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Ostrovityanka [42]
3 years ago
10

What is the magnitude of the electric field at a distance 60 cm from the center of the sphere? The radius of the sphere 30 cm, t

he charge on the sphere is 1.56 × 10−5 C and the permittivity of a vacuum is 8.8542 × 10−12 C 2 /N · m2 . Answer in units of N/C.

Physics
2 answers:
Serga [27]3 years ago
5 0

Explanation:

Below is an attachment containing the solution.

Nastasia [14]3 years ago
3 0

Answer

3.9*10^{5}N/C

Explanation:

from the expression for determing the magnetic field strength

E=\frac{q}{4\pi e_{0}d^2 }\\

since the charge is given as

q=1.56*10^{-5}c\\

and the distance is

d=60cm=0.6m

We can calculate the constant k

K=\frac{1}{4\pi e_{0}}\\ K=\frac{1}{4\pi *8.8542*10^{-12}}\\k=8.98*10^{9}\\

if we substitute values, we arrive at

E=\frac{8.98*10^9 *1.56*10^{-5}}{0.6^2} \\E=389133.33\\E=3.9*10^{5}N/C

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Mekhanik [1.2K]
The question is a bit ambiguous, but sound can be heard at different frequencies and amplitudes.
5 0
2 years ago
Read 2 more answers
An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When
sleet_krkn [62]

Answer:

2dQ²/2εA = 2U₀

Explanation:

The energy stored in the ideal parallel plate capacitor U = Q²/2C. For a parallel plate capacitor, C = εA/d where A is the area between the plates and D is the distance between them. So

U = Q²/2C = Q²/2εA/d = dQ²/2εA. At distance d₁ = d, U = U₀,

U₀ = dQ²/2εA,

When d₂ = 2d, U₁ = d₂Q²/2εA = 2dQ²/2εA = 2U₀

3 0
3 years ago
An airplane is flying in a horizontal circle at a speed of 480 km/h (). If its wings are tilted at angle =40° to the horizontal
german

Answer:

R = 2162 m

Explanation:

When wings of the airplane makes an angle of 40 degree with the horizontal so here we can say that force due to air is having two components

F_y = mg

F_x = \frac{mv^2}{R}

now we know that

F_y = F cos40

F_x = F sin40

also we know that

v = 480 km/h

v = 133.3 m/s

now plug in all data in above equations

tan 40 = \frac{v^2}{Rg}

R = \frac{v^2}{g tan40}

R = \frac{133.3^2}{9.8 tan40}

R = 2162 m

6 0
2 years ago
Gold has a specific heat of 0.130 J/g*C. If 195 joules of heat are added to 15 grams of gold how much does the temperature of th
Anna71 [15]

The correct answer to the question is :  100\ ^0C

EXPLANATION :

As per the question, the specific heat of gold is given as c = 0.130\ J/g^0C

The heat given to the gold dQ = 195 J

The mass of the gold is given as m = 15 gram.

We are asked to calculate the change in temperature.

Let the change in temperature is dT.

We know that dQ = mcdT

                      dT=\ \frac{dQ}{mc}

                             =\ \frac{195}{15\times 0.130}

                             =\ 100\ ^0C                   [ANS]

Hence, the change in temperature is 100 degree celsius.

5 0
3 years ago
A graduated beaker with 375 mL of water is sitting on a scale which measures the weight of the glass and water to be 7.60 N. Whe
Allisa [31]

Answer:

Volume of water displaced = 450 - 375 = 75 ml

Vr = volume of rock = 75 ml

Wr = 9.22 - 7.60 = 1.62 N  weight of 75 ml of rock

Density of rock = 1.62 N / 75 ml = .0216 N / ml

Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml

Density of rock / density of water = .0216 / .0098 = 2.20

8 0
2 years ago
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