what is measurement ?​

Answer:

By observing the time between transits, we know the orbital period. Kepler's Third law can be used to determine the orbital radius of the planet if the mass of the orbiting star is known (R3=T2−Mstar/Msun, the radius is in AU and the period is in earth years).

Explanation:

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Complete Question  

The complete question is shown on the first and second uploaded image  

Answer:

When water is used the reading is  [tex]    R =  2281.6 \  cm  [/tex]

When mercury is used the reading is [tex]   R =  23.83 \ cm [/tex]  

The best fluid to use is mercury because for water a slight change in toluene level will cause a  large change in height .

Explanation:

From the question we are told that  

    The length of the leg of the manometer to the top of the tank is  d =  500cm

   The toluene level where in the tank where the height of the manometer fluid level in the open arm is equal to the height where the manometer is connected to the tank is  h =150 cm  

   The manometer reading is  R  

Generally at the point where the height of the open arm is equal to the height of the of the point connected to the tank ,  

   The pressure at the height of the both arms of the manometer corresponding to the base of the tank are equal  

     i.e   [tex]P_1 = P_2[/tex]

Here [tex]P_1[/tex] is the pressure of the manometer at the point corresponding to the base of the tank and this is mathematically represented as  

       [tex] P_{atm} + P_1 =  P_{atm} + P_t[/tex]

Here [tex]P_t[/tex] is the pressure due to the toluene level in the tank and in the arm of the manometer connected to the tank and this is mathematically represented as  

         [tex]P_t  =  \rho_t  * g  * h_i[/tex]

Here  

      [tex]\rho_t [/tex] is the density of toluene with value  [tex]\rho_t =  867 kg/m^3 [/tex]

 [tex]h_i[/tex] is the height of the connected arm above the point equivalent to the base of the tank , this mathematically represented as

          [tex]h_i =  d - h + R[/tex]

  and  [tex] P_2 [/tex] is the the pressure at the open arm of the manometer at the point equivalent to the base of the base of the tank and this is mathematically represented as

         [tex] P_2 =  \rho_f * g *  h_f [/tex]

Here  

       [tex]\rho_f[/tex] is the density of the fluid in use , if it is water the density is  

       [tex]\rho_w =  1000 \  kg /m^3 [/tex]

and  if it is  mercury the density is  

        [tex]\rho_m =  13600 \  kg /m^3 [/tex]

[tex]h_f[/tex] is the height of the  fluid in the open arm of the manometer from the point equivalent to the base of the tank which is equivalent the manometer reading R

So when the fluid is water we have

      [tex] P_{atm} +  \rho_t* g *(d - h + R) =  P_{atm} + \rho_f * g *  h_f[/tex]

=>   [tex]   \rho_t* (d - h + R) =   \rho_w *  h_f[/tex]        

=>   [tex]    867  (500 - 150 + R) =    1000 *  R [/tex]

=>    [tex]    R =  2281.6 \  cm  [/tex]

So when the fluid is mercury we have      

  [tex]   \rho_t* (d - h + R) =   \rho_m *  h_f[/tex]          

=>   [tex]    867  (500 - 150 + R) =   13600  *  R [/tex]  

=>   [tex]   R =  23.83 \ cm [/tex]  

The difference in the mercury reading for mercury due to the fact that they have different densities as we have seen in this calculation

So the best fluid to use is mercury because for water a slight change in toluene level will cause a  large change in height .

Answer:

8.8 hours

Explanation:

plz give me a Brainliest

Answer:

A. 50 km south

Explanation:

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Answer:

UR ANSWER IS WAVELENGTH

Answer:

Rating of percieved exertion

Answer:

on earth yes, but in space no because there is no gravity

Explanation:

plz can i have some points

The force exerted by an object depends on the mass of the object. Greater mass results in greater force. Hence, the statement is true.

Force is an external agent acts on a body to change the substance from the state of rest or motion. There are various kinds of force such as nuclear force, magnetic force, frictional force etc.

According to Newton's law of motion force in a moving body is the product of its mass and acceleration. Hence, force is directly proportional to mass and acceleration and an increase in these quantities will increase the force exerted by the body.

Thus, greater force is needed to a move a body with greater mass. If two objects of different masses exerts force each other, the greater force will be exerted by the massive body.

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Answer:

300m per minute or 5m per second

Answer:

True

Explanation:

The term 'plyometrics' is used interchangeably with the term 'jump training'. The technique can be used for training in sports that require explosive movements.

Answer:

the answer is A.) Wedge

A knife to cut a piece of bread is a A. Wedge

A simple machine, any of several devices with few or no moving parts that are used to modify motion and the magnitude of a force in order to perform work.

since , a wedge is a simple machine  with two inclined planes which when put together forms a sharped edge, which forms a triangular shaped tool which can be used to separate portion of  two objects .

hence , a knife to cut a piece of bread is a A. Wedge

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Answer:

160,934.4 cm or in other words *160,934*

Explanation:

1 mile = 5280 ft.

5280 ft. = 63360 in.

63360 in. = 160934.4

Answer:

Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more—approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater ...

Explanation:

Acceleration is the rate of change in the speed of an object. If initial and final speed are v₁ and v₂. It changes speed in time t. Its formula is given by :

[tex]a=\dfrac{v_2-v_1}{t}[/tex]

The problem is "A skater increases her velocity from 2.0 m/s to 10.0 m/s in 3.0 seconds. What is the skater’s acceleration? "

In this problem,

v₁ = 2 , v₂ = 10 m/s and t = 3 s

Acceleration of the object,

[tex]a=\dfrac{10-2}{3}\\\\a=2.67\ m/s^2[/tex]

So, the acceleration of the skater is [tex]2.67\ m/s^2[/tex].

Answer:

Physical change

Explanation:

A physical change can be reversible where the original form of the matter can be restored, or irreversible where the original form cannot be restored. Therefore, sawdust is a physical change.

Answer:

21.3V

Explanation:

Explanation

E = V + Ir

V = E- Ir

V = 24 - {(24/36) x 4}

V = 21.3V

Answer: The process is called artificial selection because people (instead of nature) select which organisms get to reproduce.

Explanation: This is evolution through artificial selection and the characteristics to reproduce, causing the evolution of farm stock.

Answer:

Mechanical energy

Explanation:

Mechanical energy is needed for movement of objects. Muscles convert chemical energy provided by the rest of the body to allow movement.

Answer:

The correct option is the third option

Explanation:

Firstly, it must be noted that chemicals/reagents that are sensitive to sunlight are stored in dark/amber container in the laboratory. Hence, the chloroform can only be found in an amber/dark bottle.

Also, reagents/chemicals that release poisonous/offensive gases are handled in the fume cupboard in the laboratory. Thus, If Malik is going to pour the chloroform, he should pour it in a fume cupboard to avoid inhaling it because of the toxicity of it's vapor.

From the above explanation, it can be deduced that Malik should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.

Answer:

He should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.

Explanation:

Answer: Approximately 302 m/s^2

================================================

Work Shown:

s = starting velocity = 0

f = final velocity  = 46

d = distance = 3.5

a = acceleration = unknown (we're solving for this)

[tex]f^2 = s^2 + 2a*d \ \ \text{ ..... one of the kinematics equations}\\\\46^2 = 0^2 + 2a*3.5\\\\2116 = 7a\\\\7a = 2116\\\\a = \frac{2116}{7}\\\\a \approx 302.28571\\\\a \approx 302[/tex]

The acceleration to three sig figs is roughly 302 m/s^2

The acceleration is so large because the ball's final velocity is incredibly fast in such a short amount of time.

Answer:

162

Explanation:

When the amplitude of the wave is 8 units, the corresponding energy of the wave is 128 units. Therefore, option C is correct.

The term amplitude is defined as the maximum distance moved by a point on a vibrating body or wave calculated from its equilibrium position.

We have given,

first amplitude, A = 1, corresponding energy = 2

second amplitude, A = 2, corresponding energy = 8

The energy of a string at  a given amplitude is calculated as follows;

E ∝ A²

Energy varies as the square of the change in amplitude.

If amplitude of one has energy of two

Amplitude of 2 (double of 1)

= ( 2 x 2 ) = 4 x 2

= 8  4 times the energy at amplitude of 1.

Then for 8 units of amplitude change

8 x 8 = 64 x 2

= 128  

Thus, When the amplitude of the wave is 8 units, the corresponding energy of the wave is 128 units, option C is correct.

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Your question is incomplete, most probably your question was

The data table shows how the amplitude of a mechanical wave varies with the energy it carries. Analyze the data to identify the mathematical relationship between amplitude and energy. Use your equation to find the energy if the amplitude is 8 units.

Amplitude

1

2

3

4

energy

2

8

18

32

A.66 units

B. 108 units

C. 128 units

D. 88 units ​

Answer:

The correct option is;

C. The temperature of the air affects the speed of sound

Explanation:

The information given are;

The speed with which sound travels in the desert air = 358 m/s

The speed with which sound travels in the polar air = 330 m/s

The major difference between the desert air and the polar air is that the temperature of the desert air is hotter than the temperature of the air in the polar region. Therefore the speed of sound is affected by the air temperature.

The equation that gives the relationship of the speed of sound in air, [tex]v_{air}[/tex], to temperature is presented as follows;

[tex]v_{air} = 331\frac{m}{s} \times \sqrt{\dfrac{T_K}{273 K} } = 331\frac{m}{s} \times \sqrt{1 + \dfrac{T_{^{\circ}C}}{273 ^{\circ} C} }[/tex]

Which shows that the speed of sound in air, [tex]v_{air}[/tex], rises as the temperature of the air rises.

Where;

[tex]T_{^{\circ}C}[/tex] = The temperature of the air in degrees Celsius

[tex]T_K[/tex] = The temperature of the air in degrees Kelvin.

Answer:

Explanation:

Time taken by jogger to travel the distance to finishing line = 2.1 / 7.3

= .28767 hr

Bird will keep flying for this time period

distance covered by bird = speed x time

= 29.2 x .28767 km

= 8.4 km .

Answer:

1) False, since constant speed means constant velocity, but with constant direction as well. Velocity is speed with direction. When you move in a circle, there are boundaries which means that a constant moving speed will cause the object to veer(move in different directions) to maintain movement, If it doesn't veer, the object will deflect(rebound) off the surface due to friction, and impact force or depending on its physical bond of malleability or density.

2a) Create position vectors by drawing a ray from the center(origin) of the circle to each of your desired points, it's magnitude will be the radius(how convenient). Both of these vectors go from the center of the circle to the position on the circle. In general, the distance from the origin to a point is called the radius vector.

2b)

Answer:

It's 31

Explanation:

add 15 and 16

Answer: C. Trial 3

Explanation:

Trial 1 and 2 equal 1.5, Trial 3 equals 1 and Trial 4 equals 3. Trial 3 is the smallest current .

The trial for which the students would measure the smallest current is the circuit is trial 2 and trial 3.

To know the trial which generates the smallest current, we need to determine the current in each trial.

Since current I = V/R where V = voltage and R = resistance.

For trial 1, V = 1.5 V and R = 200 Ω

So, I = 1.5 V/200 Ω

= 0.0075 A

= 7.5 mA

For trial 2, V = 1.5 V and R = 100 Ω

So, I = 1.5 V/100 Ω

= 0.015 A

= 15 mA

For trial 3, V = 3 V and R = 200 Ω

So, I = 3 V/200 Ω

= 0.015 A

= 15 mA

For trial 4, V = 3 V and R = 100 Ω

So, I = 3 V/100 Ω

= 0.03 A

= 3 mA

Trial 2 and trial 3 both produce a the smallest current of 15 mA.

So, the trial for which the students would measure the smallest current is the circuit is trial 2 and trial 3.

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Answer:

a.) 10Hz

b.) 0.1 s

c.) 187.4 m/s

d.) -412.6 m/s^2

Explanation:

Given that an object is moving back and forth on the x-axis according to the equation x(t) = 3sin(20πt), t> 0, where x(t) is measured in cm and t in seconds. Give decimal answers below.

(a) How many complete back-and-forth motions (from the origin to the right, back to the origin, to the left and finally back to the origin) does the object make in one second?

from the equation given,  the angular speed w = 20π

but w = 2πf

where f = frequency.

substitute w for 20π

20π = 2πf

f = 20π/2π

f = 10 Hz

(b) What is t the first time that the object is at its farthest right?

since F = 1/T

T = 1 / f

T = 1/10

T = 0.1 s

Therefore, the t of  first time that the object is at its farthest right is 0.1 s

(c) At the time found in part (b), what is the object's velocity?

The velocity can be found by differentiating the equation;

x(t) = 3sin(20πt)

dx/dt = 60πcos(20πt)

where dx/dt  = velocity V

V = 60πcos(20π * 0.1)

V = 187.4 m/s

(d) At the time found in part (b), what is the object's acceleration?

to get the acceleration, differentiate equation  V = 60πcos(20πt)

dv/dt = -1200πSin(20πt)

dv/dt = acceleration a

a = -1200πSin(20πt)

substitute t into the equation

a = -1200πSin(20π * 0.1)

a = - 412.6 m/s^2

The type of motion the object undergoes that consist of a back-and-forth

movement that is repetitive is a simple harmonic motion.

The correct responses are;

Reasons:

The equation that represents the motion of the object is x(t) = 3·sin(20·π·t)

Where;

t ≥ 0

x(t) is in cm, and t is in seconds.

(a) The general sine function is y = A·sin(B·(t - C) + D

[tex]\mathrm{The \ period \ P} = \dfrac{2 \cdot \pi}{B}[/tex]

By comparison, we have;

B = 20·π

Therefore;

[tex]P = \dfrac{2 \cdot \pi}{20 \cdot \pi} = \dfrac{1}{10} = 0.1[/tex]

The period, which is the time to complete one cycle = 0.1 seconds

The number of cycle completed per second, is the frequency, f

[tex]f = \dfrac{1}{P} = \dfrac{1}{0.1} = 10[/tex]

f = 10 Hz = 10 cycles

The number of cycle completed per second is f = 10 cycles

(b) When the object is at the farthest right, we have;

sin(20·π·t) = Maximum = 1

[tex]\mathbf{sin\left(\dfrac{\pi}{2}\right)} = 1[/tex]

Therefore,

[tex]20 \cdot \pi \cdot t = \dfrac{\pi}{2}[/tex]

[tex]t = \dfrac{\pi}{20 \cdot \pi \cdot 2} = \dfrac{1}{40} = 0.025[/tex]

The first time it is at the farthest right is t = 0.025 seconds after start

(c) [tex]\mathrm{Velocity, \ v, \ is \ given \ by \ v = \dfrac{dx(t)}{dt}}[/tex]

Therefore;

[tex]v = \dfrac{dx(t)}{dt} = \dfrac{dx(t)}{dt} \left(3 \cdot sin \left(20 \cdot \pi \cdot t \right) = 60\cdot \pi \cdot cos \left(20 \cdot \pi \cdot t \right)[/tex]

At t = 0.25 seconds, we have;

[tex]v = \dfrac{dx(t)}{dt} = 60\cdot \pi \cdot cos \left(20 \times \pi \times 0.025 \right) = 0[/tex]

The velocity of the object at the t = 0.025 seconds is v(t) = 0

(d) Acceleration, a, is given by [tex]a = \dfrac{dv(t)}{dt}[/tex]

Therefore;

[tex]a = \dfrac{dv(t)}{dt} = \dfrac{d}{dt} \left( 60\cdot \pi \cdot cos \left(20 \times \pi \times t\right) \right) =-1200 \cdot \pi ^2 \cdot sin(20 \cdot t \cdot \pi)[/tex]

a = -1200·π²·sin(20·t·π)

∴ a = -1200×π²×sin(20×0.025×π)

sin(20×0.025×π) = 1

∴ a = -1200×π²×1 ≈ -11843.53

At the time found in part b the acceleration of the object, a ≈ -11843.53 m/s²

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Answer:

This has 5 significant figures.

===========================================================

Explanation:

Consider a mass of 10 kg, so m = 10

Let's say we apply a net force of 20 newtons, so F = 20

The acceleration 'a' is...

F = ma

20 = 10a

20/10 = a

2 = a

a = 2

The acceleration is 2 m/s^2. Every second, the velocity increases by 10 m/s.

---------------

Now let's double the net force on the object

F = 20 goes to F = 40

m = 10 stays the same

F = ma

40 = 10a

10a = 40

a = 40/10

a = 4

The acceleration has also doubled since earlier it was a = 2, but now it's a = 4.

---------------

In summary, if you double the net force applied to the object, then the acceleration doubles as well.

Acceleration is directly proportional to the net force on an object, and inversely proportional to its mass.

So if an object's mass stays the same while the net force on it doubles, then its acceleration will also double.

We don't know anything about the "trials".  This sounds like it might be a follow-up to a lab experiment that was performed when we weren't there.

We also don't know anything about "question 1".