All categories

4 years ago

What is the relationship between resistance and current in a circuirt?

1 answer:

7 0

The relationship between resistance and current in a circuit is that if either resistance or current increase, then voltage increases; or vice versa for that matter. Another part in their relationship, is that resistance multiplied by current equals voltage so in order to calculate voltage, you must have resistance and current. I hope this helps!

You might be interested in

A man pushes on a trunk with a force of 250 newtons. The trunk does not move. How much positive work is done on the trunk?

Digiron [165]

Answer:

F is 250 N

d is 0 m

F x d

=250 x 0

The answer is 0.0 J.

3 0

4 years ago

While traveling along a highway a driver slows from 24 m/s to 15 m/s in 12 seconds. What is the automobile's acceleration? (Reme

slava [35]

Answer:

Acceleration = -0.75 m/s²

Explanation:

Given the following data;

Initial velocity = 24 m/s

Final velocity = 15 m/s

Time = 12 seconds

To find the acceleration of the automobile;

Acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac {final \; velocity - initial \; velocity}{time}$

Substituting into the formula, we have;

$Acceleration = \frac{15 - 24}{12}$

$Acceleration = \frac{-9}{12}$

Acceleration = -0.75 m/s²

Therefore, the automobile is decelerating because its final velocity is lesser than its initial velocity, leading to a negative value.

5 0

3 years ago

Directions: Read the following excerpts carefully, then identify whether they employ informative, persuasive, or argumentative w

Savatey [412]

Hi there!

Informative writing has the intent to inform or educate us on a particular topic or event. It gives us more information and insight onto something.

Persuasive writing has the intent of convincing us to believe in a certain idea or to perform a certain action. For instance, advertisements have a persuasive intent; they are persuading us to buy a product or service.

Argumentative writing is similar to persuasive writing in the sense that they are persuading us to believe a certain idea. However, they are often based on logic and fact rather than opinions.

Let's look at the first excerpt.

<em>This morning at 9 a.m., a school bus collided with a car at the intersection of Osmena and Cabrera streets. The passengers were not injured, but the medical personnel checked each student as well as the driver before they were transported to their school.</em>

This text doesn't try to convince us in believing something. It doesn't argue anything and it only tries to give us more insight onto the event, which is a car accident. No opinions are stated and only events are given.

Therefore, this excerpt uses an informative writing technique.

4 0

2 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

4 0

3 years ago

Options: A.) 10 N B.) 15 N C.) 25 N D.) 35N

Lady_Fox [76]

Given:

F_gravity = 10 N

F_tension = 25 N

Let's find the net centripetal force exterted on the ball.

Apply the formula:

$\sum ^{}_{}F_{\text{net}}=F_1+F_2=F_{centripetal}$

From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.

Hence, the tensional force is positive while the gravitational force is negative.

Thus, we have:

$F_{\text{net}}=F_{\text{centripetal}}=F_{tension}-F_{gravity}=25N-10N=15N$

Therefore, the net centripetal force exterted on the ball is 15 N.

ANSWER:

15 N

7 0

1 year ago