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soldi70 [24.7K]
3 years ago
13

Which situation is an example of transferring heat by means of convection?

Physics
1 answer:
Hoochie [10]3 years ago
3 0
It would be B, the weather patterns outside.
You might be interested in
How are energy, work ang power related?
jolli1 [7]

Answer:

Work is the change in energy and power is the rate of doing work.

5 0
2 years ago
Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in
sattari [20]

Answer:

49.3 N

Explanation:

Given that Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in the rope?

The weight of the bucket of water = mg.

Weight = 4.25 × 9.8

Weight = 41.65 N

The tension and the weight will be opposite in direction.

Total force = ma

T - mg = ma

Make tension T the subject of formula

T = ma + mg

T = m ( a + g )

Substitutes all the parameters into the formula

T = 4.25 ( 1.8 + 9.8 )

T = 4.25 ( 11.6 )

T = 49.3 N

Therefore, the tension in the rope is 49.3 N approximately.

8 0
2 years ago
What effect or effects would be most significant if the Moon's orbital plane were exactly the same as the ecliptic plane?
Lilit [14]

Answer:

c. Solar eclipses would be much more frequent.

Explanation:

The <u>ecliptic plane</u> is the apparent orbit that the sun describes around the earth (although it is the earth that orbits the sun), is the path the sun follows in earth's sky.

A <u>solar eclipse</u> occurs when the moon gets between the earth and the sun, so a shadow is cast on the earth because the light from the sun is blocked.

The reason why solar eclipses are not very frequent is because the moon's orbital plane is not in the same plane as the orbit of the earth around the sun, but rather that it is somewhat inclined with respect to it.

So <u>if both orbits were aligned, the moon would interpose between the sun and the earth more frequently, producing more solar eclipses.</u>

So, if the moon's orbital plane were exacly the same as the ecliptic plane solar eclipses would be more frequent.

the answer is: c.

8 0
3 years ago
There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C
liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

V_{tot} = V_{q1} + V_{q2}

V_q = \dfrac{k \cdot q}{r}

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

V_{tot} =  \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05}  = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0

The electric field at the point exactly midway between the plates, V_{tot} = 0

3) The electric field, 'E', between plates is given as follows;

E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, F_e = E × e

∴ F_e = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, F_e ≈ 1.807 × 10⁻⁷ N

4 0
2 years ago
Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells
barxatty [35]

Answer:

a

The mass of blood is m= 5.7876kg

b

The number of blood cells is  N_t=1.04*10^{13}

Explanation:

From the question we are told that

         The volume of blood  is  V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3

         The density of the blood is  \rho_b = 1060 kg/m^3

         % of blood  that is  cell is  = 45.0%

        % of the blood that is  plasma is  = 55.0%

        density of blood cell is  \rho_d = 1125kg/m^3

        % of cell that are white is  = 1%

        % of cell that is red is  = 99%

        The diameter of the red blood cell is  = 7.5 \mu m = 7.5*10^{-6}m

         The radius of the red blood cell is  = \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m

Generally the mass is mathematically  represented as

               m = \rho_b * V_b

Substituting value

            m = 1060 * 0.00546

               m= 5.7876kg

Mass of cell is m_c = 45% of m

                         = 0.45 * 5,7876

                         = 2.60442 kg

The volume of cells is V_c = \frac{m_c}{\rho_d}

                                      = \frac{2.60442}{1125}

                                      = 2.315 *10^{-3} m^3

The volume of white blood cell is V_w = 1% of volume of cells

                                                         = \frac{1}{100} * 2.315*10^{-3}

                                                       = 2.315*10^{-5}m^3

The volume of a single cell is V_s = 4 \pi r^3

                                                                        = 4*(3.142) * (3.75*10^{-6})^3

                                                                        = 2.21*10^{-16}m^3

The volume of red blood cells is V_r = V_c - V_w

                                                           =2.315*10^{-3} - 2.315*10^{-5}

                                                           = 2.29*10^{-3}m^3

The number of red blood cell is  = \frac{V_r}{V_s}

                                                     = \frac{2.29 *10^{-3}}{2.21*10^{-16}}

                                                    = 1.037*10^{13}

The Number of white blood cell is   =\frac{V_w}{V_s}

                                                          = \frac{2.315 * 10^{-5}}{2.21*10^{-16}}

                                                          = 1.04*10^{11}

The total number of blood cells is  N_t= 1.037*10^{13} + 1.04*10^{11}

                                                        N_t=1.04*10^{13}

6 0
3 years ago
Read 2 more answers
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