The highest energy waves have the

Correct displacement 4km south,2 km north, 5km south, and 5 km north

babymother [125]

Answer:

2 km due south.

Explanation:

We need to find the displacement of 4km south,2 km north, 5km south, and 5 km north.

Let south is negative direction and north is positive direction.

The shortest path covered by an object is called its displacement.

D = -4+2-5+5

D = -2 km

So, the displacement is 2 km and it is in south direction.

8 0

3 years ago

Two moles of He gas are at 25 degrees C and a pressure of 210 kPa. If the gas is heated to 48 degrees C and its pressure reduced

SpyIntel [72]

Answer: $0.0686 m^3$

Explanation:

Given

No of moles=2

$T=25^{\circ}\approx 298 K$

$P_1=210 KPa$

$T_2=48 ^{\circ}\approx 321 k$

$P_2=77.7 KPa$

PV=nRT

Substitute

$77.7\times V_2=2\times 8.314\times 321$

$V_2=0.0686 m^3$

4 0

4 years ago

A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 13 A, while that

Feliz [49]

Answer:

(a) 1560 W

(b) 576 W

(c) 1.01

Explanation:

Voltage, V = 120 V

Current in dryer, I = 13 A

current in vacuum cleaner, i' = 4.8 A

(a) Power consumed by dryer,

P = V I = 120 x 13 = 1560 W

(b) Power consumed by vacuum cleaner

P' = V I' = 120 x 4.8 = 576 W

(c) Energy consumed by dryer

E = P x t = 1560 x 15 x 60 = 1404000 J

Energy consumed by the vacuum cleaner

E' = P' x t' = 576 x 40 x 60 = 1382400 J

the ratio of energies is

E : E' = 1404000 : 1382400 = 1.01

3 0

3 years ago

Earthquake information would most likely be shown on which type of map?

zmey [24]

Information about Earthquakes would normally be shown on a map called "geologic hazards"

6 0

4 years ago

1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of

laiz [17]

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

<u>The given parameters include:</u>

constant velocity of the elevator, u₁ = 10 m/s

initial velocity of the ball, u₂ = 20 m/s

height of the boy above the elevator floor, h₁ = 2 m

height of the elevator above the ground, h₂ = 28 m

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

$v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m$

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m + 45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

$h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s$

To learn more about projectile calculations please visit: brainly.com/question/14083704

6 0

3 years ago

The highest energy waves have the _ frequency