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katovenus [111]

3 years ago

7

(a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How

long does it take 1.00 C of charge to flow through a calculator if a 0.300 mA current is flowing?

1 answer:

OLEGan [10]3 years ago

8 0

Answer:

(a) Current flowing through truck battery is 180 A

(b) Time taken in calculator is 333.33 s

Explanation:

(a) Given:

The charge on the truck battery,q = 720 C

Time, t = 4.00 s

Consider I be the current flowing through truck battery.

The relation between current, charge and time is:

I = q/t

Substitute the suitable values in the above equation.

$I=\frac{720}{4}$

I = 180 A

(b) Given:

The charge on the calculator,q = 7.00 C

The current flowing through calculator, I = 0.3 mA = 0.3 x 10⁻³ A

Consider t be the time.

The relation between current, charge and time is:

t = q/I

Substitute the suitable values in the above equation.

$t=\frac{1}{0.3\times10^{-3} }$

I = 333.33 s

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ki77a [65]

Answer:

θ = 22.2

Explanation:

This is a diffraction exercise

a sin θ = m λ

The extension of the third zero is requested (m = 3)

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sin θ = m λ / a

sin θ = 3 630 10⁻⁹ / 5 10⁻⁶

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to better see the result let's find the angle in radians

θ = 0.3876 rad

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θ = 22.2º

4 0

3 years ago

Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres

inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )$

Here, $v_{1}$ is the velocity of water through wide ends of cylindrical pipe and $v_{2}$ is the velocity of water through narrow ends of cylindrical pipe.

Given, $v_{1} =1.4 m/s$

Now from equation continuity,

$v_{1} A_{1} = v_{2} A_{2}$ .

Here, $A_{1}$ and $A_{2}$ are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

$v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}$ .

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

$v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}$

$v_{2} = 4 \times 1.4 = 5.6 m/s$

Substituting these values with the density of water is $1000 \ kg/m^3$ in pressure difference formula we get.

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa$

3 0

3 years ago

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As part of a safety investigation, two 1900 kg cars traveling at 20 m/s are crashed into different barriers. Part A Find the ave

DedPeter [7]

Answer:

- $29.2\times 10^{3} N$

Explanation:

We are given that

Mass of cars= m=1900 kg

Initial speed of car=u=20 m/s

Final speed of car=v=0

Time= $\Delta t$ =1.3 s

We have to find the average force exerted on the car.

Average force= $\frac{change\;in\;momentum}{\Delta t}$

$F_{avg}=\frac{mv-mu}{1.3}$

$F_{avg}=\frac{1900(0)-1900(20)}{1.3}$

$F_{avg}=\frac{-38000}{1.3}=-29.2\times 10^{3} N$

Hence, the average force exerted on the car that hits a line of water barrels=- $29.2\times 10^{3} N$

8 0

3 years ago

The force between two charges, q, and 92, is F. If the distance between the

Sholpan [36]

Answer:

4F

Explanation:

F = kQ₁Q₂/d²

F' = k2Q₁2Q₂/d²

F' = 4(kQ₁Q₂/d²)

F' = 4F

4 0

3 years ago

Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o

Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

$F_{23} = \frac{k*q_{2}*q_{3}}{x^2}$ (Electric force of q₂ over q₃)

$F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2}$ (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

$\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2}$ (We cancel k and q₃)

$\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}$

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0

3 years ago

Read 2 more answers