Porfavor ayuden que es pa mañana :v

A long straight wire has fixed negative charge with a linear charge density of magnitude 4.6 nC/m. The wire is to be enclosed by

Semmy [17]

Answer: -33.3 * 10^9 C/m^2( nC/m^2)

Explanation: In order to solve this problem we have to use the gaussian law, the we have:

Eoutside =0 so teh Q inside==

the Q inside= 4.6 nC/m*L + σ *2*π*b*L where L is the large of the Gaussian surface and b the radius of the shell.

Then we simplify and get

σ= -4.6/(2*π*b)= -33.3 nC/m^2

7 0

3 years ago

Mars has twice the mass of Mercury and is 4 times further away from the Sun. Calculate theratio of the gravitational force from

svetoff [14.1K]

Answer:

F(Mars) = 2 G m M / (4 R)^2 force of Sun on Mars

F(Merc) = G m M / R^2 force of force of Sun on Mercury

R = distance of Sun from Mercury, m = mass of Mercury

F(Merc) / F(Mars) = 4^2 / 2 = 8

6 0

3 years ago

My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin

Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0 ⇒ K₀

and h₁ = 0 ⇒ U₁ = 0

we get

U₀ = K₁

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0

3 years ago

What evidence do you that suggest water waves are transverse wave

svp [43]

Answer:

If you throw a pebble into a pond, ripples

spread out from where it went in. These

ripples are waves travelling through the

water. The waves move with a transverse

motion.

Explanation:

5 0

3 years ago

3 Draw energy transfer diagrams

timofeeve [1]

Answer:

The outline of the energy transfer are;

a) Kinetic energy → Clockwork spring → Potential energy

b) Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run

c) Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy

Please find attached the drawings of the energy transfer created with MS Visio

Explanation:

The energy transfer diagrams are diagrams that can be used to indicate the part of a system where energy is stored and the form and location to which the energy is transferred

a) The energy transfer diagram for the winding up a clockwork car is given as follows;

Mechanical kinetic energy is used to wind up (turn) the clockwork car such that the kinetic energy is transformed into potential energy and stored in the wound up clockwork as follows;

Kinetic energy → Clockwork spring → Potential energy

b) Letting a wound up clockwork car run results in the conversion of mechanical potential energy into kinetic (energy due tom motion) energy as follows;

Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run

c) The energy stored in the battery of a battery powered car is chemical potential energy. When the battery powered car runs, the chemical potential energy produces an electromotive force which is converted into kinetic energy as electric current flows from the batteries

Therefore, we have;

Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy

6 0

3 years ago