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rjkz [21]
3 years ago
9

A solid spherical conductor has a radius of 12 cm. The electric field at 24 from the center of the sphere has a magnitude of 640

N/C. What is the charge density (in C/m2) on the sphere
Physics
1 answer:
hoa [83]3 years ago
4 0

Answer:

Charge density on the sphere = 2.2 × 10⁻⁸ C/m²

Explanation:

Given:

Radius of sphere (r) = 12 cm = 0.12 m

Distance from the electric field R = 24 cm = 0.24 m

Magnitude (E) = 640 N/C

Find:

Charge density on the sphere

Computation:

Charge on the sphere (q) = (1/K)ER²            (K = 9 × 10⁹)

Charge on the sphere (q) = [1/(9 × 10⁹)](640)(0.24)²

Charge on the sphere (q) = 4 × 10⁻⁹ C

Charge density on the sphere = q / [4πr²]

Charge density on the sphere = [4 × 10⁻⁹] / [4(3.14)(0.12)²]

Charge density on the sphere = [4 × 10⁻⁹] / [0.18]

Charge density on the sphere = 2.2 × 10⁻⁸ C/m²

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. A 2.0-kg block is on a perfectly smooth ramp that makes an angle of 30° with the horizontal. (a) What is the block’s accelerat
AfilCa [17]

Answer:

a) a = 4.9 m / s²,  N = 16.97 N   and b)   F = 9.8 N

Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

        sin 30 = Wx / W

        cos 30 = Wy / W

        Wx = W sin30

        Wy = W cos 30

Let's write the equations on each axis

X axis

        Wx = ma

Y Axis  

       N- Wy = 0

       N = Wy = mg cos 30

       N = 2.0 9.8 cos 30

       N = 16.97 N

We calculate the acceleration

       a = Wx / m

       a = mg sin 30 / m

       a = g sin 30

       a =9.8 sin 30

       a = 4.9 m / s²

b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

      F -Wx = 0

      F = Wx

      F = m g sin 30

      F = 2.0 9.8 sin 30

      F = 9.8 N

5 0
3 years ago
I can't seem to get the right angular acceleration and also not sure how to do part b. Help will be much appreciated.
Tcecarenko [31]

Answer:It’s 5 I believe

Explanation: it says to round to the nearest thousandths, so it’ll be 5.

8 0
3 years ago
A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.
kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, h_{max}, is given by the following equation;

h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}

Therefore, substituting the known values for the pebble, we have;

h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439

Therefore, the maximum height of the pebble projectile, h_{max} ≈ 6.4 m.

3 0
3 years ago
A gas is compressed at constant temperature from a volume of 5.68 L to a volume of 2.35 L by an external pressure of 732 torr. C
Vlada [557]

Answer: The work done in J is 324

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

W=-P\Delta V=-P(V_2-V_1)

W = amount of work done = ?

P = pressure = 732 torr = 0.96 atm    (760torr =1atm)

V_1 = initial volume = 5.68 L

V_2 = final volume = 2.35  L

Putting values in above equation, we get:

W=-0.96atm\times (2.35-5.68)L=3.20L.atm

To convert this into joules, we use the conversion factor:

1L.atm=101.33J

So, 3.20L.atm=3.20\times 101.3=324J

The positive sign indicates the work is done on the system

Hence, the work done for the given process is 324 J

5 0
3 years ago
you ride a bike for 2 hours at 25 km/hr then 3 hours at 34 km/hr. what is your average velocity (in km/hr)?
mamaluj [8]
<h3>Answer:</h3>

30.4 km/hr

<h3>Explanation:</h3>

<u>We are given</u>;

  • Speed in the first 2 hours as 25 km/hr
  • Speed in the next 3 hours as 34 km/hr

We are required to determine the average velocity in km/hr

  • To get the average velocity we divide total distance by total time.
  • Thus, we need to determine the total distance

Distance = Speed × time

Distance covered in the first 2 hours;

               = 25 km/hr × 2 hours

               = 50 km

Distance in the next 3 hours

                = 34 km/hr × 3 hours

                = 102 km

Therefore, total distance = 50 km + 102 km

                                        = 152 km

Total time = 2 hrs + 3 hrs

                = 5 hours

Therefore;

Average speed = 152 km ÷ 5 hours

                          = 30.4 km/hr

Thus, the average speed is 30.4 km/hr

4 0
3 years ago
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