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aniked [119]
2 years ago
9

What inferences can you make about the melting points of the different substances and the motion of their particles, based on th

e data?

Physics
1 answer:
miv72 [106K]2 years ago
7 0
Here’s the answer now hand over the brainilyiest

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When two objects collide their momentum after the collision is explained by what
pashok25 [27]
The statement to every reacting there is, there is a opposite and same reacting. 
hope it helps
8 0
3 years ago
Write an equation that expresses the first law of thermodynamics in terms of heat and work.
OleMash [197]

Answer:

Heat Input = Work Output       (at 100% efficiency)

ΔQ = ΔW

(you cannot get something for nothing)

3 0
2 years ago
a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector
Delicious77 [7]

Answer:

magnitude=34.45 m

direction=55.52\°

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance d between two points:

d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}} (1)

d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}} (2)

d=\sqrt{1186.81 m^{2}} (3)

d=34.45 m (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

tan \theta=\frac{Y2-Y1}{X2-X1}  (5)

tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}  (6)

tan \theta=\frac{24.8}{19.5}  (7)

Finding \theta:

\theta= tan^{-1}(\frac{24.8}{19.5})  (8)

\theta= 55.52\°  (9) This is the direction of the vector

3 0
3 years ago
A 0.5-kilogram ball is thrown vertically upward with an initial kinetic energy of 25 joules. Ap-proximately how high will the ba
tatuchka [14]
It's gravitational potential energy at the top will roughly equal it's kinetic energy when it was released (a little is lost to air resistance).  Note this will assume the release point is zero potential energy.  (we are free to define it that way, just letting you know).  Gravitational potential energy is mgh.
mgh=25J
h=25J/(0.5kg x 9.81m/s^2) = 5.097m
So it goes about 5.1 meters above the point where it was released
4 0
3 years ago
Explain why a Merry-Go-Round and a Ferris Wheel have a constant acceleration when they are moving.
luda_lava [24]

Explanation:What is centripetal acceleration?

Can an object accelerate if it's moving with constant speed? Yup! Many people find this counter-intuitive at first because they forget that changes in the direction of motion of an object—even if the object is maintaining a constant speed—still count as acceleration.

Acceleration is a change in velocity, either in its magnitude—i.e., speed—or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the speed might be constant. You experience this acceleration yourself when you turn a corner in your car—if you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion. What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we'll examine the direction and magnitude of that acceleration.

The figure below shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation—the center of the circular path. This direction is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration

a

c

a

c

​

a, start subscript, c, end subscript; centripetal means “toward the center” or “center seeking”.

5 0
3 years ago
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