Answer:
A. The particles will begin to move enough
that they slide past each other.
Explanation:
When the plastic cup is heated, the Kinetic energy of its particles starts increasing. As the temperature rises, the kinetic energy keeps increasing. With the increase of K.E, the particles start moving faster and faster. When the temperature finally reaches the melting point, the K.E of the molecules is enough to break the bonds and slide past each other.
Complete Question
A wave is described by y(x,t) = 0.1 sin(3x + 10t), where x is in meters, y is in centimetres and t is in seconds. The angular wave frequency is
Answer:
The value is 
Explanation:
From the question we are told that
The equation describing the wave is y(x,t) = 0.1 sin(3x + 10t)
Generally the sinusoidal equation representing the motion of a wave is mathematically represented as

Where w is the angular frequency
Now comparing this equation with that given we see that

Lindsay has to fly this plane towards this direction [W 12.5° S] to get to Hamilton.
From this question, the plane is still up in the air.
We have wind blowing in [W 60° N ]
To solve the problem we have to make use of the sine rule

We put the values in the equation, we have:
50/Sinθ = 200/sin60°
The next step is to cross multiply
50 x sin60° = 200Sinθ
50 x 0.8660 = 200sinθ
We make Sin θ the subject
Sine θ = 43.30/200
sine θ = 0.2165
we find the value of θ
θ = sine⁻¹(0.2165)
θ = 12.50
So Lindsay has to fly this plane towards this direction
[W 12.5° S]
Here is a similar question brainly.com/question/13338067?referrer=searchResults
Answer:
When have passed 3.9[s], since James threw the ball.
Explanation:
First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.
We'll use the kinematics equations to find these two unknowns.
![y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\](https://tex.z-dn.net/?f=y%3Dy_%7B0%7D%20%2Bv_%7B0%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cy%3D%20elevation%20%5Bm%5D%5C%5Cy_%7B0%7D%3Dinitial%20height%20%5Bm%5D%5C%5Cv_%7B0%7D%3D%20initial%20velocity%20%5Bm%2Fs%5D%20%3D41.67%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20passed%20%5Bs%5D%5C%5Cg%3D%20gravity%20%5Bm%2Fs%5E2%5D%3D9.81%5Bm%2Fs%5E2%5D%5C%5CNow%20replacing%3A%5C%5Cy%3D0%2B41.67%20%2A%282%29-%5Cfrac%7B1%7D%7B2%7D%20%2A%289.81%29%2A%282%29%5E%7B2%7D%20%5C%5C%5C%5Cy%3D63.72%5Bm%5D%5C%5C)
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can find the velocity after 2 seconds.
![v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bo%7D%20%2Bg%2At%5C%5Creplacing%3A%5C%5Cv_%7Bf%7D%20%3D41.67-%289.81%29%2A%282%29%5C%5C%5C%5Cv_%7Bf%7D%3D22.05%5Bm%2Fs%5D)
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

As we can see the equation is based on Time (t).
Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t
![y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%20%2Bv_%7Bo%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cv_%7Bo%7D%20%3D55.56%5Bm%2Fs%5D%20%3D%20initial%20velocity%5C%5Cy_%7Bo%7D%20%3D0%5Bm%5D%5C%5Cnow%20replacing%5C%5C63.72%20%2B22.05%20%2At-%284.905%29%2At%5E%7B2%7D%20%3D0%20%2B55.56%20%2At-%284.905%29%2At%5E%7B2%7D%20%5C%5C63.72%20%2B22.05%20%2At%20%3D0%20%2B55.56%20%2At%5C%5C63.72%20%3D%2033.51%2At%5C%5Ct%3D1.9%5Bs%5D)
Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.
Time = 2 + 1.9 = 3.9[s]
Answer:
acceleration 8 km/h/s south
Explanation:
First of all, let's remind that a vector quantity is a quantity which has both a magnitude and a direction.
Based on this definition, we can already rule out the following two choices:
distance: 40 km
speed: 40 km/h
Since they only have magnitude, they are not vectors.
Then, the following option:
velocity: 5 km/h north
is wrong, because the car is moving south, not north.
So, the correct choice is
acceleration 8 km/h/s south
In fact, the acceleration can be calculated as

where
v = 40 km/h is the final velocity
u = 0 is the initial velocity
t = 5 s is the time
Substituting,

And since the sign is positive, the direction is the same as the velocity (south).