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
26.
Which one of the following is not a vector quantity?
(2)
A) acceleration
C) displacement
E) instantaneous velocity
B) average speed
D) average velocity
Answer:
Answer: Speed is not a vector quantity. It has only magnitude and no direction and hence it is a scalar quantity.
Write the relation connecting Celsius scale and Fahrenheit scale of temperature
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
how do you use the coefficient to calculate the number of atoms in each molecule?
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:
A ball is thrown with an initial velocity of 30.0 m/s and makes an angle of
30.0° with the ground. Find the
A.Horizontal Distance
B.Maximum Height
C.Total Time The Ball is Traveling
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
3. What is the equation for the mechanical advantage of a lever?
MA =
length of effort arm / length of resistance arm
MA = length of effort arm * length of resistance arm
MA = length of resistance arm/length of effort arm
MA = length of effort arm + length of resistance arm
PLEASE HELPP!!!!!
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
[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
How long will be required for a car to go from a speed of 27.0m/s to a speed of 32.0m/s if the acceleration is 2 3.0m / s ?
Answer:
0.217seconds is the answer
Calculate the Combined resistance of the Circuit voltage across each resistor Current Passing through each resistor of 6,8,12ohms
Answer:
Sorry I don't know the answer
A circus performer stretches a tightrope between two towers. He strikes one end of the rope and sends a wave along it toward the other tower. He notes that it takes 0.9 s for the wave to travel the 26 m to the opposite tower. If one meter of the rope has a mass of 0.28 kg, find the tension in the tightrope.
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
Two horizontal forces are acting on a box. The box moves only along the x axis. There is no friction between the box and the surface. Suppose that = +5.9 N and the mass of the box is 3.6 kg. Find the magnitude and direction of when the acceleration of the box is +7.1 m/s^2.
Answer:
sorry I don't know I am only in 7th grade
Help! plz!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Now!!!!
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.
The energy wasted in using a machine is 600j. if the machine is 70% efficient. calculate the volume of water pumb by the machine through a height of 15m.
Answer:
yhgigy6ftu5cg l8vbbnnbbgtccccvhklhaywje nc 62bbnzmakbdbvfvdbf93bdldmffmfkqhdv
Question 6 of 10
A 2 kg vase is sitting on a 1 m high table. What is the vase's potential energy?
A. 196 J
B. 2J
C. 0.5 J
D. 19.6 J
SUBMIT
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.
Work and the Dot Product
A variable 1D force acts on an object of mass 2 kg, which is initially moving at 5 m/s to the right (along the positive x direction). The net force is given by:
F x = 20x2-10 i Newtons x
The force acts on the object as it displaced from x = 1 m to x = 4 m .
a) Findthespeedoftheobjectatx=4m.
b) Is there a gain or loss in kinetic energy or no loss in kinetic energy in the
displacement of the object? Explain.
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.
The archerfish uses a remarkable method for catching insects sitting on branches or leaves above the waterline. The fish rises to the surface and then shoots out a stream of water precisely aimed to knock the insect off its perch into the water, where the archerfish gobbles it up. Scientists have measured the speed of the water stream exiting the fish's mouth to be 3.7 m/s. An archerfish spots an insect sitting 18 cm above the waterline and a horizontal distance of 28 cm away. The fish aims its stream at an angle of 39° from the waterline.
Required:
Determine the height above the waterline that the stream reaches at the horizontal position of the insect.
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]
scripture union was founded by who in what year
Answer:
Josiah Spiers in 1867 was when scripture union was founded
If you blow air between a pair of closely-spaced Ping-Pong balls suspended by strings, the balls will swing
A) toward each other.
B) apart from each other.
C) away from the air stream.
Answer:
c
Explanation:
Your hand and wrist curl in toward the center of your body (chest and stomach) to prepare to throw the frisbee.
O True
O False
True
Hope this helps! :)
Answer:
true because when trow the frisbee gives u level
more pts first 2 so hurry up
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 !
The asteroid belt circles the sun between the orbits of Mars and Jupiter. One asteroid has a period of 5.4 earth years.
A- What is the asteroid's orbital radius?
B-What is the asteroid's orbital speed?
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]
In air an object weighs 15N, when immersed in water it weighs 12N, when immersed in another liquid, it weighs 13N, Calculate the density of the object and that of the other liquid?
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.
Kilometer is a unit of length where as kilogram is a unit of mass
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 ?
The following 1H NMR absorptions were obtained on a spectrometer operating at 200 MHz and are given in Hz downfield from TMS. Convert the absorptions to δ units. a) 416 Hz = δ b) 1.97×103 Hz = δ c) 1.50×103 Hz = δ
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
For (a):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]
For (b):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]
For (c):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]
The liquid and gaseous state of hydrogen are in thermal equilibrium at 20.3 K. Even though it is on the point of condensation, model the gas as ideal and determine the most probable speed of the molecules (in m/s). What If? At what temperature (in K) would an atom of xenon in a canister of xenon gas have the same most probable speed as the hydrogen in thermal equilibrium at 20.3 K?
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
A uniformly charged, straight filament 4.95 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder 1.65 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find:
a. the electric field at the surface of the cylinder
b. the total electric flux through the cylinder
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]
i didnt want my question public i made a mistake i want it taken down
Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed in m/s of a satellite in an orbit 980 km above the Earth's surface.
Answer:
564
Explanation:
How much work is done
In moking a charge of 2
2
Coloumbs from a point a
Ilsu to a point at 1284?
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.
como calcular la velocidad un atleta en los 100 metros planos?
Answer:
Explanation:
9ooooo
Hydrogen carried in light phase
Answer:
because it is helpful to human beings I think
Why is it that, when we observe an extragalactic source whose diameter is about one lightday, we are unlikely to see fluctuations in light output in times shorter than about one day
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.
How to observe extragalactic sources whose diameter is about one light day?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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