What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?

Answers

Answer 1

Answer:

To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as

The potential energy, $U = 13.0 J$

The volume, $V = 6.00 mm^3 = 6.00*10^(-9)m^3$

The potential energy per unit volume is defined as the energy density.

$u = (U)/(V)$

$u= ((13.0 J))/((6.00*10^(-9) m^3))$

$u= 2.167109 J/m^3$

The energy density related with electric field is given by

$u = (1)/(2) \epsilon_0 E^2$

Here, the permitivity of the free space is

$\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2$

Therefore, rerranging to find the electric field strength we have,

$E = \sqrt{(2u)/(\epsilon_0)}$

$E = \sqrt{(2(2.167109))/(8.85*10^(-12))}$

$E = 2.211010 V/m$

Therefore the electric field is 2.21V/m

Answer 2

Answer:

Final answer:

To calculate the electric field strength that would store 12.5 Joules of energy in every 6.00 mm^3 of space, we use the energy density formula. We firstly find the energy density and input it into the formula to solve for the electric field strength. The result is approximately 6.87 X 10^6 N/C.

Explanation:

The energy stored in an electric field is given by the formula U = 1/2 ε E^2. Here, U is the energy density (energy per unit volume), E is the electric field strength, and ε is the permittivity of free space.

Given that the energy stored U is 12.5 joules, and the volume is 6.00 mm^3 or 6.00X10^-9 m^3, the energy density (U) can be computed as 12.5 J/6.00X10^-9 m^3 = 2.08X10^12 Joule/meter^3.

We can solve the formula for E (electric field strength): E = sqrt ((2U)/ε). Substituting the value of ε (8.85 × 10^-12 m^-3 kg^-1 s^4 A^2), we can find E to be approximately 6.87 X 10^6 N/C.

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In an 8.00km race, one runner runs at a steady 11.8 km/hr and another runs at 15.0 km/hr. How far from the finish line is the slower runner when the faster runner finishes the race?

Answers

Answer:

The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

Explanation:

Given;

the speed of the slower runner, u₁ = 11.8 km/hr

the speed of the fastest runner, u₂ = 15 km/hr

distance, d = 8 km

The time when the fastest runner finishes the race is given by;

$Time = (Distance )/(speed)\n\nTime = (8)/(15) \n\nTime = 0.533 \ hr$

The distance covered by the slower runner at this time is given by;

d₁ = u₁ x 0.533 hr

d₁ = 11.8 km/hr x 0.533 hr

d₁ = 6.29 km

Additional distance (x) the slower runner need to finish is given by;

6.29 km + x = 8km

x = 8 k m - 6.29 km

x = 1.71 km

Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom rises from pin A to B, at angle phi from horizontal. A cable goes up from the load to B and then left to a pulley. Determine the greatest weight of the crate that can be supported without causing the cable to fail if ϕ=30∘. Neglect the size of the winch.

Answers

Explanation:

Let us assume that forces acting at point B are as follows.

$\sum F_(x) = 0$

$T + F_(AB) Sin 60$ = 0 ...... (1)

$\sum F_(y)$ = 0

$F_(AB) Cos 60 + W$ = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

$\sigma_(allow) = (T)/(A)$

T = $(20 * 1000) (\pi)/(4) * (0.5)^(2)$

= 3925 kip

From equation (1), $F_(AB)Sin (60^(o))$ = -3925

$F_(AB) * -0.304 8$ = -3925

$F_(AB)$ = 12877.29 kip

From equation (2), -12877.29 (Cos 60) + W = 0

$-12877.29 kip * (1)/(2) + W$ = 0

W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

To determine the greatest weight of the crate that can be supported without causing the cable to fail, calculate the normal stress on the cable using σ = F/A, where σ is the normal stress, F is the force on the cable, and A is the cross-sectional area of the cable. Then, compare the calculated normal stress to the allowable normal stress. Consider the angle phi in this calculation by using the equation F = W / sin(ϕ), where F is the force on the cable, W is the weight of the crate, and ϕ is the angle with respect to the horizontal.

To determine the greatest weight of the crate that can be supported without causing the cable to fail, we need to calculate the normal stress on the cable and compare it to the allowable normal stress. The normal stress can be calculated using the formula σ = F/A, where σ is the normal stress, F is the force on the cable, and A is the cross-sectional area of the cable. In this case, the force on the cable is equal to the weight of the crate, and the cross-sectional area of the cable can be calculated using the formula A = π*(d/2)^2, where d is the diameter of the cable.

Given that the diameter of the cable is 0.5 in and the allowable normal stress is 21 ksi, we can substitute these values into the equations and solve for the force on the cable:

Calculate the cross-sectional area of the cable: A = π*(0.5/2)^2 = π*(0.25)^2 = 0.1963 in^2

Plug the cross-sectional area and the allowable normal stress into the formula for normal stress: σallow = F/A → 21 ksi = F/0.1963 in^2

Solve for the force on the cable: F = 21 ksi * 0.1963 in^2 = 4.1183 ksi*in^2

Therefore, the greatest weight of the crate that can be supported without causing the cable to fail is equal to the force on the cable, which is 4.1183 ksi*in^2. However, it's important to note that we also need to consider the angle phi (ϕ) in this calculation. Since the cable goes up from the load to point B and then left to a pulley, the weight of the crate will create a vertical component and a horizontal component. To determine the weight of the crate that corresponds to the calculated force on the cable, we need to consider the trigonometric relationship between the force and the weight at an angle. In this case, the angle is 30 degrees, so we can use the equation F = W / sin(ϕ), where F is the force on the cable, W is the weight of the crate, and ϕ is the angle with respect to the horizontal.

Given that ϕ = 30 degrees and F = 4.1183 ksi*in^2, we can substitute these values into the equation and solve for the weight of the crate:

Plug the values into the equation: 4.1183 ksi*in^2 = W / sin(30)

Solve for the weight of the crate: W = 4.1183 ksi*in^2 * sin(30)

Therefore, the greatest weight of the crate that can be supported without causing the cable to fail and at an angle of 30 degrees is equal to the force on the cable, which is 4.1183 ksi*in^2, multiplied by the sine of 30 degrees.

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Two non-conducting slabs of infinite area are given a charge-per-unit area of σ = -16 C/m^2 and σb =+6.0 C/m^2 respectively. A third slab, made of metal, is placed between the first two plates. The charge density σm on the metal slab is 0 (i.e., the slab is uncharged).Required:
Find the magnitude and direction of the electric field.

Answers

Answer:

$E_(C) = -5.65 * 10^(11) \hat{x}$ N/C

$E_(A) = -1.24 * 10^(12) \hat{x}$ N/C

Explanation:

The charge per unit area of the two non-conducting slabs are given by:

$\sigma_(a) = -16 C/m^2$

$\sigma_(b) = 6 C/m^2$

The charge density on the metal $\sigma_(m) = 0$

ε0 = 8.854 x 10-12 C2/N m2

Note that the electric field inside the conductor is zero because it is an equipotential surface.

The diagram attached to this solution typifies the description given in the question:

The electric field in the region C can be calculated by:

$E_(C) = ( |\sigma_(b) |- |\sigma_(a)| )/(2 \epsilon_(0) ) \nE_(C) = (6 - 16 )/(2 * 8.854 * 10^(-12) ) \nE_(C) = -5.65 * 10^(11) \hat{x}$

The electric field in the region A can be calculated by:

$E_(A) = (- |\sigma_(a) |- |\sigma_(b)| )/(2 \epsilon_(0) ) \nE_(A) = (-16 - 6 )/(2 * 8.854 * 10^(-12) ) \nE_(A) = -1.24 * 10^(12) \hat{x}$

On a map each inch represents 25 miles what is the length of a highway if it is 6 inches long on a map

Answers

Answer:

150 hope this helps

Explanation:

Answer:

150

Explanation:

Please provide an explanation.

Thank you!!

Answers

Answer:

(a) 22 kN

(b) 36 kN, 29 kN

(d) 43 kN

Explanation:

(a) When the truck is off the bridge, there are 3 forces on the bridge.

Reaction force F₁ pushing up at the first support,

reaction force F₂ pushing up at the second support,

and weight force Mg pulling down at the middle of the bridge.

Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)

∑τ = Iα

(Mg) (0.3 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L)

F₁ = ½ Mg

F₁ = ½ (44.0 kN)

F₁ = 22.0 kN

(b) This time, we have the added force of the truck's weight.

Using the same logic as part (a), we sum the torques about the second support:

∑τ = Iα

(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)

F₁ = ½ Mg + ⅔ mg

F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)

F₁ = 36.0 kN

Now sum the torques about the first support:

∑τ = Iα

-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)

F₂ = ½ Mg + ⅓ mg

F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)

F₂ = 29.0 kN

Alternatively, sum the forces in the y direction.

∑F = ma

F₁ + F₂ − Mg − mg = 0

F₂ = Mg + mg − F₁

F₂ = 44.0 kN + 21.0 kN − 36.0 kN

F₂ = 29.0 kN

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)

As x increases, F₁ decreases and F₂ increases.

(d) Using our equation from part (c), when x = 0.6 L, F₂ is:

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)

F₂ = ½ Mg + mg

F₂ = ½ (44.0 kN) + 21.0 kN

F₂ = 43.0 kN

Answer:

a. Left support = Right support = 22 kN

b. Left support = 36 kN

Right support = 29 kN

c. Left support force will decrease

Right support force will increase.

d. Right support = 43 kN

Explanation:

given:

weight of bridge = 44 kN

weight of truck = 21 kN

a) truck is off the bridge

since the bridge is symmetrical, left support is equal to right support.

Left support = Right support = 44/2

Left support = Right support = 22 kN

b) truck is positioned as shown.

to get the reaction at left support, take moment from right support = 0

∑M at Right support = 0

Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0

Left support = 44 (0.3) + 21 (0.4)

0.6

Left support = 36 kN

Right support = weight of bridge + weight of truck - Left support

Right support = 44 + 21 - 36

Right support = 29 kN

as the truck continues to drive to the right, Left support will decrease

as the truck get closer to the right support, Right support will increase.

d) truck is directly under the right support, find reaction at Right support?

∑M at Left support = 0

Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0

Right support = 44 (0.3) + 21 (0.6)

0.6

Right support = 43 kN

How do you perceive the importance of online safety security ethics and etiquette

Answers

Answer:

is called online etiquette, or netiquette for short.

...

Here are some tips to help you get started:

Create Complex Passwords.

Boost Your Network Security.

Use a Firewall.

Click Smart.

Be a Selective Sharer.

Protect Your Mobile Life.

Practice Safe Surfing & Shopping.

Keep up to date.

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