ASAP I’m timed on this so help would really be appreciated :)

24 A uniform electric field of magnitude 1.1×104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the elect

Tatiana [17]

Answer:

$44,000 Nm^2/C$

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

$\Phi = EA cos \theta$

where:

E is the magnitude of the electric field

A is the area of the surface

$\theta$ is the angle between the direction of the field and of the normal to the surface

In this problem, we have:

$E=1.1\cdot 10^4 N/C$ is the electric field

L = 2.0 m is the side of the sheet, so the area is

$A=L^2=(2.0)^2=4.0 m^2$

$\theta=0^{\circ}$ , since the electric field is perpendicular to the surface

Therefore, the electric flux is

$\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C$

4 0

2 years ago

A crane lifts a 545 kg piano up into a high-rise apartment 75.0 meters above the ground. In order to do so, 5350 Newtons were ap

vodka [1.7K]

Answer:

Work done to pull the piano upwards is 401250 J

Explanation:

Work is done against the gravity to pull the piano upwards

So here we can say that work done is

$W = mgH$

here we know that

$mg = 5350 N$

also we know that

H = 75 m

now we have

$W = 5350 \times 75$

$W = 401250 J$

7 0

3 years ago

A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

6 0

3 years ago

You hang a heavy ball with a mass of 30 kg from a tungsten rod 2.8 m long by 1.5 mm by 2.6 mm. You measure the stretch of the ro

guajiro [1.7K]

Answer:

Young's modulus (Y) = 3.56×10^11 N/m^2

The speed of sound in tungsten = 6166.4 m/s

Explanation:

Young's modulus (Y) = stress/strain

Stress = force/area

Force = mg = 30×9.8 = 294 N

Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2

Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2

Strain = extension/length

Extension = 0.000594 m

Length = 2.8 m

Strain = 0.000594/2.8 = 2.12×10^-4

Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2

Y = h × rho × g

rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3

h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m

From the equations of motion

v^2 = u^2 + 2gh =

Initial speed (u) = 0 m/s

v = sqrt (2×9.8×1.94×10^6)

v = 6166.4 m/s

7 0

3 years ago

Read 2 more answers

A uniform, solid, 2000.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.10 kgkg poi

mars1129 [50]

Answer:

$0.0110284391534\ N$

$0.0653784219002\ N$

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of sphere = 2000 kg

$m_2$ = Mass of other sphere = 2.1 kg

r = Distance between spheres

Force of gravity is given by

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(5.04\times 10^{-3})^2}\\\Rightarrow F=0.0110284391534\ N$

The gravitational force is $0.0110284391534\ N$

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(2.07\times 10^{-3})^2}\\\Rightarrow F=0.0653784219002\ N$

The gravitational force is $0.0653784219002\ N$

6 0

3 years ago