Which direction do the particles of the medium move as compared to the energy in a transverse wave?

The index of refraction of light varies from color to color. True or False?

lys-0071 [83]

The index of refraction of light varies from color to color. TRUE.

8 0

3 years ago

Even if there were some friction on the ice, it is still possible to use conservation of momentum to solve this problem, but you

hjlf

The problem referred to in this question is missing and it is;

Two hockey pucks of identical mass are on a flat, horizontal ice hockey rink. The red puck is motionless; the blue puck is moving at 2.5 m/s to the left. It collides with the motionless red puck. The pucks have a mass of 15 g. After the collision, the red puck is moving at 2.5 m/s, to the left. What is the final velocity of the blue puck?

Answer:

The condition is that p_f - p_i which is the change in momentum will not be equal to zero but equal to the impulse (Ft).

Explanation:

In the problem described, by inspection, we can say that since there is no friction, we have a closed system and thus momentum is conserved.

Since momentum is conserved, we can say that;

Initial momentum(p_i) = final momentum(p_f)

Now, in this question we are told that some friction wants to be introduced on the ice and it's possible to still use conservation of momentum.

From impulse - momentum theory, we know that;

Impulse = change in momentum

Impulse is zero when no force is acting on the ice and we have; 0 = p_f - p_i

This will yield initial momentum = final momentum.

Now, since a force is applied, we know that impulse is; J = F × t

Thus;

Ft = p_f - p_i

Where F is the force due to friction.

Thus, the condition is that p_f - p_i will not be equal to zero

6 0

3 years ago

Diagram of an atom with labels

lina2011 [118]

See this. I hope you find your answer

6 0

3 years ago

Using Excel, or some other graphing software, plot the values of y as a function of x. (You will not submit this spreadsheet. Ho

Evgesh-ka [11]

Answer:

a) > x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept) x

1.10 0.98

b) $y = 0.98 x +1.10$

And if we compare this with the general model $y = mx +b$

We see that the slope is m= 0.98 and the intercept b = 1.10

Explanation:

Part a

For this case we have the following data:

x: 1,2,3,4,5

y: 1.9,3.5,3.7,5.1, 6

For this case we can use the following R code:

> x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept) x

1.10 0.98

Part b

For this case we have the following trend equation given:

$y = 0.98 x +1.10$

And if we compare this with the general model $y = mx +b$

We see that the slope is m= 0.98 and the intercept b = 1.10

7 0

3 years ago

. A 1.50kg mass on a spring has a displacement as a function of time given by the equation: x(t) = (7.40cm)cos[(4.16s-1)t – 2.42

rusak2 [61]

Answer:

Solution:

we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]

Position, velocity and acceleration are all based on the equation of motion.

The equation represents the position. The first derivative gives the velocity and the 2nd derivative gives the acceleration.

x(t)=8sint

x'(t)=8cost

x"(t)=-8sint

now at time t=2pi/3,

position, x(t)=8sin(2pi/3)=4*squart(3)cm.

velocity, x'(t)=8cos(2pi/3)==4cm/s

acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2

so at present the direction is in y-axis.

5 0

3 years ago