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Karo-lina-s [1.5K]
3 years ago
8

DUE BY MIDNIGHT

Physics
1 answer:
olga2289 [7]3 years ago
8 0

Answer:

Option D. 1000 J.

Explanation:

From the question given above, the following data were obtained:

Force (F) applied = 200 N

Distance (s) = 5 m

Time (t) = 10 s

Workdone (Wd) =?

Workdone (Wd) is simply defined as the product of force (F) and distance (s) moved in the direction of the force. Mathematically, it is expressed as:

Wd = F × s

With the above formula, we can calculate the Workdone as illustrated below:

Force (F) applied = 200 N

Distance (s) = 5 m

Workdone (Wd) =?

Wd = F × s

Wd = 200 × 5

Wd = 1000 J

Thus, the Workdone is 1000 J

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D. a concave mirror

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A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica
Evgen [1.6K]

This question is incomplete, the complete question is;

A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick. What is the maximum charge?

(The dielectric constant of mica is 5.4, and its dielectric strength is 1.00×10⁸ V/m)

Answer: the maximum charge q is 716.85 μF

Explanation:

Given data;

with = 3.0 cm = 0.03

breathe = 5.0 m

Area = 0.03 × 5 = 0.15 m²

dielectric strength E = 1.00 × 10⁸

∈₀ = 8.85 × 10⁻¹²

constant K = 5.4

maximum charge = ?

the capacitor C = KA∈₀ / d

q = cv so c = q/v

now

q/v = KA∈₀ / d

q =  vKA∈₀/d = EKA∈₀

we substitute

q = (1.00 × 10⁸) × 5.4 × 0.15 ×  8.85 × 10⁻¹²    

q = 716.85 × 10⁻⁶ F

q = 716.85 μF

the maximum charge q is 716.85 μF

7 0
3 years ago
A ten loop coil of area 0.23 m2 is in a 0.047 T uniform magnetic field oriented so that the maximum flux goes through the coil.
Nana76 [90]

Answer:

0.32 V

Explanation:

N = 10, A = 0.23 m^2, B = 0.47 T, t = 0.34 s

The average induced emf is given by

e = - N dФ / dt

Where, dФ be the change in magnetic flux in time dt.

dФ / dt = d / dt (B A) = A dB/dt

So,

e = - 10 x 0.23 x 0.047 / 0.34 = - 0.32 V

The negative sign shows the direction of induced emf.

6 0
3 years ago
A 8.0-cm-diameter horizontal pipe gradually narrows to 5.0 cm . When water flows through this pipe at a certain rate, the gauge
adelina 88 [10]

Answer:

A 8.0 cm diameter horizontal pipe gradually narrows to 5.0 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

The flow rate is 3.1175×10⁻³ m³/s

Explanation:

To solve the question we rely on Bernoulli's principle as follows P_{1} +\frac{1}{2}\rho v^{2} _{1} + \rho gz_{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2} + \rho gz_{2}

thus where the pipe is  horizontal we have

z₁ = z₂ hence the above equattion becomes

P_{1} +\frac{1}{2}\rho v^{2} _{1}  = P_{2} +\frac{1}{2}\rho v^{2} _{2}

since the flow rate is constant then

Q = v₁A₁ = v₂A₂

Where is the area of the two sections given by A₁ = π·D₁²÷4 and

A₂ = π·D₂²÷4

Thereffore A₁ = π·0.08²÷4 = 5.02×10⁻³ m²

and A₂ = π·0.05²÷4 = 1.96×10⁻³ m²

v₁ = v₂A₂/A₁ =0.391×v₂

The given pressures are P₁ = 31.0 kPa and P₂ = 24.0 pKa and

ρ = 1000 kg/m³

Plugging the values into the above equation we get

31.0 kPa +0.5× 1000 kg/m³× (0.391×v₂)² = 24.0 pKa +0.5×1000 kg/m³×v₂²

= 31000+76.3·v₂² =24000+500·v₂²

or 423.706·v₂² = 7000

v₂² = 7000/423.706 = 16.52 or  v₂ = 4.065 m/s and  v₁ 0.391×4.065 = 1.59 m/s

The flow rate = v₂A₂ = 1.59×1.96×10⁻³ = 3.1175×10⁻³ m³/s

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What is the power of 10 when 0.00503 is written in scientific notation
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