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grin007 [14]
3 years ago
6

An infinite line charge of linear density λ = 0.30 μC/m lies along the z axis and a point charge q = 6.0 µC lies on the y axis a

t y = 2.0 m. The x component of the electric field at the point P on the x axis at x = 3.0 m is approximately . A) 1.8 kN/C B) 4.2 kN/C C) 0.96 kN/C D) 5.2 kN/c E) 0.64 mN/C
Physics
2 answers:
ira [324]3 years ago
6 0
The line charge E-field Ec = λ/(2πr*e0),
where λ = charge/length and e0 is the permittivity constant = 8.8542E-12 F/m. 
<span>The point charge E-field Ep = kq/r^2 where electrostatic constant k = 1/(4π*e0) = 8.99E9 N-m^2/C^2.</span>
Ivahew [28]3 years ago
6 0

Answer:

D) 5.2 kN/c

Explanation:

Electric field due to infinity line charge

λ /2πε₀r = .3 x 10⁻⁶ x 2 x 9 x 10⁹ / 3 = 1.8 x 10³. It will be towards x axis.

Electric field due to point charge = k.q / r² = 9 x 10⁹ x 6 x 10⁻⁶ / 13 = 4.15 x 10³

x component of this field = 4.15 x 10³ x cos Ф = 4.15 x 10³ x 3 / √13

3.458 x 10³

Total electric field = 5.2 kN / C.

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Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da
valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

3 0
3 years ago
A spherical balloon has a radius of 6.55 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a
Ksju [112]

Answer:

M_1 = 317.7 kg

Explanation:

Mass of the helium gas filled inside the volume of balloon is given as

m = \rho V

m = 0.179(\frac{4}{3}\pi R^3)

m = 0.179(\frac{4}{3}\pi 6.55^3)

m = 210.7 kg

now total mass of balloon + helium inside balloon is given as

M = 210.7 + 990

M = 1200.7 kg

now we know that total weight of balloon + cargo = buoyancy force on the balloon

so we will have

(M + M_1)g = \rho_{air} V g

(1200.7 + M_1) = (\frac{4}{3}\pi 6.55^3) (1.29)

1200.7 + M_1 = 1518.4

M_1 = 317.7 kg

3 0
3 years ago
Bernie drove home from a vacation in 1.5 hours. He traveled a total distance of 60 miles. What was his average speed in miles pe
mafiozo [28]

Answer: 40

Explanation:

I believe this is correct. I did 60/1.5 to get 40/mph

6 0
3 years ago
Read 2 more answers
What does altered mean
Elina [12.6K]
Altered means to change something.
5 0
3 years ago
Read 2 more answers
A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can
GenaCL600 [577]

Answer:

0.02 m

Explanation:

R₁ = initial distance jumped by jumper = 7.4 m

R₂ = final distance jumped by jumper = ?

θ₁ = initial angle of jump = 45°

θ₂ = final angle of jump = 42.9°

v = speed at which jumper jumps at all time

initial distance jumped is given as

R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}

final distance jumped is given as

R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}

Dividing final distance by initial distance

\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}

\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}

R_{2} =7.38

distance lost is given as

d = R_{1} - R_{2}

d = 7.4 - 7.38

d = 0.02 m

8 0
3 years ago
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