Question number 6 :

Is my answer right?? Plzz help

Answers

Answer 1

Answer: No. You have the units wrong.

Example with numbers:

When you multiply (1/9) by 2, you get 2/9 .

With units:

When you multiply (m/s) by (kg), you get (kg-m/s).

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Draw a circuit of a kettle

Answers

Answer:

That is a standard circuit diagram of a kettle

To halt the decline in biodiversity, we must do which of the following?

Answers

adopt ecological conservation practices

In an engine governor, the two spheres (total mass of 1.0kg) are at 0.05m and rotating at 37rad/s If the engine increases the angular velocity to 58rad/s what is the new location of the spheres?

Answers

Here we can say that there is no external torque on this system

So here we can say that angular momentum is conserved

so here we will have

$I_1\omega_1 = I_2\omega_2$

now we have

$I_1 = mr^2$

$I_1 = (1kg)(0.05^2)$

$I_1 = 25* 10^(-4) kg m^2$

similarly let the final distance is "r"

so now we have

$I_2 = mr^2$

$I_2 = 1r^2$

now from above equation we have

$(25* 10^(-4))37 = (r^2)(58)$

$r = 0.04 m$

so final distance is 0.04 m between them

How can i find the acceleration?(rope and grinder have no weight) *** sorry for my english

Answers

The main formula to be used here is

Force = (mass) x (acceleration).

We'll get to work in just a second. But first, I must confess to you that I see
two things happening here, and I only know how to handle one of them. So
my answer will be incomplete, but I believe it will be more reliable than the
first answer that was previously offered here.

On the right side ... where the 2 kg and the 3 kg are hanging over the same
pulley, those weights are not balanced, so the 3 kg will pull the 2kg down, with
some acceleration. I don't know what to do with that, because . . .

At the same time, both of those will be pulled up by the 10 kg on the other side
of the upper pulley.

I think I can handle the 10 kg, and work out the acceleration that IT has.

Let's look at only the forces on the 10 kg:

-- The force of gravity is pulling it down, with the whatever the weight of 10 kg is.

-- At the same time, the rope is pulling it UP, with whatever the weight of 5 kg is ...
that's the weight of the two smaller blocks on the other end of the rope.

So, the net force on the 10 kg is the weight of (10 - 5) = 5 kg, downward.

The weight of 5 kg is (mass) x (gravity) = (5 x 9.8) = 49 newtons.

The acceleration of 10 kg, with 49 newtons of force on it, is

Acceleration = (force) / (mass) = 49/10 = 4.9 meters per second²

Which example describes constant acceleration due ONLY to a change in direction?a. increasing speed while traveling around a curve
b. an object at rest
c. traveling around a circular track
d. an object in free fall

Answers

Answer:

c. traveling around a circular track

Explanation:

Acceleration is defined as the rate of change of velocity per unit time:

$a=(\Delta v)/(\Delta t)$

However, velocity is a vector quantity. This means that acceleration can be due to:

- A change in the speed of an object, or

- A change in the direction of the object

Let's analyze each choice:

a. increasing speed while traveling around a curve --> in this case, both speed and direction are changing (the direction is changing since the object is moving around a curve), so this is not the correct choice

b. an object at rest --> here neither the speed nor the direction are changing (in fact, the object is not accelerating), so this is not the correct choice

c. traveling around a circular track --> here the direction is changing (because the track is circular), while we don't know anything about the speed. If we assume the speed to be constant, then this is the correct choice

d. an object in free fall --> here the speed is changing, while the direction not, so this is not the correct choice

'Traveling around a circular track' can be a description of constant
acceleration due only to changes in direction. But if it is, then the
progress around the circular track must be at constant speed.

The galactic center of the milky way is located in the direction of the constellation?A. Cassiopeia
B. Aquarius
C. centaurus
D. Sagittatius

How do stars locate in the galactic disk orbit the galactic center?

A. they do not orbit the galactic center.
B. they move in the circular orbits around the galactic center.
C. they bob up and down as they orbit the galactic center.
D. they orbit in elliptical orbits with random orientations.

Answers

Final answer:

The center of the Milky Way is in the direction of the constellation Sagittarius. The stars in the galactic disk move in a rosette pattern as they orbit the galactic center, which is neither completely circular nor random elliptical orbits.

Explanation:

The galactic center of the Milky Way is located in the direction of the constellation Sagittarius. The stars in the galactic disk orbit the galactic center much like how planets in our solar system orbit the sun. The movement of these stars is not completely circular, nor do they bob up and down or orbit randomly in elliptical paths. The actual orbit is more like a rosette pattern, not perfect circles but also not random orientations.

Learn more about Galactic center here:

brainly.com/question/32819189

#SPJ6

Final answer:

The galactic center of the Milky Way is located in the direction of the constellation Sagittarius. Stars in the galactic disk orbit the galactic center in circular orbits.