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A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of the tuning fork?

Answers

Answer 1

Answer:

We are given:

The tuning fork vibrates at 15660 oscillations per minute

Period of one back-and forth movement:

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations (1 minute = 60 seconds)

dividing the values

0.0038 seconds / Oscillation

Therefore, one back and forth vibration takes 0.0038 seconds

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Crude oil is a mixture of many different components. The extraction of crude oil from the Earth is important, but its refinement into different substances is a key piece to obtaining as many uses as possible from the crude oil. Using the diagram, justify the source of data used to develop the technology for refining the crude oil.

Answers

Crude oil is a mixture of nitrogen, oxygen, sulphur, and hydrogen components

What is Crude oil?

A combination of hydrocarbons known as crude oil is one that is found in naturally occurring subsurface reservoirs in the liquid phase and continues to be liquid at atmospheric pressure after passing through surface separation equipment.

Refineries transform crude oil into useful products including gasoline, diesel, and aviation fuels for transportation. Gasoline: A fuel used in both personal and commercial vehicles that are made for internal combustion engines.

In addition to some nitrogen, sulphur, and oxygen, crude oil is a combination of very flammable liquid hydrocarbons (compounds mostly made of hydrogen and carbon).

More about the Crude oil link is given below.

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Answer: Different fuel components boil at different temperatures, allowing them to be separated.

Explanation:

Which of these objects are constantly in motion? Select all that apply.A.
Earth

B.
Planes

C.
Trains

D.
Blood

E.
Sun

F.
Cars

Answers

The object Earth,Sun, and Blood are constantly in motion. The correct option is A, D, and E.

What is motion?

if a body changes its position with respect to its surroundings in a given interval of time, Then the body is said to be in motion

Motion is generally classified as follows.:

i) Rectilinear motion.

ii) Circular motion.

iii) Rotational motion.

iv) Periodic motion.

The Earth is continuously in motion because it continuously revolves around The Sun in an elliptical orbit, due to which a year is 365 days. and also The earth rotates about its own axis once a day.

The Sun also revolves around the galactic center of our Milkyway galaxy. and it also rotates about its own axis continuously. so that the sun is also continuously in motion.

The Human blood is continuously in motion Because our blood is continuously circulating whole over the body with the help of our heart. The heart continuously pumps our blood and circulates it inside the human body.

Hence the Earth, Sun, and blood are continuously in motion.

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earth, blood, and the sun

A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.

Answers

Answer:

vf = 30 m/s : (the magnitude of the velocity of the stone just before it hits the ground)

Explanation:

Because the stone moves with uniformly accelerated movement we apply the following formulas:

vf²=v₀²+2*g*h Formula (1)

Where:

h: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

g: acceleration due to gravity in m/s²

Free fall of the stone

Data

v₀ = 10 m/s

vf = 30.0 m/s

g = 9,8 m/s²

We replace data in the formula (1) to calculate h:

vf²=v₀²+2*g*h

(30)² = (10)² + (2)(9.8)*h

(30)²- (10)²= (2)(9.8)*h

h =( (30)²- (10)²) /( 2)(9.8)

h = 40.816 m

Semiparabolic movement of the stone

Data

v₀x = 10 m/s

v₀y = 0 m/s

g = 9.8 m/s²

h= 40.816 m

We replace data in the formula (1) to calculate vfy :

vfy² = v₀y² + 2*g*h

vfy² = 0 + (2)(9.8)( 40.816)

$v_(fy)=√(2*9.8*40.816) = 28.284 (m)/(s)$

$v_(f)=\sqrt{v_(ox)^2+v_(fy)^2}=√((10)^2+(28.284)^2) = 30(m)/(s)$

The magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

The given parameters;

initial vertical velocity of the stone, $v_y_0$ = 10 m/s

final vertical velocity of the stone, $v_y_f$ = 30 m/s

The height traveled by the stone before it hits the ground is calculated as;

$v_y_f^2 = v_y_0^2 + 2gh\n\nh = (v_y_f^2- v_y_0^2)/(2g) \n\nh = ((30)^2 - (10)^2)/(2* 9.8) \n\nh = 40.82 \ m$

If the the stone is projected horizontally with initial velocity of 10 m/s;

the initial vertical velocity = 0

Final vertical velocity of the stone is calculated as follow;

$v_y_f^2 = v_y_0^2 + 2gh\n\nv_y_f^2 = 0 + 2* 9.8* 40.82\n\nv_y_f^2 = 800.07\n\nv_y_f = √(800.07) \n\nv_y_f = 28.28 \ m/s$

The horizontal velocity doesn't change.

the final horizontal velocity, $v_x_f$ = initial horizontal velocity = 10 m/s

The resultant of the final velocity of the stone before it hits the ground;

$v _f= √(v_x_f^2 + v_y_f^2) \n\nv_f = √(10^2 + 28.28^2) \n\nv_f= 29.99 \ m/s \approx 30 \ m/s$

Thus, the magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

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A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?

Answers

Answer:

a n c

Explanation:

Final answer:

The volume rate of flow can be determined using the equation Q = Av, where Q is the volume rate of flow, A is the cross-sectional area of the pipe, and v is the average speed of the water. Given the diameter of the wider section of the pipe is 6.0 cm and the gauge pressure is 32.0 kPa, we can calculate the volume rate of flow using the provided information. The volume rate of flow is found to be 0.0018 m³/s.

Explanation:

The volume rate of flow can be determined using the equation Q = Av, where Q is the volume rate of flow, A is the cross-sectional area of the pipe, and v is the average speed of the water.

Given that the diameter of the wider section of the pipe is 6.0 cm, the radius is 3.0 cm and the gauge pressure is 32.0 kPa. Similarly, for the narrower section with a diameter of 4.0 cm, the radius is 2.0 cm and the gauge pressure is 24.0 kPa.

Using the equation Q = Av and the fact that the flow rate must be the same at all points along the pipe, we can set up the equation A₁v₁ = A₂v₂. Solving for v₂, we have v₂ = A₁v₁ / A₂ = πr₁²v₁ / πr₂², where r₁ is the radius of the wider section and r₂ is the radius of the narrower section.

Substituting the values, we get v₂ = (3.14)(3.0 cm)²(32.0 kPa) / [(3.14)(2.0 cm)²] = 18.0 cm/s. Since v = d/t, we can convert cm/s to m³/s by multiplying by 0.0001, so the volume rate of flow is 0.0018 m³/s.

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Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?

Answers

Answer:

a) $v=5.6725\,m.s^(-1)$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^(-2)$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^(-1)$

putting the values in eq. (1)

$v^2=0^2+2* 3.62 * 10$

$v=8.5088\,m.s^(-1)$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20* 8.5088$

$p=170.1764\,kg.m.s^(-1)$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)* v'=p$

$(20+10)* v'=170.1764$

$v'=5.6725\,m.s^(-1)$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$(1)/(2) (M+m).v'^2=(M+m).g.h$

$(1)/(2) (20+10)* 5.6725^2=(20+10)* 9.8* h$

$h\approx 1.6420\,m$

above the normal hanging position.

NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider such a spacecraft that has a diameter of d = 148 m. What is the speed v, in meters per second, the spacecraft must rotate at its outer edge to replicate the force of gravity on earth?