cos 55 N
Sum= (15sin55)m E, (-12 + 15 cos 55)m N

4 0

2 years ago

Help pleasee

agasfer [191]

Answer:

i/f = i/o + i/i f = focal, o = object, i = image

1 / i = 1 / f - 1 / o = (o - f) / o f

i = o * f / ( o - f) image distance

i = 12.5 * 22 / (12.5 - 22) = -28.9 cm

Image is real

Image is 28.9 cm to left of lens

M = - i / o = = 28.9 / 12.5 = 2.3 magnification (convex lens)

8 0

2 years ago