at the beach, describe where or when you would see a wavefronts. How are wave fronts useful to surfers?

muminat

Power P is the rate at which energy is generated or consumed and hence is measured in units that represent energy E per unit time t. This is:

P = E/t

Solving for t:

t = E/P

t = 6007 J / 500 W

t = 12.014 s

7 0

3 years ago

Name two examples of deposition.

Rashid [163]

The reverse of deposition is sublimation and hence sometimes deposition is called desublimation. One example of deposition is the process by which, in sub-freezing air, water vapor changes directly to ice without first becoming a liquid.

8 0

4 years ago

Read 2 more answers

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

(c) 2.86s

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

<u>The answer for (a) is 1.6m</u>

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }

(b) At what vertical distance above the surface of the water is the diver just then?

The time is the same as in the last question: $t=0.8s$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)$

$y(0.8)=10m-\frac{1}{2}(6.2784m)$

$y(0.8)=10m-(3.1392)$

$y(0.8)=6.86m$

<u>the answer to (b) is 6.86m</u>

(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using $y(t)=y_{0}-\frac{1}{2}gt^2$

if $y(t)=0$

$0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t$

and since $g=9.81m/s^2$

$t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s$

At $t=1.43s$ the car hits the water, so the horizontal distance at that time is:

$x(t)=x_{0}+v_{0}t$

$x(1.43)=(2m/s)(1.43s)=2.86m$

<u>the answer to (c) is 2.86s</u>

6 0

3 years ago

A farsighted girl has a near point at 2.0 m but has forgotten her glasses at home. The girl borrows eyeglasses that have a power

lisov135 [29]

Find the pictures in the attachment

6 0

3 years ago

Someone answer these questions PLEASEE

lara31 [8.8K]

7. First write down all the known variables while separating the values for each direction:

x-direction:

vix = 20m/s

vfx = 20m/s

x = 39.2m

y-direction:

viy = 0m/s

ay = -9.8m/s^2

y = ?

Based on the knowns, the first step is to calculate the time of flight from the x-direction as it will be the same as value for the y-direction. Find the correct kinematic equation to do so:

x = (1/2)(vix+vfx)t

(39.2) = (1/2)(20+20)t

1.96s = t

Now that we have the time of flight, we can use the kinematic equation that will relate the known variables in the y-direction:

y = viy*t + (1/2)ay*t^2

y = (0)(1.96) + (1/2)(-9.8)(1.96)^2

y = -18.82m (Value is negative because gravity constant was negative. It is the height reference that from the top of the building down, which is why it is negative. The sign can be ignored for this question.)

8. First write down all the known variables while separating the values for each direction:

x-direction:

x = 12m

vfx = 0m/s

vix = ?

y-direction:

y1 = 1.2m

y2 = 0.6m

viy = 0m/s;

ay = -9.8m/s^2

First find time in the y-direction as it would be the same value for the x-direction.

(y2 - y1) = viy*t + (1/2)ay*t^2

(-0.6) = (0)t + (1/2)(-9.8)t^2

t = 0.35s

Now that we have the time of flight, we can use the kinematic equation that will relate the known variables in the x-direction:

x = (1/2)(vix+vfx)t

(12) = (1/2)(vix+(0))(0.35)

68.6m/s = vix

6 0

4 years ago