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ANEK [815]
3 years ago
5

A point charge (5.0 pC) is located at the center of a spherical surface (radius = 2.0 cm), and a charge of 3.0 pC is spread unif

ormly upon this surface. Determine the magnitude of the electric field 1.0 cm from the point charge.
Physics
1 answer:
Alla [95]3 years ago
8 0

Answer:

E = 0.45 kN/C

Explanation:

From; the question, we can opine that the electric field inside the hollow sphere as a result of the  charge at the surface is zero.

Using the equation:

\mathbf{E = kq/r^2}  to determine the magnitude of the electric field with the radius 1.0 cm from the point charge.

where ;

constant k = 8.98*10^9

point charge  q = 5.0 pC = 5.0 × 10⁻¹²; we have :

\mathbf{E = kq/r^2}

\mathbf{E =\dfrac{8.98*10^9*5*10^{-12}}{1.0^2}}

E = 0.45 kN/C

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In an energy pyramid, the total energy available for consumption from one trophic level to the next one it is
Flura [38]

Answer: Conserved

Explanation:

5 0
3 years ago
A. Which of the four thermal processes describes the pressure-volume relationship at a constant temperature?
lara31 [8.8K]
The thermal process that occurs when pressure and volume are variable, while the temperature remains constant is known as an isothermal process. Due to the change in temperature being zero, the internal energy of the system does not change. Isothermal processes can be carried out adiabatically or non-adiabatically. (An adiabatic process is where the heat energy moving through the boundary of a system is 0)
4 0
3 years ago
A metal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction.
kherson [118]

Answer:

a) k = 120 N / m

, b)    f = 0.851 Hz

, c)  v = 1,069 m / s

, d)  x = 0

, e)  a = 5.71 m / s²

, f)   x = 0.200 m

, g)  Em = 2.4 J

, h) v = -1.01 m / s

Explanation:

a) Hooke's law is

         F = k x

         k = F / x

          k = 24.0 / 0.200

          k = 120 N / m

b) the angular velocity of the simple harmonic movement is

        w = √ k / m

        w = √ (120 / 4.2)

        w = 5,345 rad / s

Angular velocity and frequency are related.

       w = 2π f

        f = w / 2π

        f = 5.345 / 2π

        f = 0.851 Hz

c) the equation that describes the movement is

        x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

        x = 0.2 cos wt

Speed ​​is

       v = dx / dt

       v = -A w sin wt

The speed is maximum for sin wt = ±1

       v = A w

       v = 0.200 5.345

       v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

       x = A cos wt = 0

       x = 0

e) the acceleration is

       a = d²x / dt² = dv / dt

       a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

       a = A w²

        a = 0.2   5.345

        a = 5.71 m / s²

f) the position for this acceleration is

       x = A cos wt

       x = A

       x = 0.200 m

g) Mechanical energy is

        Em = ½ k A²

        Em = ½ 120 0.2²

       Em = 2.4 J

h) the position is

         x = 1/3 A

Let's calculate the time to reach this point

         x = A cos wt

        1/3 A = A cos 5.345t

         t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed ​​is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

      a = -1.91 m / s²

5 0
3 years ago
On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze
sp2606 [1]

Answer:

0.78333 m/s in the opposite direction

1.566 m/s in the same direction

Explanation:

m_1 = Mass of penny = 0.0025 kg

m_2 = Mass of nickel = 0.005 kg

u_1 = Initial Velocity of penny = 2.35 m/s

u_2 = Initial Velocity of nickel = 0 m/s

v_1 = Final Velocity of penny

v_2 = Final Velocity of nickel

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.0025-0.005}{0.0025+0.005}\times 2.35+\frac{2\times 0.5}{0.4005+0.5}\times 0\\\Rightarrow v_1=-0.78333\ m/s

The final velocity of the penny is 0.78333 m/s in the opposite direction

v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.0025}{0.0025+0.005}\times 2.35+\frac{0.005-0.0025}{0.005+0.0025}\times 0\\\Rightarrow v_2=1.566\ m/s

The final velocity of the nickel is 1.566 m/s in the same direction

6 0
3 years ago
1. Cam Newton can sprint 40 meters in 5.79 seconds! How fast can he run?
Vesnalui [34]

This is a Physics question where we need to figure out how many meters Cam can run per second. To figure this out we divide the distance by the change in time.

40/5.79 = 6.9 meters per second approximately.

3 0
3 years ago
Read 2 more answers
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