All categories

3 years ago

What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?

Physics

1 answer:

katrin [286]3 years ago

4 0

Answer:

q = 8.85 x 10⁻¹¹ C

Explanation:

given,

Electric field, E = 1.18 N/C

distance, r = 0.822 m

Charge magnitude = ?

using formula of electric field.

$E = \dfrac{kq}{r^2}$

k is the coulomb constant

$q= \dfrac{Er^2}{k}$

$q= \dfrac{1.18\times 0.822^2}{9\times 10^9}$

q = 8.85 x 10⁻¹¹ C

The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C

You might be interested in

A 800hz tuning fork is vibrating, producing a sound wave in the air.

blsea [12.9K]

Answer:

The frequency of the sound wave is 800Hz

The speed of sound in a is about 340m/s.

Velocity = frequency x wavelength

making wavelength the subject formula

wavelength = Velocity/frequency.

wavelength = 340/800

wavelength = 0.425m.

3 0

2 years ago

If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy

ruslelena [56]

The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

<em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

$T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\ \times \ years}{3600 \ s \ \times\ 24\ h\ \times \ 365.25 \ days} \\\\T = 1.397 \times 10^{10} \ years\\\\T = 13.97 \ billion \ years$

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0

2 years ago

An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe

kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0

3 years ago

Describe a good level of body fat

m_a_m_a [10]

Answer: a good level of body fat can be found using your weight and height as a reference.

Explanation:

6 0

3 years ago

A campground official wants to measure the distance across an irregularly-shaped lake, but can't do so directly. He picks a poin

sleet_krkn [62]

Answer:

149 m

Explanation:

The distances across the lake is forming a triangle.

let the distance between the point and the left side be 'x'

and the distance between the point and the right be 'y'

and the distance across the lake be 'z' and the angle opposite to 'z' be 'Z' given:

∠Z = 83°

x = 105 m

y = 119 m

Now, applying the Law of Cosines, we get

z² = x² + y² - 2xycos(Z)

Substituting the values in the above equation, we get

z² = 105² + 119² - 2×105×119×cos(83°)

z = √22140.48

z = 148.796 m ≈ 149 m

The point is 149 m across the lake

8 0

3 years ago