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dimulka [17.4K]
3 years ago
14

Las placas de un capacitor de placas paralelas están en forma horizontal. En el espacio se coloca una losa de material dieléctri

co con que llena la mitad inferior del espacio entre las placas. ¿Cuál es la nueva capacitancia resultante
Physics
1 answer:
snow_lady [41]3 years ago
5 0
I’m sorry I do not understand what this says
You might be interested in
Which two types of simple machines can be found in a bicycle? lever and wheel and axle pulley and wedge wedge and wheel and axle
insens350 [35]

Answer:

A). A Lever and Wheel and Axle.

Explanation:

1. A wheel axle is most obvious, as the wheels are connected via a chain and axle with spokes. This helps keep the wheels moving forward with less effort than walking.

2. A Lever is used to help Balance you as you ride the bike. Without it, you would tip over, and lose all the momentum you had gained up until that point.

3. (I took the test on e2020)

5 0
4 years ago
Read 2 more answers
Pls help with this question i’ll give brainliest
bearhunter [10]

Answer:

5m/s to the right

Explanation:

Momentum = mass * velocity

Momentum before = momentum after

m₁u₁+m₂u₂ = m₁v₁+m₂v₂

3000*10 + 1000*0 = 3000*v₁ + 1000*15

30000-15000=3000v₁

15000=3000v₁

v₁=5m/s to the right (to the right because answer is positive)

6 0
3 years ago
The total mass of the wheelbarrow and the road is 80 kg calculate the weight of the wheelbarrow and the road
Alja [10]

Answer:

The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.

Explanation:

Given that,

The total mass of the wheelbarrow and the road is 80 kg.

The weight of an object is given by :

W = mg

where

g is acceleration due to gravity

So,

W = 80 × 9.8

= 784 N

So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.

7 0
3 years ago
9) Of all the types of light the Sun gives off, it emits the greatest amount of light at visible wavelengths of light. If the Su
erastova [34]

Answer:

*  most of the emission would be in the infrared part, the visible radiation would be very small.

*total intensity of the semition decreases that the intensity depends on the fourth power of the temperature

Explanation:

The radiation emitted by the Sun is approximately the radiation of a black body, if the Sun were to cool, the maximum emission wavelength changes

          λ T = 2,898 10⁻³

          λ = 2,898 10⁻³ / T

if the temperature decreases the maximum wavelength the greater values ​​are moved, that is to say towards the infrared. Therefore the emission curve also moves, in this case most of the emission would be in the infrared part, the visible radiation would be very small.

Furthermore, the total intensity of the semition decreases that the intensity depends on the fourth power of the temperature according to Stefan's law

           P = σ A eT⁴

7 0
3 years ago
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
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