All categories

3 years ago

Trigonometry Help, question posted below @AL2006

Physics

1 answer:

Oxana [17]3 years ago

8 0

It's ' D '.

-- Period: The angle is (pi/2 · t).
From t=0 to t=4 sec, the angle changes by 2pi. So the period is 4 sec.

-- Amplitude: The biggest the cosine can ever be is 1 .
When the cosine part is 1, 'd' is 6, so the amplitude of the model is 6 .

-- Position when t=0:
When t=0, the angle in the cosine is zero.
The cosine of zero is 1 .
So when t=0, d=6 . There you are.

You might be interested in

Suppose an acorn with a mass of 3.17 g falls off a tree. At a particular moment during the fall, the acorn has a kinetic energy

denis-greek [22]

Potential and kinetic energy both decrease with the acorn's falling potential and kinetic energy.

The acorn's potential energy is at its peak when it reaches the top of the tree, yet its kinetic energy is zero (i.e., it is not accelerating).

The height of the ball reduces along with the potential energy as the acorn tumbles down the tree, but the kinetic energy rises (energy due to motion)

The height will be 0 and the kinetic and potential energy will be zero at the ground. This demonstrates that as an item falls, both potential and kinetic energy are lost.

Learn more about Energy here

brainly.com/question/13881533

#SPJ4

6 0

2 years ago

A river flows at 2m/s. The velocity of a ferry relative to the shore is 4m /s at right angle to the current. What is the velocit

monitta

The velocity of the ferry relative to the current is 4.5 m/s.

<h3>Relative velocity</h3>

Relative velocity is the velocity of a body as observed from the reference point of another body either stationary or in motion.

Since the river is flowing parallel to the shore and the ferry is moving perpendicular to the shore, their velocities are at right angles to each other.

The two velocities form a right angled-triangle of sides 2, 4 and a hypotenuse which gives the relative velocity of the ferry to the current.

Using Pythagoras rule:

c² = a² + b²

Let c be the hypotenuse
a = velocity of the ferry, and
b = the velocity of the current, and

c² = 4² + 2²

c² = 16 + 4

c = 20

c = √20

c = 4.47 m/s

c ≈ 4.5 m/s

Therefore, the velocity of the ferry relative to the current is 4.5 m/s.

Learn more about relative velocity and Pythagoras rule at: brainly.com/question/25617868

5 0

3 years ago

Beats are the difference in (a) Frequency (b) Amplitude (c) Intensity (d) None

sasho [114]

Answer:

Beats are the difference in frequency.

(a) is correct option.

Explanation:

Beat :

Beat is the difference of the frequency of two waves.

The difference in frequency is equal to the number of beat per second.

Amplitude :

Amplitude of the wave is the maximum displacement.

Frequency :

Frequency is the number oscillations of wave in per second.

Intensity :

Intensity is the power per unit area.

Hence, Beats are the difference in frequency.

7 0

3 years ago

After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0

3 years ago

Dhorpatan hunting reserve doesn't affect rare animals.

Eddi Din [679]

Answer:

Pheasants and partridge are common and their viable population in the reserve permits controlled hunting. Endangered animals in the reserve include Musk deer, Wolf, Red panda, Cheer pheasant and Danphe. The hunting license is issued by the Department of National Parks and Wildlife Conservation.

HOPE THIS HELPS, HAVE A GREAT DAY!!

Explanation:

4 0

2 years ago