All categories

3 years ago

I would like to know why this is the correct answer

Physics

1 answer:

Marta_Voda [28]3 years ago

4 0

Answer:

see below

Explanation:

First, the obvious, as you press the gas pedal harder the acceleration goes up as well. Conversely, is you do not press the pedal, you will not accelerate. This determines that is I press the gas pedal, it will CAUSE the car to accelerate. This proves causation.

Now, correlation. The definition of correlation in statistics is any statistical relationship between two random variables or data. This simply means that these two events are connected to one another. A POSITIVE correlation is when two correlated events move in the same direction as one another. I have added a graph to help visualize this. In this problem as the gas is pressed harder, the acceleration increases. If the pressure on the pedal was decreased, then the acceleration also decreases. If the pressure on the pedal is constant, the the acceleration is constant.

I hope this helps!

You might be interested in

How do you know that potassium an alkali metal is highly reactive

Pavel [41]

Potassium is in the most reactive group of elements, the alkali metals, but it's not the most reactive metal within the group. The alkali metals, Group 1A, are the most reactive metals because they have one valence or outer electron. They lose this electron very easily, forming ions with a charge of +1.

7 0

3 years ago

nikitadnepr [17]

The answer is
B:because

8 0

3 years ago

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

3 0

3 years ago

Kathy 82 kg performer standing on a diving board at the carnival dive straight down into a small pool of water. Just before stri

mixas84 [53]

Solution :

Given weight of Kathy = 82 kg

Her speed before striking the water, $V_o$ = 5.50 m/s

Her speed after entering the water, $V_f$ = 1.1 m/s

Time = 1.65 s

Using equation of impulse,

$dP = F \times dT$

Here, F = the force ,

dT = time interval over which the force is applied for

= 1.65 s

dP = change in momentum

dP = m x dV

$= m \times [V_f - V_o]$

= 82 x (1.1 - 5.5)

= -360 kg

∴ the net force acting will be

$F=\frac{dP}{dT}$

$F=\frac{-360}{1.65}$

= 218 N

8 0

3 years ago

You have two points in a soil. Point A is at 75 cm; point B is at 25 cm above the reference. The capillary potential energy at p

user100 [1]

Answer:

The potential energy at point A is 17.1675 J

Explanation:

The capillary potential is the work expended to bring up a unit mass of liquid to a point in a capillary region from a level liquid surface. It is the capillary potential that facilitates the movement of moisture within soil capillaries

In meteorology it is used to describe the level of saturated soil above the water table

Potential energy is the energy inherent in a body by virtue of its position, therefore the potentials of both point A and B are

Point A, elevation = 75 cm capillary potential = -100 cm

Point B, elevation = 25 cm capillary potential = -200 cm

The total potential energy at point A is

Elevation above reference - capillary potential =75-(-100) = 175 cm

which gives per unit mass

PE = m × g × h = 1 kg × 9.81 m/s ² × 1.75 m = 17.1675 kg·m²/s² = 17.1675 J

8 0

3 years ago