Ask question

Ask question

All categories

3 years ago

9

A metal sphere of radius 2.0 cm carries an excess charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sph

ere? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

1 answer:

Nonamiya [84]3 years ago

5 0

Answer:

The electric field is $E = 7.5 *10^{6} \ N/C$

Explanation:

From the question we are told that

The radius of the metal sphere is $R = 2.0 \ cm = 0.02 \ m$

The excess charge which the metal sphere carries is $q = 3.0 \mu C = 3.0*10^{-6} \ C$

The distance of the position being to the center is $D = 6.0 \ cm = 0.06 \ m$

The coulomb constant is $k =9*10^{9} \ N \cdot m^2 /C^2$

Generally the electric field is mathematically represented as

$E = \frac{k * q}{D^2}$

substituting values

$E = \frac{9*10^{9} * 30.*10^{-6}}{(0.06)^2}$

$E = 7.5 *10^{6} \ N/C$

You might be interested in

A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of

ololo11 [35]

Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

$v^2=u^2+2gh$

$h =\dfrac{v^2}{2g}$ ....(I)

For second half,

We need to calculate the time

Using equation of motion

$h =ut+\dfrac{1}{2}gt_{2}^2$

$h=0+\dfrac{1}{2}gt_{2}^2$

$t_{2}=\sqrt{\dfrac{2h}{g}}$

Put the value of h from equation (I)

$t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}$

$t_{2}=\dfrac{v}{g}$

According to question,

$t_{1}+t_{2}=10$

$t_{1}=t_{2}$

Put the value of t₁ and t₂

$\dfrac{v}{g}+\dfrac{v}{g}=10$

$\dfrac{2v}{g}=10$

$v=\dfrac{10\times g}{2}$

Here, g = 10

The initial velocity is

$v=\dfrac{10\times10}{2}$

$v=50\ m/s$

Hence, The initial velocity is 50 m/s.

3 0

3 years ago

Initial velocity 10 m/s accelerates at 5 m/s for 2 seconds whats the final velocity

stiks02 [169]

Answer:

<em>The final velocity is 20 m/s.</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

$v_f=v_o+at$

The provided data is: vo=10 m/s, $a=5\ m/s^2$ , t=2 s. The final velocity is:

$v_f=10~m/s+5\ m/s^2\cdot 2\ s$

$v_f=20\ m/s$

The final velocity is 20 m/s.

8 0

3 years ago

Most of the Earth's volcanoes occur

timofeeve [1]

Answer:

A.

Explanation:

NEAR THE CENTER OF TECTONIC PLATES.

3 0

3 years ago

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

umka21 [38]

Answer:

Explanation:

s = s₀ + v₀t + ½at²

s = 0 + 0(15) + ½(6)(15²)

s = 675 m

Not sure what the free fall acceleration is needed for, but if the object is dropped from a high enough point, it will travel in 15 seconds

s = ½10(15²) = 2250 m if air resistance is ignored

7 0

3 years ago

A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ

valentinak56 [21]

Explanation:

The given data is as follows.

mass = 0.20 kg

displacement = 2.6 cm

Kinetic energy = 1.4 J

Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

Total energy = Kinetic energy + spring potential energy

= 1.4 J + 2.2 J

= 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So, K.E = Total energy

= 3.6 J

Also, we know that

K.E = $\frac{1}{2}mv^{2}_{m}$

or, v = $\sqrt{\frac{2K.E}{m}}$

= $\sqrt{2 \times 3.6 J}{0.2 kg}$

= $\sqrt{36}$

= 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0

3 years ago