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2 years ago

Hello people ~

Physics

1 answer:

vlabodo [156]2 years ago

3 0

Answer:

graph of cosine function

Explanation:

f(x) = sin(x)

The curve of the sine function crosses the y-axis at the origin, heads up to y = 1 at x = π/2 and down to y = -1 at x = 3π/2.

Domain: (-∞, ∞)

Range: [-1, 1]

f(x) = cos(x)

The curve of the cosine function crosses the y-axis at (0, 1) which is its maximum, heads down to y = -1 at x = π then heads up again.

Domain: (-∞, ∞)

Range: [-1, 1]

Given information:

Antoine and Adriane begin their journey at the top of the Ferris wheel.

After one revolution, they will again be at the top of the wheel.

If the wheel starts when time t = 0 seconds, and they begin their journey at the top of the wheel (the maximum point), the most appropriate graph to model their ride is the graph of the cosine function since it is at its highest point when x = 0.

You might be interested in

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"

Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

4 0

3 years ago

How do you work out how many atoms there are in a isotope?

GalinKa [24]

206Pb = 1.342 x10^22 atoms

To find the number of atoms, you must first find the number of moles. If 238U is 238.029g/mol, and we have 1.75 grams, how many moles is that? 1.75 divided by 238.029 = 0.007352045 moles. To find the number of atoms in 0.007352045 moles, you multiply by a mole: 
0.007352045 x 6.02 x 10^23 = 4.426 x10^21 atoms. 

Same procedure for 206Pb: 
4.59 divided by 205.97446 = 0.022284316 moles 
0.022284316 x 6.02 x 10^23 = 1.342 x10^22 atoms. 

Hope that helps you!
https://answers.yahoo.com/question/index?qid=20100331153014AAoMXcu

4 0

3 years ago

Three resistances 2 ohm ,3ohm and 5 ohm are connected in parallel and a

hammer [34]

The potential difference across 3 Ohm resistor is 20V.

The resistors are connected in parallel which means all the three resistances have a fully potential difference of 20V.

7 0

3 years ago

In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor

leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

$W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}$ .....(I)

Where, $F(x)$ = net force

We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

$W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

We need to find the value of $\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

Using newton's second law

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}$ ...(II)

We know that,

Acceleration is rate of change of velocity.

$a=\dfrac{dv}{dt}$

Put the value of a in equation (II)

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

Now, the work done by the net force on the block is,

$W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0

3 years ago

How far will an object move in 6 s if its average

Dafna11 [192]

Given:-

Time taken by the particle (t) = 6 s
Average speed (v) = 40 m/s

To Find: Distance (s) travelled by the particle.

We know,

s = vt

where,

s = Distance travelled,
v = Speed &
t = Time taken.

Putting the values,

s = (40 m/s)(6 s)

→ s = 240 m ...(Ans.)

6 0

3 years ago