All of the following are examples of positive direction except:

DiKsa [7]

This has to do with independent and dependent variables. The independent variable is what affects the dependant variable, so the question is, does a students mark on a test affect their hours of sleep or does their hours of sleep affect their test results? Which one makes more sense? I would say the latter.

6 0

2 years ago

A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 2.50 10

Katen [24]

Answer:

B = 6.18 10⁻⁶ T

the magnetic field is in the negative direction of the y axis

Explanation:

The magnetic force is given by

F = q v x B

as in the exercise indicate that the velocities perpendicular to the magnetic field,

F = q v B

Newton's second law is

F = m a

let's substitute

q v B = m a

B = m a / q v

let's calculate

B = 9.1 10⁻³¹ 2.50 10¹³ / (1.6 10⁻¹⁹ 2.30 10⁷)

B = 6.18 10⁻⁶ T

The direction of the field can be obtained with the right hand rule, where the thumb points in the direction of the velocity, the fingers extended in the direction of the magnetic field and the palm in the direction of the force for a positive charge.

In the exercise indicate that the velocity is the z axis

the acceleration and therefore the force in the x axis

therefore the magnetic field is in the negative direction of the y axis

7 0

2 years ago

Instead, suppose that it was ﬁred upward at 60◦ with respect to a horizontal line. Then its horizontal component of velocity is

larisa [96]

Answer:

50 m/s

Explanation:

Angle = 60 degree

Horizontal component of velocity = 50 m/s

A projectile motion is the motion of an object in two dimensions under the influence of gravity.

In this case, the object has no acceleration along horizontal direction, it has acceleration in vertical direction which is equal to the acceleration due to gravity of earth.

When the projectile reaches at the maximum height it travels only along the horizontal and thus it has only horizontal velocity at that instant.

Thus, the velocity of teh projectile at maximum height is same as horizontal component of velocity that meas 50 m/s.

5 0

2 years ago

How is newtons third law of motion demonstrated on a roller coaster?

butalik [34]

<span>Newton's Third Law of Action-Reaction is that for each and every action that happens, there is an equal and opposite reaction to it. In the scenario of a roller coaster, this is when you push down on the seat of the roller coaster as it flies along and the seat pushes back against you.</span>

8 0

2 years ago

I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......

pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100 $cm^{2}$ x 10cm = 1000 $cm^{3}$

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

$\frac{1g}{cm^{3} }$ x 1000 $cm^{3}$ = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ $\frac{8g}{cm^{3} }$ = 2.5 $cm^{3}$

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 $cm^{3}$ overflowed. So now we the same process as in number a) just with a few adjustments.

$\frac{1g}{cm^{3} }$ x (1000 $cm^{3}$ - 2.5 $cm^{3}$ ) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0

2 years ago