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1 year ago

Where is there kinetic energy in this system?

Physics

1 answer:

Alika [10]1 year ago

6 0

Answer:

So kinetic means to move, something like that right, so the two balls that go in the air are where the kinetic energy is.

Explanation:

Hope it helps.

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Which statement about images is correct? a) A virtual image cannot be formed on a screen. b) A virtual image cannot be viewed by

ankoles [38]

Answer:

A). A virtual image cannot be formed on a screen.

Explanation:

A virtual image can not be formed on a screen.

For image:

1.A virtual image can be viewed by the unaided eye.

2. A real image must be erect or maybe inverted.

3.Mirrors can produce virtual as well as real image ,it depends on which type of mirror is.

4.A virtual image can be photographed.

So the option A is correct.

5 0

2 years ago

Jim stands beside a wide river and wonders how wide it is. he spots a large rock on the bank directly across from him. he then w

LuckyWell [14K]

To solve this problem, we must imagine that Jim’s initial position, the position of the rock, and Jim’s final position all connects to form a triangle. Now we can imagine that the triangle is a right triangle with the 90° angle on the initial position.

The angle of 30° is directly opposite to the length of his total stride while the width of the river is the side adjacent to the angle. Therefore can use the tan function to solve for the width of the river:

tan θ = opposite side / adjacent side

tan 30 = total stride distance / width of river

where total stride distance = 65 * 0.8 = 52 m

width of river = 52 m / tan 30

<span>width of river = 90.07 m</span>

7 0

1 year ago

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

$\frac{1}{2}mv^2=\ mgh$

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

1 year ago

the brightest , hottest, and most massive stars are the brilliant blue stars designated as spectral class O. if a class O star w

4vir4ik [10]

The speed is 0.956 m / s.

<u>Explanation</u>:

The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.

K.E = 1/2 $\times$ m $\times$ $v^{2}$

where K.E represents the kinetic energy,

m represents the mass,

v represents the velocity.

K.E = 1/2 $\times$ m $\times$ $v^{2}$

1.10 $\times$ 10^42 = 1/2 $\times$ 3.26 $\times$ 10^31 $\times$ $v^{2}$

$v^{2}$ = (1.10 $\times$ 10^42 $\times$ 2) / (3.26 $\times$ 10^31)

v = 0.956 m / s.

6 0

2 years ago

Lucy has three sources of sound that produce pure tones with wavelengths of 60cm, 100cm, and 124cm.

denis23 [38]

Answer:

a) We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength

b) There is resonance for the lengths 25 and 75 cm

c) Resonance occurs for tubes with length 31 and 93 cm

Explanation:

To find the length of the tube that has resonance we must find the natural frequencies of the tubes, for this at the point that the tube is closed we have a node and the open point we have a belly; in this case the fundamental wave is

λ = 4L

The next resonance called first harmonic λ₃ = 4L / 3

The next fifth harmonic resonance λ₅ = 4L / 5,

WE see that the general form is λ ₙ= 4L / n n = 1, 3, 5 ...

Let's use these expressions for our problem

Let's start with the shortest wavelength.

a) Lam = 60 cm

Let's look for the tube length that this harmonica gives

L = λ n / 4

To find the shortest tube length n = 1

L = 60 1/4

L = 15 cm

For n = 3

L = 60 3/4

L = 45 cm

For n = 5

L = 60 5/4

L = 75 cm

For n = 7

L = 60 7/4

L = 105cm

We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength, in different harmonics 1, 3 and 5

.b) λ = 100 cm

For n = 1

L = 100 1/4

L = 25 cm

For n = 3

L = 100 3/4

L = 75 cm

For n = 5

L = 100 5/4

L = 125 cm

There is resonance for the lengths 25 and 75 cm in the fundamental and third ammonium frequency

c) λ= 124 cm

L = 124 1/4

L = 31 cm

For the second resonance

L = 124 3/4

L = 93 cm

Resonance occurs for tubes with length 31 and 93 cm in the fundamental harmonics and third harmonics

8 0

2 years ago