In a line graph the dependent variable is plotted on the

What types of natural phenomena could serve as time standards?

Delvig [45]

Answer: The rising and setting of the Sun.

The appearance and disappearance of the Moon.

Changes in season.

Appearance and disappearance of the Stars.

Explanation: The rising and setting of the Sun helps can be used as a standard to determine time since the rissing and setting of the Sun occurs at similar time and interval in a given geographical ocation.

The appearance and disappearance of the Moon which occurs during the night and early hours of the morning is similar in a given geographical locationation so it can be used as a standard for Time.

Change in season is another natural phenomenon which occurs at similar times in a given geographical location like Winter/summer, rainy/dry etc can be used as standards for time.

The appearance and disappearance of the Stars can also b used as standards for time in a given geographical location.

5 0

4 years ago

A 5200w electric oven

OLga [1]

Answer:

104 kWh

Explanation:

To calculate the kWh you multiply power with the amount of hours.

So for this we take 5200 W × 20 = 104 000 Wh

Then change Wh to kWh so that makes 104 kWh

8 0

2 years ago

When a perfume bottle is opened, some liquid changes to gas and the fragrance spreads around the room. Which sentence explains t

shutvik [7]

The molecules of the perfume diffuse through the room at a certain critical concentration and later are detected by our noses.

4 0

3 years ago

Two samples of water are mixed together.

horrorfan [7]

Let the cold water go up x degrees.

Let the hot water go down 100 - x degrees.

The formula for heat exchange is m*c*delta t

Givens

Ice

deltat = x

m = 0.50 kg

c = 4.18

Hot water

deltat = 100 - x

m = 1.5 kg

c = 4.18

Formula

The heat up = heat down

0.50 * c * x = 1.5 * c * (100 - x) Divide both sides by c

Solution

0.50 *x = 1.5*(100 - x) Remove the brackets.

0.5x = 150 - 1.5x Add 1.5x to both sides.

0.5x + 1.5x = 150 - 1.5x + 1.5x Combine like terms

2x = 150 Divide by 2

x = 75

Answer

A

6 0

4 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

b)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

c)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is: