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3 years ago

a charge of 6.4x10^-7 C experiences an electric force of 1.8x10^-1 N what is the electric field strength

1 answer:

5 0

Field strength = force ÷ charge (E=F/Q)

E= 1.8×10^-1 ÷ 6.4×10^-7... to obtain ur answer

the formula is derived from the definition
Electric field : A region in space where a charged body experiences a force.

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Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o

Verdich [7]

Answer:

V = 20 miles /sec

Explanation:

We have remaining distance = d = 96 miles

Lets call Pascal velocity V in miles per hour

Now if he increases his velocity by 50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t

1.5* V = d/ t₁ ⇒ 1.5 * V = 96 /t₁

And in the case of reducing his velocity

(V / 4) = d/ (t₁ + 16 ) ⇒ V * (t₁ + 16 ) = 4*d ⇒ V*t₁ + 16*V = 384

So we a 2 equation system with two uknown variables

1.5*V = 96/t₁ (1)

V*t₁ + 16*V = 384 (2)

We solve from equation (1) t₁ = 64/V

And by substitution in equation (2)

V * (64/V) + 16* V = 384

64 + 16 *V = 384 ⇒ 16*V = 320 ⇒ V= 320/16

V = 20 miles /sec

6 0

3 years ago

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

= $\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]$

∴NOTE: Graph is attached

8 0

4 years ago

A 72-kg skydiver is falling from 10000 feet. At an instant during the fall, the skydiver

MissTica

Answer:

Approximately $2.31\; \rm m \cdot s^{-2}$ (assuming that the acceleration due to gravity is $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Assuming that $g = 9.81\; \rm m \cdot s^{-2}$ the weight on this 72-kg skydiver would be $W = m \cdot g = 72 \; \rm kg \times 9.81\; \rm m \cdot s^{-2} = 706.32\; \rm N$ (points downwards.)

Air resistance is supposed to act in the opposite direction of the motion. Since this skydiver is moving downwards, the air resistance on the skydiver would point upwards.

Therefore, the net force on this skydiver should be the difference between the weight and the air resistance on the skydiver:

$\begin{aligned}F(\text{net force}) &= W - F(\text{air resistance})\\ &= 706.32\; \rm N - 540\; \rm N =166.32\; \rm N \end{aligned}$ .

Apply Newton's Second Law of motion to find the acceleration of this skydiver:

$\begin{aligned}a &= \frac{F(\text{net force})}{m} \\ &= \frac{166.32\; \rm N}{72\; \rm kg} = 2.31\; \rm m \cdot s^{-2} \end{aligned}$ .

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3 years ago

The value of a scientific variable ____. A) can change

gregori [183]

The answer is A: can change

5 0

3 years ago

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Stan does 178 J of work lifting himself 0.5 m. What is Stan’s mass? The acceleration of gravity is 9.8 m/s 2 . Answer in units o

Rom4ik [11]

Work= (force)(distance)
178= m(9.81)x0.5
178=m(4.905)
178/4.905=m

His mass is 36.3 kg

4 0

3 years ago

Read 2 more answers