k = [tex] \dfrac{ (\dfrac{h}{ \lambda} )^{2} }{2m} [/tex]
k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷
k = (1.12 × 10-³⁰)^2/3.346×10-²⁷
k = 1.25 × 10-⁶⁰ /3.346×10-²⁷
k = 0
ldk why, my answer is coming this :(
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.
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
If the distance between the center of two objects is quadrupled. The gravitational
force between the two objects will change by a factor of:
1) 16
2) 0.25
3) 4
4) 0.0625
Answer:
F' = F/16
Explanation:
The gravitational force between masses is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
If the distance between the center of two objects is quadrupled, r' = 4r
New force will be :
[tex]F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(4r)^2}\\\\F'=\dfrac{Gm_1m_2}{16r^2}\\\\F'=\dfrac{1}{16}\times \dfrac{Gm_1m_2}{r^2}\\\\F'=\dfrac{F}{16}[/tex]
So, the new force will change by a factor of 16.
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
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.
define a system who's momentum is observed
product of force and perpendicular distance
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
45. Pressure in air undergoes a decrease when the air
a) rises to higher altitudes.
b) accelerates to higher speed.
c) fills a greater space.
d) All of these.
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.
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
12. A car is travelling at 30 m/s when the driver sees a red light in the distance and immediately applies the brakes. The car comes to a stop 1.5 s later. How far did the car move from the time the driver applied the brake to when it came to a stop?
Answer:
22.5 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 30 m/s
Time (t) = 1.5 s
Final velocity (v) = 0 m/s
Distance (s) =?
The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:
s = (u + v)t/2
s = (30 + 0)1.5 / 2
s = (30 × 1.5) / 2
s = 45 / 2
s = 22.5 m
Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.
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 !
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.
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
semiconductor have negative temperature coefficient of resistance why
Answer:
As the number of free electrons increases, the resistance of this type of non-metallic material decreases with increasing temperature.
Explanation:
Which phenomenon occurs when one wave is superimposed on another?
A. Interference
B. Refraction
C. Diffraction
D. Polarization
Answer:Alternativa A. Damos o nome de interferência a superposição de efeitos que ocorre ao ser produzido dois pulsos de onda, que serão propagados e acabarão inevitavelmente por se encontrar. No instante em que os pulsos se cruzarem, há então, uma superposição de efeitos individuais de cada um deles. Se durante o cruzamento, houver um reforço das ondas, estará ocorrendo a este fenômeno.
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:
A rectangular field is of length 42 cm and breadth 25 m. Find the area of the field in SI unit. EXPLAIN STEP BY STEP
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
Hydrogen carried in light phase
Answer:
because it is helpful to human beings I think
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.
To learn more about the mechanical advantage, refer to the link;
https://brainly.com/question/7638820
#SPJ1
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 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:
The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.
Explanation:
The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum
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 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]
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
A lumberjack is trying to drag a small tree that he cut down. If the static
coefficient of friction of the tree on the ground is 0.5 and the tree weighs 430
N, what is the minimum amount of horizontal force that he will need to apply
so that the tree will start moving?
A. 215 N
B. 430 N
C. 365 N
D. 500 N
Answer:
A
Explanation:
weight of the tree =normal force
Horizontal force =coefficient of friction x Fnormal
0.5×430=215
Suppose that 2 J of work is needed to stretch a spring from its natural length of 32 cm to a length of 46 cm. (a) How much work (in J) is needed to stretch the spring from 37 cm to 41 cm
Answer:
the work required is 0.163 J
Explanation:
Given;
Energy applied to the spring, E = 2 J
initial length of the spring, x₀ = 32 cm
final length of the spring, x₁ = 46 cm
Extension of the spring, Δx = x₁ - x₀ = 46 cm - 32 cm = 14 cm = 0.14 m
The spring constant is calculated as follows;
E = ¹/₂kΔx²
[tex]k = \frac{2E}{(\Delta x)^2} = \frac{2\times 2}{(0.14)^2} = 204.1 \ N/m^2[/tex]
The extension of the spring when it is stretched from 37 cm + 41 cm:
Δx = 41 cm - 37 cm = 4 cm = 0.04 m
The work required:
W = ¹/₂kΔx²
W = ¹/₂ x (204.1) x (0.04)²
W = 0.163 J
Therefore, the work required is 0.163 J
Which term describes friction that acts on a stationary object?
O A. Static friction
B. Sliding friction
C. Kinetic friction
D. Resistance friction
Answer:
Static friction
Explanation:
Answer:
Static friction
Explanation:
Use the simulation to compare the masses of the three colored and unlabeled weights of different sizes. To do so, set the spring constant of both springs to the same value. Hang known weights on the left spring and an unknown weight on the right spring, and compare the two. Use as many known weights as necessary to determine the unknown masses, and then place each into the appropriate mass bins in the ranking task below. A. M<50 g
B. M = 50 g
C. 50 g D. M = 100 g
E. 100 g F. M = 250 g
G. M> 250 g
1. Blue medium sized weight
2. Magenta small sized weight
3. Gold large sized weight
Answer:
The answer is given as follows,
Explanation:
Gold large-sized weight 100 g < M < 250g
50 g < Magenta small-sized weight < 100g
100g < Blue medium-sized weight < 250g
Hence,
100g < Blue medium-sized weight < 250g
50 g < Magenta small-sized weight < 100g
100 g < Gold large-sized weight < 250g.
The correct formula for finding the relative velocity of an object is:
WILL MARK BRAINLIEST TO THE CORRECT ANSWER!!
Answer:
[tex]V_{a/c} = V_{a/b} + V_{b/c}[/tex]
Explanation:
The relative velocity of an object is the velocity of the object relative to the observer or frame of reference.
The velocity of particle "A" with respect to particle "B" is written as [tex]V_{A/B} = V_{A} - V_{B}[/tex]
From the given options, the second option is the correct answer.
[tex]V_{a/c} = V_{a/b} + V_{b/c}\\\\Re-arrange \ the \ above \ equation;\\\\V_{a/c} - V_{b/c}= V_{a/b}\\\\or\\\\V_{a/b}= V_{a/c} - V_{b/c}[/tex]