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2 years ago

9. P waves move faster than S waves A. True B. False

Physics

1 answer:

ankoles [38]2 years ago

8 0

Answer: true
(I need 20 characters to submit the answer so I wrote this not to be confused with the answer)

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Distance meters Time (seconds) Trial Tria 2 Trials Average 3 053 059 0.52 055 6 106 719 104 Uning the values trom the data table

kumpel [21]

Jsisisusuagahannananana this is worth 22 points g that’s crazy 25

5 0

2 years ago

Psychophysics is a discipline in psychology that focuses solely on the identification of stimuli. T F

Rzqust [24]

Hi There,

This is False.
Hope this helped!

3 0

3 years ago

Read 2 more answers

If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int

vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the system

$Q$ is the heat absorbed by the system

$W$ is the work done by the system on the surrounding

In this problem, the work done by the system is

$W=-225 kcal=-941.4 kJ$

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

$Q=-5 \cdot 10^2 kJ=-500 kJ$

with a negative sign as well because it is released by the system.

Therefore, by using the initial equation, we find

$\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ$

8 0

3 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0∘ below the positive x axis. A second vector, B⃗, has a magn

jolli1 [7]

The magnitude of the vector C is 96.32m

<h3>How to solve for the magnitude of vector c</h3>

Ax = AcosθA

= 40 cOS 20

= 37.59

Ay = AsinθA

-40sin20

= -13.68

Bx = B cos θ B

= 75Cos50

= 48.21

By = BsinθB

= 75sin50

= 57.45

Cx = AX + Bx

= 37.59 + 48.21

= 85.8

Cy = Ay + By

= -13.65 + 57.45

= 43.77

The magnitude is solved by

|c| = $\sqrt{Cx^{2}+Cy^{2} }$

= √85.8² + 43.77²

= 96.32m

The magnitude of the vector c is 96.32m

Read more on the magnitude of a vector here:

brainly.com/question/3184914

#SPJ1

6 0

1 year ago