Answer:
The basic difference is wavelength (frequency)
ultraviolet < visible < infrared where the visible spectrum lies at
4000 - 7000 Angstroms = 400 - 700 millimicrons (10^-9 meters)
Tjhe shorter wavelengths (ultraviolet) carry more energy / photon
A particle is moving in a circle of radius 2 meters according to the relation (please refer to picture), where Vo is measured in radians and in seconds. The speed of the particle at = 4 seconds is
(A) 13 m/s
(B) 16 m/s
(C) 26 m/s
(D) 52 m/s)
(E) 338 m/s
Answer:
ω = dΘ / dt
dΘ / dt = d (3 t^2 + 2 t) / dt = 6 t + 2
At 4 sec ω = 6 * 4 + 2 = 26 / sec
V = ω * R = 26 / sec * 2 m = 52 m/s
A ball that has a mechanical energy of 65 J haas 10 j of kinetic energy. The ball has j of potential energy
Answer:
55 j
Explanation:
Mechanical energy = kinetic + potential
65 = 10 + P.E.
55 j = PE
1. A hockey puck has a mass of 0.115-kg and strikes the pole of the net at (37 m/s)x. It bounces off in the opposite direction at (-25 m/s)x +(5 m/s)y. (A) What is the impulse on the puck? (B) If the collision takes .005s what is the force acted on the puck?
(a) The impulse on the puck is 1.323 kgm/s.
(b) The magnitude of the force that acted on the puck is 264.5 N.
Impulse of the puckThe impulse on the puck is equal to the change in momentum of the puck.
The change in th momentum of the puck is calculated as follows;
ΔP = m(u - v)
where;
v is the final velocity = (-25x + 5y) m/su is the initial velocity = 37 m/sMagnitude of the final velocity
|v| = √[(-25)² + (5)²] = 25.5 m/s
ΔP = 0.115(37 - 25.5)
ΔP = 1.323 kgm/s
Thus, the impulse on the puck is 1.323 kgm/s.
Force on the puckThe magnitude of the force that acted on the puck is calculated as follows;
F = ΔP/t
F = 1.323/0.005
F =264.5 N
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A person is dragging a packing crate of mass 100 kg across a rough horizontal floor where the coefficient of kinetic friction is 0. 400. He exerts a force f sufficient to accelerate the crate forward. At what angle above horizontal should his pulling force be directed in order to achieve the maximum acceleration?.
For a person dragging a packing crate of mass 100 kg, the angle above the horizontal is mathematically given as
theta = 21.8 °
What angle above horizontal should his pulling force be directed in order to achieve the maximum acceleration?.Generally, the equation for the Force is mathematically given as
[tex]Fnet = F_{costheta} - F_{friction}[/tex]
Therefore
Fnet = Fcostheta - mu*((mg) - Fsintheta)
sin theta = mu*costheta
mu = tantheta
theta = tan^-1(mu)
theta = tan^-1(.400)
theta = 21.8 °
In conclusion, the angle above horizontal
theta = 21.8 °
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A net constant force of 1500 n gives a toy rocket an acceleration of 2. 5 m/������ 2. What is the mass of the rocket?
Answer:
600Kg
Explanation:
Force= mass × acceleration
mass= 1500/2.5
600kg
The mass of the rocket will be 600 kg.In the International System of Units (SI), the kilogram is the unit of mass.
What is mass?Mass is a numerical measure of inertia, which is a basic feature of all matter. It is, in effect, a body of matter's resistance to a change in speed or position caused by the application of a force.
In the International System of Units (SI), the kilogram is the unit of mass.
The given data in the problem is;
The net constant force is, F= 1500 N
The acceleration is,a=2. 5 m/s²
The mass of the rocket is,m=?
[tex]\rm F= mass \times acceleration \\\\ m = \frac{F}{a} \\\\ m= \frac{1500}{2.5}\\\\ m = 600 \ kg[/tex]
Hence the mass of the rocket will be 600 kg.
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What is the Atomic Mass / Weight for silver? *Round your answer to the nearest whole number
A metal bar is in the xy-plane with one end of the bar at the origin. a force f⃗ =( 6.88 n )i^+( -2.62 n )j^ is applied to the bar at the point x= 2.88 m , y= 3.70 m .
A metal bar is in the xy-plane with one end of the bar at the origin, [tex]\[r = (3.87 \, \text{m}) \mathbf{i} + (3.21 \, \text{m}) \mathbf{j}\][/tex], [tex]\theta\approx 55.16^\circ\][/tex], magnitude of torque is 22.58 Nm.
a) Position Vector r can be calculated as:
Given:
[tex]\(x = 3.87 \, \text{m}\)[/tex]
y = 3.21 m
The position vector r for the point where the force is applied is:
[tex]\[r = (3.87 \, \text{m}) \mathbf{i} + (3.21 \, \text{m}) \mathbf{j}\][/tex]
b) Magnitude of Torque can be calculated as:
Magnitude of r:
[tex]\[|r| = \sqrt{(3.87 \, \text{m})^2 + (3.21 \, \text{m})^2} \approx 5.019 \, \text{m}\][/tex]
Magnitude of F:
[tex]\[|F| = \sqrt{(6.82 \, \text{N})^2 + (-3.24 \, \text{N})^2}\\\\ \approx 7.610 \, \text{N}\][/tex]
Calculate the dot product of r and F:
[tex]\[r \cdot F = (3.87 \, \text{m}) (6.82 \, \text{N}) + (3.21 \, \text{m}) (-3.24 \, \text{N})\\\\ \approx 16.658 \, \text{N} \cdot \text{m}\][/tex]
Calculate the angle [tex]\(\theta\)[/tex] using the dot product and magnitudes:
[tex]\[\cos(\theta) = \frac{r \cdot F}{|r| \cdot |F|} \\\\= \frac{16.658 \, \text{N} \cdot \text{m}}{(5.019 \, \text{m})(7.610 \, \text{N})}\\\\ \approx 0.5526\][/tex]
[tex]\[\theta \approx \cos^{-1}(0.5526)\\\\ \approx 55.16^\circ\][/tex]
Now calculate the magnitude of the torque using the sine of the angle:
[tex]\[|\tau| = |r| \cdot |F| \cdot \sin(\theta)\\\\ = (5.019 \, \text{m}) (7.610 \, \text{N}) \sin(55.16^\circ)\\\\ \approx 22.58 \, \text{N} \cdot \text{m}\][/tex]
c) Direction of Torque:
According to the right-hand rule, the torque points either into or out of the plane formed by r and F. Its direction is perpendicular to that plane. The sign of the angle determines the precise orientation.
In this case, since [tex]\(\theta \approx 55.16^\circ\)[/tex] is positive (counterclockwise rotation), the torque points out of the plane formed by r and F.
Thus, the direction of the torque is outward from the xy-plane.
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Your question seems incomplete, the probable complete question is:
A metal bar is in the xy-plane with one end of the bar at the origin. A force F=(6.82)i+ -3.24j is applied to the bar at the point x=3.87m , y=3.21m .
What is the position vector r for the point where the force is applied?
a Express your answer in terms of the unit vectors i and j .
b What are the magnitude of the torque with respect to the origin produced by F?
c What are direction of the torque with respect to the origin produced by F?
Question 3
The FRC robot is rushing to pick up a game ball to shoot. The robot
is at rest and then it travels 30m in 1.96s. What is the robot's
acceleration in this time?
Answer:
15.61 m/s^2
Explanation:
From 3 rd equation of motion
s=ut+1/2at^2
given s = 30
u= 0 ( as body was at rest)
t= 1.96s
30=0*1.96+1/2a(1.96)^2
30=0+1/2a(1.96)^2
a=2*30÷(1.96)^2
a= 15.61 m/s^2 (approx)
What is current of 12 ohm? help me pleasee
1. 0.25 A
2. 0.5 A
3. 1 A
4. 1.5 A
5. 2 A
Answer:
the current of 12 ohm is 2A
Why wouldn’t you move during an earthquake?
Answer:
So you maybe don't fall, have something fall on you, and for a bunch of different reasons. EARFQUAKES are really dangrous.
Explanation:
Hope this helps you!
Which of the following was correctly predicted by Einstein's theory of gravity?
(A) the origin of mass
(B) law of refraction
(C) the measure of G
(D) light deflected around large objects because of their gravity
Answer:
(D) light deflected around large objects because of their gravity
I hope this helps you
:)
How many stars are in a typical spiral galaxy, like our own?.
Answer:
200 billion to 400 billion stars in a typical spiral
Explanation:
hope this helps
have an awesome day -TJ
Write an expression for the coefficient of kinetic friction between the block and the horizontal surface. Your answer should be in terms of x, vB, and g (the acceleration due to gravity)
An expression for the coefficient of kinetic friction between the block and the horizontal surface is given as μk = Ff/mg.
What is coefficient of kinetic friction?
The coefficient of kinetic friction is the force of frcition between a moving object and the horizontal surface.
The expression for the coefficient of kinetic friction between the block and the horizontal surface is given as;
Ff = μkFₙ
Ff = μkmg
μk = Ff/mg
where;
μk is the coefficient of kinetic frictionFf is the kinetic frictional forcem is mass of the blockg is acceleration due to gravityThus, an expression for the coefficient of kinetic friction between the block and the horizontal surface is given as μk = Ff/mg.
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2. What would be the effect on your body if you did not have kidneys? Page
112
A. My body would not absorb nutrients.
B. Waste would collect in the blood and become toxic
C. I could not eat gluten
D. Saliva could not be produced
Answer:
b
Explanation:
the kidney eliminates waste from the blood hence no kidney will result in accumulation of waste
If the barge is made out of 4.8- cm -thick steel plate on each of its four sides and its bottom, what mass of coal can the barge carry in freshwater without sinking
The maximum mass of coal that the barge will carry in freshwater without sinking is 0.111 kg.
Volume of the bargeThe volume of the barge is calculated as follows;
V = L³
V = (4.8 cm)³
V = 110.6 cm³
Density of the bargeFor the barge to float in freshwater, its density must be less or equal to density of water.
density(barge) = density(water)
density(barge) = 1 g/cm³
Maximum mass to be carried by the barge without sinkingmass = density x volume
mass = 1 g/cm³ x 110.6 cm³
mass = 110.6 g
mass = 0.111 kg
Thus, the maximum mass of coal that the barge will carry in freshwater without sinking is 0.111 kg.
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A laser beam is used to levitate a metal disk against the force of earth's gravity.
(a) derive an equation giving the required intensity of light, i, in terms of the mass m of the disk, the gravitational acceleration g, the speed of light c, and the cross-sectional area of the disk a. assume the disk is perfectly reflecting and the beam is directed perpendicular to the disk. (do not substitute numerical values; use variables only.)
i =
(b) if the disk has mass 3.89 g and radius 2.40 cm, find the necessary light intensity.
(c) give two reasons why using light pressure as propulsion near earth's surface is impractical.
The necessary light intensity of the disk if the disk has mass 3.89 g and radius 2.40 cm is 6.483 x 10^12cd.
Calculations and Parameters:a. Given that:
i= intensity of light m= mass of disk g= gravitational acceleration c= speed of light a= cross-sectional area of the diskI = P/A
= Force x Speed/Area
Gravitational force, F= GM1M2/r^2
F= ma, F= mg
I= mg x c/q
That is,
i= mass x gravitational acceleration x speed/area
b.
Mass= 3.89g
Radius, r= 2.40cm
i=?
i= mg x c/a
Speed of light= 3.0 x 10^8 m/s
Area, a= /pi r^2
= 3.142 x (0.024)^2
= 3.142 x 0.000576
= 0.0018m^2.
i= 3.89 x 10 x 3.0 x 10^8/0.0018
i= 6.483 x 10^12cd.
c. Light has no mass, therefore, it lacks momentum and cannot exert pressure for propulsion of materials.
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What is current of 12 ohm? help me pleasee
Answer:
the current of 12 ohm is 2Amps
4. The splashing of a liquid resembles an oscillation. The restoring force in this scenario will be due to which of the following?
Potential energy
Kinetic energy
Gravity
Mechanical energy
Please show all of your work.
Gravity
when liquid splashed and is at the top, the force bringing the liquid back down to the midpoint is gravity. This brings the liquid back to its equilibrium level.
Answer:
Gravity
Explanation:
The splashing of liquid resembles like an oscillation .
The force acting on liquid to pull it down is gravity and the force bouncing it back is caused by normal reaction
Hence gravity is the answer
20. Free-Body Diagram You are sitting in the back seat of a
car going around a curve to the right. Sketch motion
and free-body diagrams to answer these questions:
a. What is the direction of your acceleration?
b. What is the direction of the net force on you?
Centripetal acceleration
#1
Direction towards centre#2
Direction of force also towards centre
Find the number of permutations when 7 objects are taken 1 at a time
Using the formula for permutation, there are 5040 ways that 7 objects are taken 1 at a time.
What is permutation?The permutation of objects refers to the number of ways that an object can be selected from a number of objects.
Given that;
nPr = n!/(n - r)!
Hence;
7P1 = 7!/(7 - 1)!
7P1 = 7 * 6 !/6!
7P1 = 7 * 6 * 5 * 4 * 3 * 2 *1
7P1 = 5040
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What causes changes in the orbit of objects in the solar system.
Which planets are generally referred to as the ice giants?.
Answer:
The correct answer is Uranus and Neptune
Brody drives 180 km in two hours. How fast is he travelling?
should be in km/hour)
Answer:
90 km/h
Explanation:
we just divide the distance by the time to get the speed.
By what means can the internal energy of a closed system increase?.
Answer:
when work is done on the system or heat comes into the system
Which planet moves the fastest and which planet moves slowest? a, b, c, d, e, or f. Explain
A 1. 5-kω resistor and 27. 7-mh inductor are connected in series to a vrms = 120 v ac power source oscillating at a frequency of f = 60 hz. The voltage as a function of time is given by
The voltage as a function of time for the power source oscillating at the given frequency is determined as V = 169.7sin(120πt).
Peak voltageThe peak voltage of the source is calculated as follows;
Vrms = 0.7071V₀
120 = 0.7071V₀
V₀ = 120/0.7071
V₀ = 169.7 V
Angular frequency of the waveω = 2πf
where;
f is frequency = 60 Hz
ω = 2π x 60
ω = 120π rad/s
Voltage equation as function of timeV = V₀sin(ωt)
V = 169.7sin(120πt)
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The voltage equation is a function of a time is V = 169.7sin(120πt) if the 1. 5-kω resistor and 27. 7-mh inductors are connected in series to a vrms = 120 v ac power source oscillating at a frequency of f = 60 Hz.
What is induced voltage?Electromagnetic induction causes the induced voltage. Electromagnetic induction is the process of creating emf (induced voltage) by exposing a conductor to a magnetic field.
We have:
[tex]\rm V{rms} = 120[/tex]
We know:
[tex]\rm V_{rms} = 0.7071V_o[/tex]
[tex]\rm 120 = 0.7071V_o[/tex]
[tex]\rm V_o = 169.70[/tex] Volts
Expression for the angular frequency is given by:
ω = 2πf
Where f is frequency
ω = 2π x 60
ω = 120π rad/s
Voltage equation as a function of time is given by:
V = V₀sin(ωt)
V = 169.7sin(120πt)
Thus, the voltage equation is a function of a time is V = 169.7sin(120πt) if the 1. 5-kω resistor and 27. 7-mh inductors are connected in series to a vrms = 120 v ac power source oscillating at a frequency of f = 60 Hz.
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A bicycle wheel of radius 40. 0 cm and angular velocity of 10. 0 rad/s starts accelerating at 80. 0 rad/s2. What is the centripetal acceleration of the wheel at this time point?
The magnitude of the centripetal acceleration of the bicycle wheel is 40 m/s².
Centripetal acceleration of the wheel
The centripetal acceleration of the bicycle wheel is the inward or radial acceleration of the bicycle on the circular path.
The magnitude of the centripetal acceleration of the bicycle wheel is calculated as follows;
a = ω²r
where;
ω is angular speed
r is the radius of the circular path
a = (10)² x 0.4
a = 40 m/s²
Thus, the magnitude of the centripetal acceleration of the bicycle wheel is 40 m/s².
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The acceleration of a body travelling in a circular route is known as centripetal acceleration. The centripetal acceleration of the wheel at this time point is 40 m/sec².
What is centripetal acceleration?The acceleration of a body travelling in a circular route is known as centripetal acceleration. It is given by the formula,
[tex]a_c = \dfrac{v^2}{r} = \omega^2 r[/tex]
The centripetal acceleration of the wheel with angular velocity 10 rad/sec and velocity 40cm (0.4 m) can be written as,
[tex]a_c = \omega ^2\times r\\\\a_c = (10)^2\times 0.4\\\\a_c = 40\rm\ m/sec^2[/tex]
Hence, the centripetal acceleration of the wheel at this time point is 40 m/sec².
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Which Tesla would win in a drag race?
(A) 2009 Tesla Roadster
(B) 2016 Tesla Model X
Answer:B
Explanation:
A has about 248 hp but B has about 683 hp. 2016 is younger too. good day
What increases the transfer of the wind's energy to the water?
Question 4 options:
A. flat water
B. large waves
C. large rocks
PLEAS HELP ME FAST
Answer:
A, flat water
Explanation:
Scientists observed a rocky, rotating object orbiting in space. It was 250 km wide. The rocky object was located several times farther away from the sun than the earth
The rocky object orbiting in space is most likely referred to as an asteroid.
What is an Asteroid?This is referred to a rocky object which revolve around the sun and are considered too small to be called a planet.
Asteroids was observed with the use of a telescope and is around 250 km wide thereby making it the most appropriate choice.
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