A scientist who studies chemicals and the effects of pollution in the air, water, and soil .

Sound waves, water waves, and light waves are all alike in that they all

Paha777 [63]

Sound and water waves are longitudinal waves, they require a medium to travel through and occilate particles 90 degrees to the wave motion
Light is a transverse wave. It doesnt require a medium to travel through.
All three reflect, refract and diffract

Light is difficult to think of because it acts in ways which waves cannot explain in some cirumstances. It acts like a particle (called photons) in some conditions, but acts like a normal sound or water wave does in others. Try not to get too caught up in light being a wave or a particle because even physists dont know how to explain it yet.

7 0

3 years ago

The speed of an arrow fired from a compound

san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

$X=v_{ox}*t$

$v_{0x} =v_0cos(\alpha)$

Y axis:

$Y= Y_0 +v_{y0} t - \frac{g}{2} t^2$

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, $Y=0$ , then

$0= Y_0 +v_{y0} t - \frac{g}{2} t^2$

$0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

Caculate the segcond degree equation to obtain the two possible values for t:

$t_1= 10.18 \\t_2= -0.04046$

But, in physics, time it could not be negative, so we take $t_1= 10.18$

Caculate now:

$X=79m/s*cos(\39)*10.18s= 624.996 m$

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

$v_0= 79m/s+13m/s= 92m/s$

Using the same procedure that item A, caculate X

First, we need to know the new time

$0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

And we obtain:

$t_1=11.845s\\t_2=-0.041s$

One more time, we take the positive time: $t_1=11.845s$

Finally:

$X=92m/s *cos(39)*11.845s=846.887 m$

6 0

3 years ago

Which statement about cellulose is true?

zlopas [31]

Is there any answers? Or is it asking you to choose?

6 0

3 years ago

Read 2 more answers

Imagine that two balls, a basketball and a much larger exercise ball, are dropped from a parking garage. If both the mass and ra

pashok25 [27]

Here if we assume that there is no air friction on both balls then we can say

$F = mg$

now the acceleration is given as

$F = ma = mg$

$a = g$

so here both the balls will have same acceleration irrespective of size and mass

so we can say that to find out the time of fall of ball we can use

$y = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2y}{g}}$

now from above equation we can say that time taken to hit the ground will be same for both balls and it is irrespective of its mass and size

3 0

3 years ago

A string is tied between two posts separated by 2.4 m. When the string is driven by an oscillator at frequency 567 Hz, 5 points

Alex787 [66]

Explanation:

The given data is as follows.

Length (l) = 2.4 m

Frequency (f) = 567 Hz

Formula to calculate the speed of a transverse wave is as follows.

f = $\frac{5}{2l} \times v$

Putting the gicven values into the above formula as follows.

f = $\frac{5}{2l} \times v$

567 Hz = $\frac{5}{2 \times 2.4 m} \times v$

v = 544.32 m/s

Thus, we can conclude that the speed (in m/s) of a transverse wave on this string is 544.32 m/s.

5 0

3 years ago