The _____ of a mechanical wave is a direct measure of its energy.

A hill that has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction. At

Andreas93 [3]

The angle of incline of the hill above the horizontal is 8.81°.

Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.

<h3>Tangent of the angle of the incline of the hill,</h3>

The tangent of the angle of the incline of the hill, Ф is

tanФ = vertical rise/horizontal distance = grade of hill

Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m

So, substituting the values of the variables into the equation, we have

tanФ = vertical rise/horizontal run

tanФ = 15.5 m/100.0 m

tanФ = 0.155

<h3>Angle of incline of the hill</h3>

Taking inverse tan of both sides, we have

Ф = tan⁻¹(0.155)

Ф = 8.81°

So, the angle of incline of the hill above the horizontal is 8.81°.

Learn more about angle of incline of a hill here:

brainly.com/question/10056962

6 0

2 years ago

an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco

Semenov [28]

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

$a_c = \frac{v^2}{r}$

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

$a_c = \frac{(12\ m/s)^2}{2\ m}$

<u></u>

Now, the centripetal force applied by the athlete on the hammer will be:

$F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)$

6 0

2 years ago

You have 1.2kg of (AU)gold (that's quite a bit of money at $1839/ounce). Using a heat source you apply 3096 J to the gold and re

melamori03 [73]

Answer:

129 J/Kg°C

Explanation:

Given :

Mass of gold, m = 1.2kg

Quantity of heat applied, Q = 3096 J

Temperature, t2 = 40°C

Temperature, t1 = 20°C

Change in temperature, dt = (40-20)°C = 20°C

Using the relation :

Q = mcdt

Where, C = specific heat capacity of gold

3096 = 1.2kg * C * 20°C

3096 J = 24kg°C * C

C = 3096 J / 24 kg°C

C = 129 J/Kg°C

7 0

2 years ago

B. Jerome plays middle linebacker for South's varsity football team. In a game against

weqwewe [10]

Answer:

Option D

670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

Final momentum is also mv but v being westward direction, we take it negative

Final momentum=82*-2.5= -205 Kg.m/s

Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s

Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s

4 0

3 years ago

Why does a third class lever cannot magnify force?

maw [93]

Explanation:

The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.

5 0

2 years ago