Answer:
16.9000000000000001 J
Explanation:
From the given information:
Let the initial kinetic energy from point A be [tex]K_A[/tex] = 1.9000000000000001 J
and the final kinetic energy from point B be [tex]K_B[/tex] = ???
The charge particle Q = 6 mC = 6 × 10⁻³ C
The change in the electric potential from point B to A;
i.e. V_B - V_A = -2.5 × 10³ V
According to the work-energy theorem:
-Q × ΔV = ΔK
[tex]-Q \times ( V_B - V_A) = (K_B - K_A)[/tex]
[tex]-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)[/tex]
[tex]15 = (K_B - 1.9000000000000001 \ J)[/tex]
[tex]K_B = 15+ 1.9000000000000001 \ J[/tex]
[tex]\mathbf{K_B =1 6.9000000000000001 \ J}[/tex]
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
The question is incomplete, the complete question is;
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
A) The atom moves to a state of lower energy
B) The atom is ionized
C) One of the electrons leaves the atom
D) The atom can be excited to a higher energy state
Answer:
The atom can be excited to a higher energy state
Explanation:
According to the Bohr model of the atom, electrons in an atom can be excited from a lower to a higher energy level when energy is absorbed by the atom.
If electrons having an energy of 10.5ev are incident on a hydrogen atom, this energy is transferred to the atom by collision. Since the energy transferred is less than the ionization energy of hydrogen atom in its ground state(13.6ev), the atom is not ionized.
Rather, the atom is excited from ground state to a higher energy level.
A camera lens with focal length f = 50 mm and maximum aperture f>2
forms an image of an object 9.0 m away. (a) If the resolution is limited
by diffraction, what is the minimum distance between two points on the
object that are barely resolved? What is the corresponding distance
between image points? (b) How does the situation change if the lens is
“stopped down” to f>16? Use λ= 500 nm in both cases
Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff? ( EASY QUESTION.. PLZZ HELPPP MEEE I WILL MARK YOU THE BRAINLIEST PLZZ)
Answer:
144 meters
Explanation:
the ball is thrown with a speed of 24 meters per second right so if the ball reaches the ground in 6 seconds. the hight of the cliff must be S=v.t
S (height cliff)=24m/s×6s=144
g the total mechanical energy of the satellite-Earth system when the satellite is in its current orbit is E. In order for the satellite to orbit Earth in a new stable circular orbit at an altitude of 12RE, the energy of the satellite-Earth system must be
Answer:
The correct answer is "[tex]\frac{4E}{3}[/tex]".
Explanation:
According to the question,
Energy of satellite,
⇒ [tex]E_s=-\frac{GM_sM_E}{2r}[/tex]
For the very 1st case:
[tex]r = R_E+R_E[/tex]
[tex]=2R_E[/tex]
or,
⇒ [tex]E=-\frac{GM_sM_E}{4R_E}[/tex]...(1)
For the new case:
[tex]r = R_E+\frac{R_E}{2}[/tex]
[tex]=\frac{3R_E}{2}[/tex]
then,
⇒ [tex]E'=-\frac{GM_sM_E}{2 \frac{3R_E}{2} }[/tex]
[tex]=-\frac{GM_sM_E}{3R_E}[/tex]...(2)
From equation (1) and (2), we get
⇒ [tex]E'=\frac{1}{3}(4E)[/tex]
[tex]=\frac{4E}{3}[/tex]
an object that has momentum must have
Answer:
Mechanical energy
Explanation:
If an object does not have any momentum, then it doesn't have mechicanl energy.
A 55 kg pole vaulter falls from rest from a height of 5.4 m onto a foam rubber pad. The pole vaulter comes to rest 0.24 s after landing on the pad.
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision.
Answer:
a) 10.3 m/s
b) 566 N
Explanation:
[tex]v {}^{2} = {u}^{2} + 2as \\ v {}^{2} = 0 {}^{2} + 2(9.81)(5.4) \\ v = 10.3 \: ms {}^{ - 1} [/tex]
[tex]force \: = \frac{d(mv)}{dt} \\ = 55(10.293) \\ = 566 \: newtons[/tex]
The athelete velocity will be 10.3 and constant force 566 N.
What is velocity?The displacement that an object or particle experiences with respect to time is expressed vectorially as velocity. The meter per second (m/s) is the accepted unit of velocity magnitude (also known as speed).
Alternately, the magnitude of velocity can be expressed in centimeters per second (cm/s). Depending on how many dimensions are included, there are numerous ways to indicate the direction of a velocity vector.
The car's velocity in relation to your body is zero when you are driving. The speed of the car in relation to you if you were to stand by the side of the road is 20 m/s northward.
Therefore, The athelete velocity will be 10.3 and constant force 566 N.
To learn more about velocity, refer to the link:
https://brainly.com/question/18084516
#SPJ5
A mass is suspended on a spring. The spring is compressed so that the mass is located 5 cm above its rest position. The mass is released at time t= 0 and allowed to oscillate. It is observed that the mass reaches its lowest point 1/2s after it is released.
Required:
Find an equation that describes the motion of the mass.
Answer:
y = 5 cos 2πt
Explanation:
We will use the formula for simple harmonic motion curve where;
y = a cos ωt
Where;
a is amplitude
t is period
ω is angular frequency with the formula; ω = 2π/t
We are told that when the spring is compressed, the mass is located 5 cm above its rest position.
Thus;
a = 5 cm
it's highest point is 5 cm, but we are told that after 1/2 second of being released, it reaches its lowest point.
Since highest point is 5, then lowest point will be -5.
The difference in time between the highest and lowest point is ½ s. Which is half of the period.
Thus;
t/2 = ½
Thus, t = 1 s
Now, we know that;
t = 1/f = 2π/ω
Since t = 1, then 1 = 1/f
f = 1
Thus;
2π/ω = 1
Thus, ω = 2π
Thus, the equation is;
y = 5 cos 2πt
The equation that describes the motion of the mass is y = 5 cos 2πt.
The given parameters;
maximum displacement of the spring, A = 5 cmtime taken for the mass to reach the lowest point (half period), t = 0.5 sThe general equation of the wave is given as;
[tex]y = A\ cos\ \omega t[/tex]
where;
A is the amplitude of the vibrationω is the angular speed of massThe angular speed of the mass is calculated as;
[tex]\omega = 2\pi f\\\\[/tex]
The period of the oscillation is calculated as;
[tex]T = 2t \\\\T = 2(0.5 s) = 1 \ s[/tex]
The frequency of the wave is calculated as;
[tex]f = \frac{1}{T} \\\\f = \frac{1}{1} \\\\f = 1\ Hz[/tex]
The equation that describes the motion of the mass is calculated as;
[tex]y = A\ cos \ \omega t\\\\y = A\ cos \ 2\pi ft\\\\y = 5\ cos \ 2\pi (1) t\\\\y = 5 \ cos \ 2\pi t[/tex]
Thus, the equation that describes the motion of the mass is y = 5 cos 2πt.
Learn more here:https://brainly.com/question/14017869
An arrow is shot from a height of 1.5 m toward a cliff of height H. It is shot with a velocity of 30 m/s at an angle of 60 degrees above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff
Answer:
The height of the cliff is 27.02 m
Explanation:
Given;
height above the ground from which the arrow is shot, h₁ = 1.5 m
initial velocity of the projectile, v = 30 m/s
angle of projection, θ = 60⁰
time taken to reach top of the cliff, t = 4.0 s
The vertical component of the velocity is calculated as;
[tex]v_y = v \times sin(\theta)\\\\v_y = 30 \times sin(60)\\\\v_y = 25.98 \ m/s[/tex]
The height attained by the projectile at the given time is calculated as;
[tex]h_2= v_y t - \frac{1}{2}gt^2\\\\h_2 = 25.98\times 4 \ - \ \frac{1}{2} \times 9.8\times 4^2\\\\h_2= 103.92 \ - \ 78.4 \\\\h_2 = 25.52 \ m[/tex]
The height of the cliff is calculated as;
H = h₁ + h₂
H = 1.5 m + 25.52 m
H = 27.02 m
For each value of the principal quantum number n, what are the possible values of the electron spin quantum number m_s?
a. 0
b. -3/2
c. +3/2
d. +1/2
e. -1/2
Answer:
d) and e)
Explanation:
For any value of the principal quantum n, the only possible values of the electron spin quantum number m_s are + 1/2 and -1/2.Since this number is related with the angular momentum of the electron, it has a magnitude (1/2) and an orientation given by the sign (+ or -)So, the only right answers are d) and e).What is the efficiency of a machine that uses 102 kJ of energy to do 98 kJ of work?
what is measured by the ammeter
Answer:
amperes
Ammeter, instrument for measuring either direct or alternating electric current, in amperes. An ammeter can measure a wide range of current values because at high values only a small portion of the current is directed through the meter mechanism; a shunt in parallel with the meter carries the major portion.
Explanation:
hope it helps
small plastic container, called the coolant reservoir, catches the radiator fluid that overflowswhen the automobile engine becomes hot. The radiator is made of copper, and the coolant has a coefficient of volume expansion of 410 x 10-6 [1/ oC]. If the radiator is filled to its 15 [quart] capacity when the engine is cold at 6.0 [oC], how much overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 [oC]
Answer:
0.53 quart
Explanation:
The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C
Since, ΔV = VγΔθ
substituting the values of the variables into the equation, we have
ΔV = VγΔθ
ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C
ΔV = 528900 × 10⁻⁶ quart
ΔV = 0.528900 quart
ΔV ≅ 0.53 quart
Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.
I really don’t know the answer for this
what is electricity ?
Answer:
it is energy resulting in charged particles
Electricity a form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current
Select the correct answer.
What are the directions of an object's velocity and acceleration vectors when the object moves in a circular path with a constant speed?
OA. The question is meanimgless, since the acceleration is zero.
ов.
The vectors point in opposite directions.
Oc.
Both vectors point in the same direction.
OD
The vectors are perpendicular,
Answer:
A
Explanation:
If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.
Answer: C.
Explanation: plato users
A bike and rider of total mass 75.0 kg moving at a velocity of 30 m/s to the right
collides with a skater of mass 60,0 kg moking at a velocity of 6.0 m/s to the left.
After they collide, the skater has a velocity of O m/s. What is the final velocity of the
bike and rider?
O A. O m/s
O B. 1.8 m/s to the left
O C. 3.0 m/s to left
O D. 1.8 m/s to the right
Answer:
[tex](75.0 \times 30) + (60.0 \times 6.0) = (75.0 \times V) + (60.0 \times 0) \\ 2250 + 360 = 75V \\ 75V = 2610 \\ V = 34.8 \: m {s}^{ - 1} [/tex]
plz answer the question
Answer:
Ray A - incident ray
Ray B - reflected ray
Which describes farsightedness? O Distant objects are blurry. O Concave lenses can correct it. O Objects appear larger when wearing corrective glasses. O Corrective glasses do not change apparent the size of objects.
Answer:
O Distant objects are blurry. describes farsightedness.
Explanation:
Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.
When it comes to the movement of air, friction
A. increases with altitude.
B. is greater near the ground surface.
C. diminishes turbulence.
D. is responsible for weaker winds aloft.
Answer: When it comes to the movement of air, friction is greater near the ground surface.
Explanation:
A resistance in motion observed by an object while on another object is called friction.
For example, a vehicle moving on road will have friction between its tires and the road.
Friction is more near the ground surface rather than away from the ground surface.
Thus, we can conclude that when it comes to the movement of air, friction is greater near the ground surface.
you are stowing items and come across an aerosol bottle of hairspray.what should you do?
Answer:
below
Explanation:
Help help help HELP AAAAA
The graph below shows the distance traveled by the skateboarder on each of the different road conditions. Using the graph, determine which of the roads was dry, wet, or muddy. Explain your answer using complete sentences.
Answer:
Road A- dry
Road B- mud
Road C- wet
Explanation:
Surface conditions do affect the ease and speed with which a skateboarder can move, on a muddy surface, the tyres of the skate boards finds it difficult to establish adequate fictional force between the skates trees and the traveling surface. Hence, the muddy surface presents a very slippery travel ground for the skate, hence leading the to skateboarder needing to apply caution.
The speed on a wet surfave is height as the amount of firece that will be applied in other to accelerate is very small. The surface is wet and hence serves as a lubricant between the contact surface.
The dry road also has a high speed but lower than a wet surface, frictional force is high here and this tend to slow the skateboarder down except in sloppy terrains.
12. What type of circuit is the diagram below?
series circuit
parallel circuit
Answer:
parallel circuit
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
Basically, the components of an electric circuit can be connected or arranged in two forms and these includes;
I. Series circuit
II. Parallel circuit: it's an electrical circuit that has the same potential difference (voltage) across its terminals or ends. Thus, its components are connected within the same common points so that only a portion of current flows through each branch.
Hence, the type of circuit that the above diagram above represents is a parallel circuit.
Answer:
parallel circuit
Explanation:
I got it right on my exam
An ideal heat engine operates between two
temperatures 367.7 K and 865.1 K. What is
the efficiency of the engine?
0.787
0.42504
None
0.213
0.5750
Answer:
As it is the most efficient heat engine, it's efficiency is [T1 - T2]/T1. It can be measured for every Carnot cycle. From the formula and diagram, we can understand that the efficiency of an ideal heat engine also depends on the difference between the hot and cold reservoirs.
Explanation:
In an experiment to measure the temperature of a Bunsen burner flame, a 250 g piece of iron is held in the flame for several minutes until it reaches the same temperature as the flame . The hot metal is then quickly transferred to 285 g of water contained in a 40.0 g copper calorimeter at 15.0 oC. The final temperature of the copper and water is 80.0 oC.
Using your answer from determine the temperature of the Bunsen flame.
Answer:
wait
Explanation:
An airplane, starting from rest, moves down the runway at constant acceleration for 23 s and then takes off at a speed of 66 m/s. What is the average acceleration of the plane (in m/s2)?
Answer:
46
Explanation:
A 2890-lb car is traveling with a speed of 58 mi/hr as it approaches point A. Beginning at A, it decelerates uniformly to a speed of 18 mi/hr as it passes point C of the horizontal and unbanked ramp. Determine the total horizontal force F exerted by the road on the car just after it passes point B.
Answer:
4592.57 lb
Explanation:
The missing diagram for this question is attached in the image below.
Given that:
the weight of the car = 2890 lb
At point A, the speed of the car [tex](V_A)[/tex] = 58 mi/hr
At point C, the speed of the car [tex](V_C)[/tex] = 18 mi/hr
To ft/s:
[tex](V_A)[/tex] = 58 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_A)[/tex] = 85.07 ft/s
[tex](V_C)[/tex] = 18 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_C)[/tex] = 26.4 ft/s
Between A to C, the total distance is;
[tex]S_{AC} = S_{AB}} + S_{BC} \\ \\ S_{AC} = 331 + \dfrac{\pi r}{2} \\ \\ S_{AC}= 331 + \dfrac{\pi \times 207}{2} \\ \\ S_{AC} = 656.154 \ ft[/tex]
Now, we need to determine the deceleration of the car using the formula:
[tex]V_C^2 = V_A^2 + 2 aS_{AC}[/tex]
[tex]26.4^2 = 85.07^2 + 2 a (654.154)[/tex]
[tex]696.96 = 7236.9049+ 2 a (654.154)[/tex]
[tex]696.96-7236.9049 = 2 a (654.154)[/tex]
[tex]-6539.9449 = 2 a (654.154)[/tex]
[tex]a= \dfrac{-6539.9449} {2(654.154)}[/tex]
a = -4.99 ft/s²
The velocity of the car as it passes via B
[tex]v_B^2 = v_A^2 + 2aS_{AB}[/tex]
[tex]v_B^2 = 85.07^2 + 2(-4.99 \times 331)[/tex]
[tex]v_B =\sqrt{ 85.07^2 + 2(-4.99 \times 331)}[/tex]
[tex]v_B =\sqrt{ 85.07^2 +3303.38}[/tex]
[tex]v_B =\sqrt{ 10540.2849}[/tex]
[tex]v_B =102.67 \ ft/s[/tex]
Along B, the car's acceleration is:
[tex]a_B = \sqrt{a^2 + (\dfrac{v_B^2}{r})^2}[/tex]
[tex]a_B = \sqrt{(-4.99)^2 + \dfrac{102.67^2}{207}^2 }[/tex]
[tex]a_B = 51.17 \ ft/s^2[/tex]
Finally, the total horizontal force F exerted = m[tex]a_B[/tex]
[tex]= (\dfrac{2890}{32.2}) \times 51.17[/tex]
= 4592.57 lb
Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).
Answer:
Total cost = 56.16 cents
Explanation:
Given the following data;
Power = 45 Watts
Time = 4 hours
Number of days = 30 days
Cost = 10.4 cents
To find how much does it cost her to watch TV for one month;
First of all, we would determine the energy consumption of the TV;
Energy = power * time
Energy = 45 * 4
Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).
Energy consumption = 0.18 Kwh
Next, we find the total cost;
Total cost = energy * number of days * cost
Total cost = 0.18 * 30 * 10.4
Total cost = 56.16 cents
Ethyl alcohol is :
a. None of the above
b. Semi polar solvent
c. Polar solvent
d. Non-Polar solvant
Answer:
D. Non- polar solvant
Explanation:
l think that's it
Answer:
I think the answer is D polar solvent
A particle of unit mass moves so that displacement after t seconds is given by x = 2 cos (t - 2). Find the acceleration and kinetic energy at the end of 3 seconds. (K.E = (1/2) m v²)
Answer:
a₃ = -1.08 m/s², K = 1.42 J
Explanation:
The particle is in a periodic motion, so the general expression is
x = A cos (wt + Ф)
let's compare the terms with the expression they give us
x = 2 cos (t - 2)
the amplitude of motion is A = 2 m, the angular velocity w = 1 rad / s, and the phase is Ф = - 2.
to find the acceleration we use its definition
v = dx / dt
a = dv / dt
a = [tex]\frac{ d^2x}{dt^2}[/tex]
let's perform the derivative
v = - A w sin (wt + Ф)
a = - A w² cos wt + Ф)
substituting the values
a = - 2 1² cos (t-2)
for t = 3 s
a₃ = 2 cos (3-2)
remember angles are in radians
a₃ = -1.08 m/s²
To calculate kinetic energy, let's find the velocity for t = 3 s
v = - 2 sin (t-2)
v = -2 sin (3-2)
v = - 1.683 m / s
body mass is m = 1 kg
we calculate
K = ½ m v²
K = ½ 1 (-1.683) ²
K = 1.42 J
