In an experiment what are all the parts of an experiment that remain unchanged are called

Which most likely indicates a chemical change has occurred?

Law Incorporation [45]

green liquid becoming a red liquid

4 0

2 years ago

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The pressure of air pushing down on a house roof is extremely large. Why is the roof not crushed?

Degger [83]

There is still air inside of a house, which is pushing the roof upwards, so the forces are equal and the roof is not crushed.

7 0

3 years ago

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is . Find the time i

a_sh-v [17]

Answer:

a) It took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

b) The minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensitie. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies:

Gamma rays

X-rays

Ultraviolet rays

Visible region

Infrared

Microwave

Radio waves.

Any radiation that belongs to electromagnetic spectrum has a speed in vacuum of $3x10^{8}m/s$ .

a) Find the time it took for his voice to reach the Earth via radio waves.

To know the time that took for Neil Armstrong's voice to reach the Earth via radio waves, the following equation can be used:

$c = \frac{d}{t}$ (1)

Where v is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

$t = \frac{d}{c}$ (2)

The distance from the Earth to the Moon is $3.85x10^{8} m$ , therefore.

$t = \frac{3.85x10^{8} m}{3x10^{8}m/s}$

$t = 1.28s$

Hence, it took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

b) Determine the minimum time that will be required for a message from Mars to reach the Earth via radio waves.

The distance from the Earth to the Mars at its closest approach is $5.76x10^{10}m$ , therefore.

$t = \frac{5.76x10^{10}m}{3x10^{8}m/s}$

$t = 192s$

Hence, the minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

3 0

3 years ago

Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at

sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

E = k*Q / r^2

Where, k: Coulomb's Constant

Q: The charge of particle

r: The distance from source

- The Electric Field due to charge 1:

E_1 = k*Q_1 / r^2

E_1 = k*Q / (2*a)^2

E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

E_2 = k*Q_2 / r^2

E_2 = k*Q_2 / (13*a)^2

E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

E_net = E_1 + E_2

2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

2Q = Q / 4 + Q_2 / 169

Q_2 = 295.75*Q

2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

2Q = Q / 4 - Q_2 / 169

Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0

2 years ago

Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary

kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

$I=mr^2$

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

$I_A=1 \times 0.5^2\ kg.m^2$

$I_A=0.25\ kg.m^2$

T= I α

0.5= 0.25 α

$\alpha = 2\ rad/s^2$

For Wheel B

m= 1 kg

d= 2 m,r=1 m

$I_B=1 \times 1^2\ kg.m^2$

$I_B=1 \ kg.m^2$

Given that angular acceleration is same for both the wheel

$\alpha = 2\ rad/s^2$

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0

3 years ago

In an experiment what are all the parts of an experiment that remain unchanged are called​

In an experiment what are all the parts of an experiment that remain unchanged are called