What is a continuous range of a single feature such as a wave lengt

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

AysviL [449]

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

7 0

3 years ago

The formula shown below is used to calculate the energy released when a specific quantity of fuel is burned. Calculate the energ

olya-2409 [2.1K]

Answer: 8400 J

Explanation:

The formula referenced in the question is:

$Q=m. c. \Delta T$

Where:

$Q$ is the thermal energy

$m=100g \frac{1 kg}{1000 g}=0.1 kg$ is the mass of the water sample

$c=4200 \frac{J}{kg\°C}$ is the specific heat capacity of water

$\Delta T=20\°C$ is the variation in temperature

Solving:

$Q=(0.1 kg)(4200 \frac{J}{kg\°C})(20\°C)$

$Q=8400 J$ This is the thermal energy released

8 0

3 years ago

4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest

yarga [219]

All the answers are:

a) The time that will it take to rise to the highest point is 3.06 seconds.

b) The ball will rise to a height of 45.87 meters.

c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.

The time when the ball has 10 m/s downward is 1.02 seconds.

d) The displacement of the ball will be zero at 6.12 seconds.

e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.

f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.

g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Let's calculate the values for each case.

a) At the highest point, the final velocity is 0, so we can use the following equation.

$v_{f}=v_{i}-gt$ (1)

Where:

v(i) is the initial velocity
v(f) is the final velocity
g is the acceleration due to gravity (9.81 m/s²)

We know that v(i) = 30 m/s.

$0=30-9.81t$

Solve it for t:

$t=3.06\: s$

Hence, the time is 3.06 s.

b) At the highest point, the final velocity is 0, so we can use the following equation.

$v_{f}^{2}=v_{i}^{2}-2gh$ (2)

$0=v_{i}^{2}-2gh$

We know that the initial velocity is 30 m/s.

$0=30^{2}-2gh$

Solving it for h we have:

$h=\frac{30^{2}}{2*9.81}$

$h=45.87 \: m$

Then, the height is 45.87 m.

c) Using equation (1) we can find the time (t).

$10=30-(9.81t)$

So, the time elapsed to get 10 m/s is:

$t_{upward}=2.04\: s$

We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.

$t_{upward}=2.04+t$

$t=1.02\: s$

This is the time when the ball has 10 m/s downward.

Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.

d) It will be when the ball returns to the ground.

$t=2t_{upward}$

$t=2*3.06$

$t=6.12\: s$

The displacement will be zero after 6.12 s.

e) Here we need to find the time when v(f) is 15 m/s

$15=30-gt$

$t=\frac{15}{9.81}$

$t=1.53\: s$

The time when the v(f) is 15 m/s is 1.53 s.

f) Here, we need to find t when h = 45.87/2 m = 22.94 m

We can use the next equation:

[tex]h=v_{i}t-0.5gt^{2}/tex]

[tex]22.94=30t-0.5*9.81*t^{2}/tex]

Solving this quadratic equation, t will be:

[tex]t=0.90\: s/tex]

Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.

g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Learn more about vertical motion here:

brainly.com/question/13966860

I hope it helps you!

3 0

2 years ago

If a 100N weight sits in an express elevator what will be the weight when the elevator is traveling at a constant speed

yarga [219]

This is a "trick" question.

If the elevator is traveling at constant speed, it means it is at rest. This means anything inside the elevator traveling at constant speed, weights the same as in an elevator not moving -also at rest-.

So the 100N weight's weight doesn't change in an elevator traveling at constant speed.

4 0

3 years ago

If an astronaut has a mass of 16 Kg on Earth, what would be his mass on the moon and on the space station

nlexa [21]

Answer:

The astronaut's mass is 16 kg.

Explanation:

Mass can be defined as a measure of the amount of matter an object or a body comprises of. The standard unit of measurement of the mass of an object or a body is kilograms.

Irrespective of the location of an object or a body at a given moment in time, the mass (amount of matter that they're made up of) is constant. This ultimately implies that, whether you're in the moon, space, earth or any other place, your mass remains the same (constant).

Therefore, if an astronaut has a mass of 16 Kg on Earth, his mass on the moon and on the space station would remain the same, as his original mass of 16 Kg because mass is indestructible.

3 0

3 years ago