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valentinak56 [21]

3 years ago

15

Use examples to explain how the geosphere interacts with two other of Earth's spheres. Explain the interaction for each using co

mplete sentences.
This is for science class

2 answers:

Orlov [11]3 years ago

7 0

The geosphere interacts with the hydrosphere when water causes rock to erode. The atmosphere provides the geosphere with heat and energy for erosion, and the geosphere reflects the sun's energy back into the atmosphere.

Artemon [7]3 years ago

3 0

Answer: Interaction

Explanation: The geosphere interacts with the hydrosphere when water causes rock to erode. The atmosphere provides the geosphere with heat and energy for erosion, and the geosphere reflects the sun's energy back into the atmosphere

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A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the

Sladkaya [172]

Answer:

(a). The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Explanation:

Given that,

Radius of sphere R=2.81 cm

Charge = +2.35 fC

Potential at center of sphere

$V = 0$

(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

$V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr$

$V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr$

$V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}$

$V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})$

$V_{r}=-0.00012190\ V$

$V_{r}=-1.219\times10^{-4}\ V$

The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). We need to calculate the potential at a distance r = R

Using formula of potential difference

$V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}$

$V_{R}=-0.0003759\ V$

$V_{R}=-3.759\times10^{-4}\ V$

The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Hence, This is the required solution.

5 0

3 years ago

The _____ of a mechanical wave is a direct measure of its energy.

Wittaler [7]

Answer: Amplitude

Explanation:

8 0

2 years ago

What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m?

polet [3.4K]

Answer:

0.044 V

Explanation:

E = Electric field = $5.5\times 10^6\ V/m$

d = Thickness of membrane = 8 nm

When the electric field strength is multiplied by the membrane thickness we get the voltage

Voltage across a gap is given by

$V=Ed\\\Rightarrow V=5.5\times 10^6\times 8\times 10^{-9}\\\Rightarrow V=0.044\ V$

The voltage across the membrane is 0.044 V

4 0

2 years ago

Two small conducting point charges, separated by 0.4 m, carry a total charge of 200 C. They repel one another with a force of 12

Lunna [17]

Answer:

200 C

Explanation:

Let C1 and C2 be their charges. According to Coulomb's law

$F_C = k\frac{C_1C_2}{R^2}$

where k = $8.99\times10^9 nm^2/C^2$ is the constant, R = 0.4m is the distance between them, F = 120 N is their resulting charge force

$120 = 8.99\times10^9\frac{C_1C_2}{0.4^2}$

$C_1C_2 = \frac{120*0.4^2}{8.99\times10^9} = 2.13\times10^{-9}$

Since their total charge is 200C:

$C_1 + C_2 = 200$ or $C_1 = 200 - C_2$

We can substitute the above equation

$C_1C_2 = (200 - C_2)C_2 = 2.13\times10^{-9}$

$-C_2^2 +200C_2 - 2.13\times10^{-9} = 0$

$C= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$C= \frac{-200\pm \sqrt{(200)^2 - 4*(-1)*(-0.00000000213)}}{2*(-1)}$

$C= \frac{-200\pm200}{-2}$

$C = 1.06 \times 10^{-11}$ or $C \approx 200$

So the larger charge is C = 200 C

8 0

3 years ago

A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in

sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$ ---------------------- (1)

in which R₁ is the resistance at temperature T₁

$R_{T(2)$ is the resistance at temperature T₂

Given that V= IR

R = $\frac{V}{I}$

Therefore, the resistance at temperature 28°C is;

$R_{28}= \frac{120V}{1.36A}$

= 88.24Ω

$R_{T(2)$ = $\frac{120V}{1.23A}$

= 97.56Ω

From (1) above;

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

$\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)$

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

$\frac{0.1056}{4.5*10^{-4}}= T-28^0C$

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

$P= I^2_2R_{T^0C$

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0

2 years ago