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4 years ago

Two conducting parallel plates 5.0 × 10−3 meter apart are charged with a 12-volt potential

Physics

1 answer:

photoshop1234 [79]4 years ago

7 0

Answer: 2.4×10^-3 v/m

Explanation: distance between plates of capacitor (d) =5.0×10^-3m

Potential difference between plates (v) = 12v

Force on electronic charge (f) = 3.8×10^-16 N

Strength of electric field (E) =?

The formulae that relates potential difference, eoectiic field strength and distance between plates is given as

v = Ed

By substituting the parameters, we have that

12 = E × 5.0×10^-3

E = 12/ 5.0 × 10^-3

E = 2.4×10^-3 v/m

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A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surfac

Ostrovityanka [42]

Answer:

Explanation:

Given

acceleration in x direction

$a_x=5.10 m/s^2$

Final velocity in x direction $v_x=3500 m/s$

time taken=150 s

let initial velocity in x direction be $u_x$

$v_x=u_x+a_x\times t$

$3500-5.10\times 150=u_x$

$u_x=2735 m/s$

Now solve for Y direction

$v_y=u_y+a_y\times t$

$4046=u_y+7.3\times 150$

$u_y=4046-1095=2951 m/s$

$tan\theta =\frac{u_y}{u-x}=\frac{2951}{2735}$

$tan\theta =1.078$

$\theta =47.14 ^{\circ}$ ACW from x-axis

5 0

3 years ago

an airplane accelerates down a runway at 5.20 m/s^2 for 32.8s until it finally lifts off the ground. determine the distance trav

hammer [34]

Answer:

7,272.7m

Explanation:

Our knowns:

a = 5.2m/s^2

t = 32.8s

$V_{i}$ = 0 m/s

Our Unknowns:

Distance (x): ?

We can use displacement formula:

ΔX = $V_{i}$ t + $\frac{1}{2}$ a $t^{2}$

Plug in knowns:

X = 0 + 1/2 ((5.2)(32.8))^2

Answer:

ΔX = 7272.7m

(I beleive this should be right) let me know if it helped!

6 0

3 years ago

One hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area 1.32 × 10-3 m2

kenny6666 [7]

Answer:

The charge flos through the coil is 0.023C

Explanation:

To solve this problem, it is necessary to apply the concepts related to Faraday's Law in which it is possible to calculate the emf Voltage induced due to a charge in a magnetic field

and Ohm's Law for the calculation of the current based on a given load over time.

Our data are given by:

$N=100$

$A= 1.32*10^{-3}m^2$

$R=15\Omega$

Where

N is the number of loops, A the area and R the Resistance.

The change in magnetic field can be calculated as,

$dB = 1.31-(-1.31)$

$dB = 2.62T$

The Faraday's law of electromagnetic induction is given by definition as,

$V = NA \frac{dB}{dt}$

In the other hand Ohm's law says:

$V = IR$

$V = \frac{dq}{dt} R$

Equating both equations we have

$\frac{dq}{dt} R = NA \frac{dB}{dt}$

We can re-arrange the equations to solve q, then

$dq = \frac{NA(dB)}{R}$

$q = \frac{(100)(1.32*10^{-3})(2.62)}{15}$

$q = 0.023C$

Therefore the charge flos through the coil is 0.023C

8 0

3 years ago

How do Swati and Banks adjust their body position during a skydiving jump so they can fall at the same rate?

Wittaler [7]

Sky diving involves free fall under gravity along with the drag due to air molecules pushing against the body slowing the rate of fall of a body. This is actually a significant amount of force. The drag force depends on the contact surface area and weight of the body. More the surface area in contact, more would be the drag. The sitting position of the skydiver would experience less drag than the chest down position. This is because of the less contact surface area of the body with the air molecules while in the former case. Since no two persons have identical body shape and weight, the rate of fall can be made nearly equal but not exactly equal. This is would be possible when they are having same body position.

3 0

3 years ago

What is the formula for electrostatic force?

Nonamiya [84]

F = k $\frac{q1 q2 }{ r^{2} }$

k - coefficient
q1 and q2 - charge
r - distance btw them

7 0

4 years ago

Read 2 more answers