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Answer:

Find answers below.

Explanation:

A graph can be defined as the graphical representation of data (informations) on horizontal and vertical lines i.e x-axis and y-axis respectively.

On the x-coordinates, we have the following points or values;

1. Smiley face = 2

2. Diamond = 2

3. Sun = 0

4. Heart = 3.75

On the y-coordinates, we have the following points or values;

1. Smiley face = 1

2. Diamond = 3.75

3. Sun = 3.75

4. Heart = 0

Note: to read the points on a graph, you would look at the exact points marked on the x-coordinates (x-axis) and y-coordinates (y-axis) respectively.

The scale used for this graph is 0.75 for each unit on the x-coordinates and y-coordinates respectively.

Answer:

because it is helpful to human beings I think

Answer:

Sorry I don't know the answer

Answer:

the area of the rectangular field is 10.5 m²

Explanation:

Given;

length of the rectangular field, L = 42 cm = 0.42 m

breadth of the rectangular field, b = 25 m

The area of the rectangular field is calculated as follows;

Area = Length x breadth

Area = 0.42 m x 25 m

Area = 10.5 m²

Therefore, the area of the rectangular field is 10.5 m²

The mechanical advantage is the ratio of the length of the effort arm to the length of the resistance arm. Option A is correct.

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtain the efficiency of the given mechanical machine.

Mechanical advantage is a measure of how much a machine multiplies the input force.

The equation for the mechanical advantage of a lever is;

MA =length of effort arm/length of resistance arm

[tex]\rm MA=\frac{L_E}{L_R}[/tex]

The mechanical advantage is the ratio of the length of the effort arm to the length of the resistance arm.

Hence, option A is correct.

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The mechanical advantage is the ratio of the length of the effort arm to the length of the resistance arm. Option A is correct.

What is the mechanical advantage?

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtain the efficiency of the given mechanical machine.

Mechanical advantage is a measure of how much a machine multiplies the input force.

The equation for the mechanical advantage of a lever is;

MA =length of effort arm/length of resistance arm

Answer:

564

Explanation:

Answer:

Josiah Spiers in 1867 was when scripture union was founded

Answer:

W = 20 J

Explanation:

Given that,

Charge, q = 2 C

It is moves from a point at 118 volt to a point at 128 volt.

We need to find the work done in moving the charge,

[tex]W=\Delta VQ[/tex]

Put all the values,

[tex]W=(128-118)\times 2\\\\W=20\ J[/tex]

So, the work done in moving the charge is 20 J.

Answer:

the tension in the tightrope is 233.68 N

Explanation:

Given the data in the question;

Time taken to reach the opposite tower t = 0.9 s

Distance between the two towers S = 26 m

mass per one meter length =  0.28 kg

First we calculate the velocity;

Velocity V = Distance / time

we substitute

Velocity V = 26 m / 0.9 s

Velocity V = 28.889 m/s

We know that Velocity V can also be expressed as;

V = √( T / m )

we make T the subject of formula

V² = T / m

T = mV²

we substitute

T = 0.28 × ( 28.889 )²

T = 233.68 N

Therefore,  the tension in the tightrope is 233.68 N

Answer:

Answer: Speed is not a vector quantity. It has only magnitude and no direction and hence it is a scalar quantity.

Answer:

a)  v_f = 5,06 m/s, b)  GAIN in kinetic energy.

Explanation:

a) For this exercise we will use the relationship between work and kinetic energy

           W = ΔK

Work is defined by

           W = F. d

bold indicates vectors

the displacement is

            d = x_f - x₀

            d = 4 -1

            d = 3i m

we calculate

          W = 20 10⁻² 3 i.i

let's remember that

         i.i = j.j = 1

         i.j = 0

          W = 6.0 10⁻¹ J

we substitute in the first equation

          W = K_f - K₀

          W = ½ m (v_f ² -v₀²)

          v_f ² = [tex]\frac{2W}{m} + v_o^2[/tex]

let's calculate

          v_f ² = 2 6.0 10⁻¹ /2 + 5²

          v_f = √25.6

          v_f = 5.06 m / s

b) we can see that the speed at the end of the movement is greater than the initial speed, therefore there is a GAIN in kinetic energy.

M1 = 15/g = 15/9.8 = 1.53 kg = mass of object in air. M2 = 12/9.8 = 1.22 kg = mass of object immersed. M1-M2 = 1.53-1.22 = 0.31 kg lost by object = mass of water displaced. ... Do = 4.94 g/cm^3 = density of object.

True

Hope this helps! :)

Answer:

true because when trow the frisbee gives u level

Answer: See explanation

Explanation:

Celsius and Fahrenheit are the scales that are used in the measurement of temperature. Celcius is also refered to as centigrade. The relation that exist between Celsius and Fahrenheit is typically proportional.

The conversion from Celsius to Fahrenheit is expressed as:

F = (9/5 × C) + 32.

The conversion from Fahrenheit to Celsius is expressed as:

C = 5/9(F - 32)

For example to convert 100°C to Fahrenheit will be:

F = (9/5 × C) + 32.

F = (9/5 × 100) + 32

F = 180 + 32

F = 212°F

Answer:

The fish gobbles the mosquito at height 18 cm.

Explanation:

Initial velocity, u = 3.7 m/s

horizontal distance, d = 28 cm

Angle, A = 39 degree

Let the time is t.

Horizontal distance = horizontal velocity x time

d =  u cos A x t

0.28 = 3.7 cos 39 x t

t = 0.097 s

Let the height is h.

Use the second equation of motion

[tex]h =u t-0.5 gt^2\\\\h= u sin A t - 0.5 gt^2\\\\h= 3.7 sin 39 \times 0.097 - 0.5\times 9.8\times 0.097\times0.097\\\\h =0.226 -0.046 \\\\h=0.18 m=18 cm[/tex]

Answer:

For (a): The chemical shift is [tex]2.08\delta[/tex]

For (b): The chemical shift is [tex]9.85\delta[/tex]

For (c): The chemical shift is [tex]7.5\delta[/tex]

Explanation:

To calculate the chemical shift, we use the equation:

[tex]\text{Chemical shift in ppm}=\frac{\text{Peak position (in Hz)}}{\text{Spectrometer frequency (in MHz)}}[/tex]

Given value of spectrometer frequency = 200 MHz

Given peak position = 416 Hz

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{416Hz}{200MHz}\\\\\text{Chemical shift in ppm}=2.08\delta[/tex]

Given peak position = [tex]1.97\times 10^3 Hz[/tex]

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{1.97\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=9.85\delta[/tex]

Given peak position = [tex]1.50\times 10^3 Hz[/tex]

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{1.50\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=7.5\delta[/tex]

Answer:

Explanation:

9ooooo

Answer:

To find out the number of atoms: MULTIPLY all the SUBSCRIPTS in the molecule by the COEFFICIENT. (This will give you the number of atoms of each element.)

Explanation:

Answer:

sorry I don't know I am only in 7th grade

Answer:

c

Explanation:

The reason why we are unlikely to see fluctuations in light output in extragalactic sources with a diameter of about one light day over timescales shorter than about one day is due to the size and distance of the source, as well as the speed of light.

When we observe an extragalactic source with a diameter of about one light day, we are essentially observing light that has traveled a very long distance through space to reach us. This light may have originated from a region of the source that is changing in brightness or emitting intense bursts of light, but by the time the light reaches us, these fluctuations are smeared out over a longer period of time due to the speed of light.

For example, if the source were emitting a burst of light that lasted for only a few hours, by the time that light travelled a distance of one light day (which is about 25 billion miles or 40 billion kilometres), the burst would be spread out over a longer period of time. This is because the light emitted at the beginning of the burst would have already traveled a significant distance away from the light emitted at the end of the burst by the time it reached us. As a result, we would observe the burst as a more gradual increase and decrease in light output over a period of several days, rather than a sharp increase and decrease over a few hours.

In addition, the turbulent interstellar and intergalactic media that the light passes through can also scatter and delay the light, further smearing out any short-term fluctuations in light output. This effect is known as interstellar scintillation and can make it even more difficult to observe short-term variations in the light output of extragalactic sources.

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Answer:

(a) Radius = 4.6 x 10^11 m

(b) speed = 16.96 km/s

Explanation:

Time period, T = 5.4 earth years

mass of sun, M = 1.989 x 10^30 kg

(a) Let the orbital radius is R.

use the formula of period

[tex]T^2 = \frac{4 \pi^2 R^3}{G M}\\\\\left ( 5.4\times 365\times 24\times 3600 \right )^2 = \frac{4\times3.14\times 3.14\times R^3}{6.67\times10^{-11}\times 1.989\times 10^{30}}\\\\R = 4.6\times 10^{11} m[/tex]

(b) Let the speed is v.

[tex]v=\frac{2 \pi\times R}{T}\\\\v=\frac{2\times 3.14\times 4.6\times 10^{11}}{5.4\times 365\times 24\times 3600}\\\\v = 16963.6 m/s =16.96 km/s[/tex]

Answer:

The electric field at the cylinder surface = 80.19 kN/C

Electric flux via the cylinder [tex]\mathbf{\phi_E = 828.63 Nm^2/C}[/tex]

Explanation:

Given that:

For the filament

The length = 4.95 m

The charge = 2.00  µC

The charge per unit length for the filament can be computed as:

[tex]\lambda = \dfrac{q}{l}[/tex]

[tex]\lambda = \dfrac{2}{4.5}\mu C/m[/tex]

Using Gauss's law:

[tex]\phi_E = \oint E^{\to}*dA^{\to}[/tex]---- (1)

where;

electric flux = [tex]\phi_E[/tex]

permittivity of free space = [tex]\varepsilon_o[/tex]

electric field = E

surface area = dA

However, the electric flux [tex]\phi_E[/tex] via the cylinder can be expressed as:

[tex]\phi_E = \dfrac{\lambda l'}{\varepsilon_o}[/tex]

Equation (1) can now be rewritten as:

[tex]\dfrac{\lambda l'}{\varepsilon_o} = \oint E^{\to}*d(2 \pi rl')[/tex]

[tex]|E| =(\dfrac{\lambda l'}{\varepsilon_o 2 \pi rl' })[/tex]

replacing the values into the above equation:

[tex]|E| =(\dfrac{(\dfrac{2}{4.5} \mu C/m) (1.65 \ cm)}{(8.825 \times 10^{-12} C^2/Nm^2)2 (3.14) (10 \ cm )(1.65 cm) })[/tex]

[tex]|E| =(\dfrac{(\dfrac{2}{4.5} \times 10^{-6} C/m) }{(8.825 \times 10^{-12} C^2/Nm^2)2 (3.14) (0.1 \ m ) })[/tex]

[tex]\mathbf{|E| =80.19 \ kN/C}[/tex]

Thus, the electric field at the cylinder surface = 80.19 kN/C

The electric flux now is calculated using the said formula:

[tex]\phi_E = \dfrac{\lambda l'}{\varepsilon_o}[/tex]

[tex]\phi_E = \dfrac{(\dfrac{2}{4.5}\times 10^{-6} \ C)(0.0165 \ m)}{8.85 \times 10^{-12} \ C^2/Nm^2}[/tex]

[tex]\mathbf{\phi_E = 828.63 Nm^2/C}[/tex]

Answer:

a) the most probable speed of the molecules is 409.2 m/s

b) required temperature of xenon is 1322 K

Explanation:

Given the data in the question;

a)

Maximum probable speed of hydrogen molecule (H₂)

[tex]V_{H_2[/tex] = √( 2RT / [tex]M_{H_2[/tex] )

where R = 8.314 m³.Pa.K⁻¹.mol⁻¹ and given that T = 20.3 K

molar mass of H₂; [tex]M_{H_2[/tex] = 2.01588 g/mol

we substitute

[tex]V_{H_2[/tex] = √( (2 × 8.314 × 20.3 ) / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = √( 337.5484 / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = 409.2 m/s

Therefore, the most probable speed of the molecules is 409.2 m/s

b)

Temperature of xenon  = ?

Temperature of hydrogen = 20.3 K

we know that;

T = (Vxe² × Mxe) / 2R

molar mass of xenon; Mxe = 131.292 g/mol

so we substitute

T = ( (409.2)² × 131.292 × 10⁻³) / 2( 8.314  )

T = 21984.14167 / 16.628

T = 1322 K

Therefore, required temperature of xenon is 1322 K

Answer: Choice D) 19.6 J

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

Work Shown:

m = 2 kg = mass

g = 9.8 m/s^2 = acceleration of gravity (approximate)

h = 1 m = height

---------

PE = potential energy

PE = m*g*h

PE = 2*9.8*1

PE = 19.6 J

The vase has approximately 19.6 Joules of potential energy.

We didn't have to make any conversions because the unit Joule is equivalent to the more complicated unit of kg*m^2/s^2 so it only involves kilograms, meters and seconds. If however the mass was given in say grams (instead of kg), then you'd need to convert to kg.

Statements imply it is thrown with velocity 30cos30° horizontally and 30sin30° vertically.

Vertically:

Total time taken = 2 x time to go up

= 2(v - u)/a

= 0 - 30sin30°)/(-g)

= 30/g

Therefore, it would travel 30/g sec in horizontal direction as well.

Horizontally :

Distance = horizontal speed x time

= 30cos30° (30/g)

= 450√3 /g

If g = 10, distance is 45√3 m.

Vertically,

Distance = vert. speed x (time of flight/2)

= 30sin30° x (30/g)/2

= 90 m.

Time taken = 30/g = 3 sec

Answer:

0.217seconds is the answer

By George, you've nailed it, Stacy !

That's a fact, uh huh.

Truer words were never written.

Your statement is one of unquestionable veracity.

The pure truthiness of it cannot be denied.

Was there a question you wanted to ask ?

Answer:

oh 50 points! how did you do it??!?!?!?! I see up to 8 points only

Answer:

Hello, thank you for giving out points, you are very kind !