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4 years ago

The chart shows the temperatures of four different substances.

Physics

2 answers:

Sliva [168]4 years ago

5 0

Answer:

Your correct answer is C 20

Explanation:

garik1379 [7]4 years ago

4 0

Answer:

C) 20

Explanation:

Happy Halloween LOL

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Capacitors C1 = 6.45 µF and C2 = 2.50 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon

denpristay [2]

Answer:

For C1, Q = 1.6125×10⁻³ C

For C2, Q = 6.25×10⁻⁴ C

Explanation:

Note: Since the capacitors are connected in parallel, The voltage across each of them is equal.

From the question,

Q = CV........................ Equation 1

Where Q = Charge on the capacitor, V = Voltage across the capacitor, C = Capacitance of the capacitor.

For the first capacitor,

Q = C1V............. Equation 2

Where C1 = 6.45 μF= 6.45×10⁻⁶ F, V = 250 V

Substitute into equation 2

Q = (6.45×10⁻⁶ )(250)

Q = 1.6125×10⁻³ C.

For the the second capacitor,

Q = C2V............. Equation 3

Given: C2 = 2.50 μF = 2.5×10⁻⁶ F, V = 250 V

Q = (2.5×10⁻⁶ )(250)

Q = 6.25×10⁻⁴ C

4 0

3 years ago

A hoop, a solid disk, and a solid sphere, all with the same mass and the same radius, are set rolling without slipping up an inc

myrzilka [38]

Answer:

The sphere

Explanation:

Because it has a smaller inertia (I) value in the explanation in the attached file

6 0

3 years ago

When the mass of a body is constant, then its acceleration is directly proportional to the force acting on it. When the force ac

Fed [463]

Answer:

True

Explanation:

8 0

3 years ago

Read 2 more answers

A jet airliner moving initially at 612 mph (with respect to the ground) to the east moves into a region where the wind is blowin

Ganezh [65]

Answer:

966.22 mph

Explanation:

Velocity of plane with respect to wind (Vp,w)= 612 mph east

velocity of wind with respect to ground, (Vw,g) = 362 mph at 15° North of

east

Write the velocities in vector form

$V_{p,w}=612\widehat{i}$

$V_{w,g}=362\left ( Cos15\widehat{i}+Sin15\widehat{j} \right )= 349.67\widehat{i}+93.69\widehat{j}$

Use the formula for the relative velocity

$V_{p,w}=V_{p,g}-V_{w,g}$

Where, V(p,w) is the velocity of plane with respect to wind

V(p,g) is the velocity of plane with respect to ground

V(w,g) is the velocity of wind with respect to ground

So, $V_{p,g}=V_{p,w}+V_{w,g}$

$V_{p,g}=\left ( 612+349.67 \right )\widehat{i}+93.69\widehat{j}$

$V_{p,g}=961.67\widehat{i}+93.69\widehat{j}$

Magnitude of velocity of lane with respect to ground

$V_{p,g} = \sqrt{961.67^{2}+93.69^{2}}$

V(p,g) = 966.22 mph

5 0

4 years ago

Oil having a specific gravity of 0.9 is pumped as illustrated with a water jet pump. The water volume flowrate is 1 m3 /s. The w

love history [14]

Answer:

The rate at which the pump moves oil is 1 m³/s

Explanation:

Assumptions:

there is steady-state flow

oil and water are incompressible

first fluid is water, second fluid is oil and third fluid is the mixture of oil and water.

$\rho_1Q_1 + \rho_2Q_2 = \rho_3Q_3 -------equation (i)$

where;

ρ is the fluid density

Q is the volumetric flow rate

$Q_1 + Q_2 = Q_3--------equation (ii)$

Substitute in Q₃ in equation i

$\rho_1Q_1 + \rho_2Q_2 = \rho_3(Q_1 +Q_2)$

divide through by ρ₁

$\frac{\rho_1Q_1}{\rho_1}+ \frac{\rho_2Q_2}{\rho_1} =\frac{ \rho_3(Q_1 +Q_2)}{\rho_1}\\\\Note; \frac{\rho_2}{\rho_1} = \gamma_2 \ and \ \frac{\rho_3}{\rho_1} = \gamma_3\\\\Q_1 + \gamma_2Q_2 = \gamma_3(Q_1+Q_2)$

Make Q₂ the subject of the formula

$Q_2 = \frac{Q_1(1- \gamma_3)}{\gamma_3-\gamma_2} = \frac{1 (\frac{m^3}{s}) (1-0.95)}{0.95-0.9} = 1 \ \frac{m^3}{s}$

Therefore, the rate at which the pump moves oil is 1 m³/s

6 0

3 years ago