Convert 75.0 degrees Celsius into Fahrenheit

Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t

katrin2010 [14]

Answer:

$1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

23.4843749996 m

Yes

Explanation:

E = Electric field = $2\times 10^5\ N/C$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of proton= $1.67\times 10^{-27}\ kg$

q = Charge of electron = $1.6\times 10^{-19}\ C$

Acceleration is given by

$a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g$

The acceleration is $1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m$

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0

3 years ago

A water discharge of 9 m3/s is to flow through this horizontal pipe, which is 0.98 m in diameter. If the head loss is given as 1

Mashutka [201]

Answer: The power required by the pump to produce a discharge of 9m³/s is 5990joules/secs.

Explanation: The given parameters from the questions are:

Flow rate Q = 9m³/s, Diameter D = 0.98m, acceleration a = 1.0, head loss(Pressure P) is given by the function 10v²/2g.

STEP 1. Find the velocity of water in the pipe from the equation:

Diameter D = (√4.Q/π.v), where v is the velocity, and Q is flow rate

Making v subject of the formula gives:

v = 4Q/π.√D =[ 4 × 9m³/s / 3.142 × (√0.98m)] = 11.69m/s.

STEP 2. Find the pressure from the relationship, P = 10v²/2g, NB. g = a

P = 10 × (11.69m/s)² / 2× 1.0m/s²

P = 683.25N/m² or Pascal.

STEP 3. Find force exerted by the pump;

Recall that Pressure P = Force/Area

But Area A = π.r², where r = D/2

Therefore, A = π.(D/2)²

A = 3.142 × [0.98m/2]² = 0.75m²

Therefore, Force = Pressure × Area

Force F = 683.25N/m² × 0.75m²

F = 512.44N.

STEP 4. Find work done

Work done W by the pump is = Force × distance d moved by the water

W = F . d

Also recall that flow rate Q = Velocity/time.

Q = v/t, we can write t = v/Q.

Time t = 11.69m/s / 9m³/s = 1.298s

Also recall that velocity v = distance d/time t, v = d/t, making d subject of formula gives v × t

Distance d = v × t = 11.69m/s × 1.298s = 15.17m.

Hence,

Work Done W = Force × distance

W = 512.44N × 15.17m = 7775.56Nm or joules.

Lastly, Power P = Work done/ time

P = 7775.56joules/1.298s

P = 5990.4joules/s.

8 0

3 years ago

Quadrilateral A has side lengths 3, 4, 5, and 6. Quadrilateral B is a scaled copy of Quadrilateral A with a scale factor of 2. S

S_A_V [24]

Answer: 6 and 8 would be side lengths because if it has a scale factor of 2 then you multiply the ones on the chart.

Explanation:Go find one

7 0

3 years ago

A sample of aluminum has a mass of 4.86 grams and a volume of 1.8 cubic centimeters. What is the density of aluminum rounded to

TEA [102]

Answer:

3g/cm³

Explanation:

2.7

The seven adds 1 to the 2 making it 3.0

6 0

3 years ago

A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to

Nostrana [21]

Answer:

a) r=4.24cm d=1 cm

b) $Q=5x10^{-10} C$

Explanation:

The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

$V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}$

$V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m$

The distance must be the separation the r distance can be find also using

$C=\frac{Q}{V_{ab}}$

But now don't know the charge these plates can hold yet so

a).

d=0.01m

$C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}$

$A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}$

$A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } } \\r=42.55x^{-3}m$

b).

$C=\frac{Q}{V_{ab}}$

$Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C$

5 0

3 years ago