Answer:
It will take 7.75 seconds for the car to go 155m
Explanation:
From the question, we can understand that the distance covered by the moving car is got by a product of its speed and the time it travels.
i.e distance = speed X time.
However, in this case, we have the distance travelled and the speed of the car, and we are looking for the time of travel
TO solve this, we will simply make the travel time the subject of the formula in the equation above.
i.e time = distance / speed
time = 155/20= 7.75 seconds.
Hence, it will take 7.75 seconds for the car to go 155m
Assume the three blocks (m. = 1.0 kg, m = 20 kg and m = 40 ko) portrayed in the figure below move on a frictionless surface and a force F: 36w acts as shown on the 4.0 kg block.
a) Determine the acceleration given this system (in m/s2 to the right). m/s2 (to the right)
b) Determine the tension in the cord connecting the 4.0 kg and the 1.0 kg blocks in N). Determine the force exerted by the 1.0 kg block on the 2.0 kg block (in N). N (a) What If How would your answers to parts (a) and (b) of this problem change if the 2.0 kg block was now stacked on top of the 1.0 kg block? Assume that the 2.0 kg block sticks to and does not slide on the 1.0 kg block when the system is accelerated.
(Enter the acceleration in m/s2 to the right and the tension in N.) acceleration m/s (to the right) tension
Answer:
a) 5.143 m/s^2
b) T = 15.43 N
c) Fr = 10.29 N
d) 5.143 m/s^2 , T = 15.43 N
Explanation:
Given:-
- The mass of left most block, m1 = 1.0 kg
- The mass of center block, m2 = 2.0 kg
- The mass of right most block, m3 = 4.0 kg
- A force that acts on the right most block, F = 36 N
Solution:-
a)
- For the first part we will consider the three blocks with masses ( m1 , m2 , and m3 ) as one system on which a force of F = 36 N is acted upon. The masses m1 and m3 are connected with a string with tension ( T ) and the m1 and m2 are in contact.
- We apply the Newton's second law of motion to the system with acceleration ( a ) and the combined mass ( M ) of the three blocks as follows:
[tex]F = M*a\\\\36 = ( 1 + 2 + 4 )*a\\\\a = \frac{36}{7}\\\\a = 5.143 \frac{m}{s^2}[/tex]
Answer: The system moves in the direction of external force ( F ) i.e to the right with an acceleration of 5.143 m/s^2
b)
- The blocks with mass ( m1 and m3 ) are connected with a string with tension ( T ) with a combined acceleration of ( a ).
- We will isolate the massive block ( m3 ) and notice that two opposing forces ( F and T ) act on the block.
- We will again apply the Newton's 2nd law of motion for the block m3 as follows:
[tex]F_n_e_t = m_3 * a\\\\F - T = m_3 * a\\\\36 - T = 4*5.143\\\\T = 36 - 20.5714\\\\T = 15.43 N[/tex]
Answer:- A tension of T = 15.43 Newtons acts on both blocks ( m1 and m3 )
c)
- We will now isolate the left most block ( m1 ) and draw a free body diagram. This block experiences two forces that is due to tension ( T ) and a reaction force ( Fr ) exerted by block ( m2 ) onto ( m3 ).
- Again we will apply the the Newton's 2nd law of motion for the block m3 as follows:
[tex]F_n_e_t = m_1*a\\\\T - F_r = m_1*a\\\\15.43 - F_r = 1*5.143\\\\F_r = 15.43 - 5.143\\\\F_r = 10.29 N[/tex]
- The reaction force ( Fr ) is contact between masses ( m1 and m2 ) exists as a pair of equal magnitude and opposite direction acting on both the masses. ( Newton's Third Law of motion )
Answer: The block m2 experiences a contact force of ( Fr = 10.29 N ) to the right.
d)
- If we were to stack the block ( m2 ) on-top of block ( m1 ) such that block ( m2 ) does not slip we the initial system would remain the same and move with the same acceleration calculated in part a) i.e 5.143 m/s^2
- We will check to see if the tension ( T ) differs or not as the two block ( m1 and m2 ) both experience the same Tension force ( T ) as a sub-system. with a combined mass of ( m1 + m2 ).
- We apply the Newton's 2nd law of motion for the block m3 as follows:
[tex]T = ( m_1 + m_2 ) *a\\\\T = ( 1 + 2 ) * 5.143\\\\T = 15.43 N[/tex]
Answer: The acceleration of the whole system remains the same at a = 5.143 m/s^2 and the tension T = 15.43 N also remains the same.
Thana reminds Alston that because the electric field is uniform, a constant electric force is exerted on the electron. Alston recognizes that, in this case, they can use the kinematic equations to describe the motion of the charged particle while it is inside the region containing the electric field. He asks Thana to write down an equation they can use to calculate the acceleration of the particle while it is inside the region containing a uniform electric field. Which of These equations is correct?
Answer:
a = - e E / m
a = - 1,758 10¹¹ E
Explanation:
For this exercise we can use Newton's second law
F = m a
where the force is electric
the forces given by the product of the charge by the electric field
F = q E
in this case tell us that the charge is the charge of the electron
q = -e = - 1.6 10⁻¹⁹ C
we substitute
- e E = m a
a = - e E / m
we calculate
a = - 1.6 10⁻¹⁹ / 9.1 10⁻³¹ E
a = - 1,758 10¹¹ E
The negative sign indicates that the acceleration is in the opposite direction to the electric field
If two twins (54 kg each) were 0.02 m apart, what is the force of gravity between them?
Answer:
Force, [tex]F=4.86\times 10^{-4}\ N[/tex]
Explanation:
We have,
Masses of two twins are 54 kg each
They are placed at a distance of 0.02 m
It is required to find the force of gravity between them. The formula used to find the gravitational force between masses is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
plugging all the known values:
[tex]F=6.67\times 10^{-11}\times \dfrac{54^2}{(0.02)^2}\\\\F=4.86\times 10^{-4}\ N[/tex]
So, the force of gravity between them is [tex]4.86\times 10^{-4}\ N[/tex].
To get up on the roof, a person (mass 69.0 kg) places a 6.40 m aluminum ladder (mass 11.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes (in N) of the forces on the ladder at the top and bottom
Answer:
N = 243.596 N ≈ 243.6 N
Explanation:
mass of person = 69 kg ( M )
mass of aluminium ladder = 11 kg ( m )
length of ladder = 6.4 m ( l )
base of ladder = 2 m from the house (d )
center of mass of ladder = 2 m from the bottom of ladder
person on ladder standing = 3 m from bottom of ladder
Calculate the magnitudes of the forces at the top and bottom of the ladder
The net torque on the ladder = o ( since it is at equilibrium )
assuming: the weight of the person( mg) acting at a distance x along the ladder. the weight of the ladder ( mg) acting halfway along the ladder and the reaction N acting on top of the ladder
X = l/2
x = 6.4 / 2 = 3.2
find angle formed by the ladder
cos ∅ = d/l
∅ = [tex]cos^{-1][/tex] 2/6.4 = [tex]cos^{-1}[/tex]0.3125 ≈ 71.79⁰
remember the net torque around is = zero
to calculate the magnitude of forces on the ladder we apply the following formula
[tex]N = \frac{mg(dcosteta)+ Mgxcosteta}{lsinteta}[/tex]
m = 11 kg, M = 69 kg, l = 6.4 , x = 3, teta( ∅ )= 71.79⁰, g = 9.8
back to equation N = [tex]\frac{11*9.8(2*cos71.79)+ 69*9.8*3* cos71.79}{6.4sin71.79}[/tex]
N = (67.375 + 633.938) / 2.879
N = 243.596 N ≈ 243.6 N
where would you expect to find vesicles of neurotransmitters
A. Synaptic gap
B. postsynaptic dendrites
C. Channels in the postsynaptic
D. Presynaptic terminal button
Answer:
D. Presynaptic terminal button
explanation:
Terminal Buttons are small knobs at the end of an axon that release chemicals called neurotransmitters. The terminal buttons form the Presynaptic Neuron
hope this helped!
At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror's radius of curvature is 0.560 m.
A) Locate the image of a patient10.6m from the mirror. B) Indicate whether the image is upright or inverted.C) Determine the magnification of the image.
Answer:
Explanation:
For a convex mirror, the value of its image distance and its focal length are negative.
using the mirror formula 1/f = 1/u+1/v
f is the focal length = Radius of curvature/2 = 0.560/2
f= 0.28m
u is the object distance = 10.6m
v is the position of the image = ?
On substitution;
1/0.28 = 1/10.6 + 1/-v
3.57 = 0.094 - 1/v
3.57 - 0.094 = -1/v
3.476 = -1/v
v = -1/3.476
v = -0.2877m
B) Since the image distance is negative, this means that the image is an upright and a virtual image. All Upright images has their image distance to be negative.
C) Magnification = Image distance/object distance
Magnification = 0.2877/10.6
Magnification = 0.0271
2. If rain is falling vertically downward, and you are running for shelter, should you hold your umbrella
vertically, tilted forward, or tilted backward to keep the driest? Please explain.
Answer:
Tilted forward to keep the driest.
Explanation:
The rain is falling vertically so there is no wind. In these circumstances the umbrella should be tilted vertically forward.
The situation is the same as if you would stand still and the rain would come under an angle from the front.
Complete the first and second sentences, choosing the correct answer from the given ones.
1. The water temperature in the dish depends on the A / B / C / D.
A. average kinetic energy of water molecules
B. total kinetic energy of water molecules
C. water mass. D. potential energy of the container with water
2. The internal energy of the water in the vessel is E / F / G.
E. potential energy of the vessel with water
F. average kinetic energy of water molecules
G. sum of kinetic energy and potential water molecules
Answer:
Hope this helps :)
Explanation:
1. A
2. G (because the basic definition of internal energy is, the sum of kinetic and potential energies of water molecules)
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At what distance from a 70.0 Watt speaker is the intensity 0.0195 W/m^2
(Treat the speaker as point of the source)
(Unit=meters)
PLEASE HELP ME!
Answer:
Distance = 16.9 m
Explanation:
We are given;
Power; P = 70 W
Intensity; I = 0.0195 W/m²
Now, for a spherical sound wave, the intensity in the radial direction is expressed as a function of distance r from the center of the sphere and is given by the expression;
I = Power/Unit area = P/(4πr²)
where;
P is the sound power
r is the distance.
Thus;
Making r the subject, we have;
r² = P/4πI
r = √(P/4πI)
r = √(70/(4π*0.0195))
r = √285.6627
r = 16.9 m
Answer:
16.9 m
Explanation:
An 89.2-kg person with a density 1025 kg/m3 stands on a scale while completely submerged in water. What does the scale read?
Answer:
89.11kg
Explanation:
Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.
Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}
The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.
The weight of the fluid=0.087m3× 1kg/me = 0.087kg
Now the weight of the fluid displaced is referred to as the upthrust.
Now the real weight - the apparent weight = the upthrust.
Hence the apparent weight = real weight - upthrust
Apparent weight = 89.2-0.087 = 89.11kg
Two parallel, vertical, plane mirrors, 38.8 cm apart, face each other. A light source at point P is 30.1 cm from the mirror on the left and 8.7 cm from the mirror on the right.
(a) How many images of point P are formed by the mirrors?
(b) Find the distance from the mirror on the right to the two nearest images behind the mirror.
first nearest image=
second nearest image=
(c) Find the number of reflections of light rays for each of these images.
first nearest image=
second nearest image=
Answer:
Explanation shown below.
Explanation:
1.The number of images formed by 2 parallel mirrors is an infinite number of images.
2. The characteristics of a plane mirror is such that the object distance equals the image distance.
Hence the object distance is 8.7cm from the right; the image formed would be 8.7cm behind the mirror.
Now a second image is going to be formed by the left mirror which is going to have an image distance of 30.1cm behind the mirror.
Now this image would be reflected on the right side to form a new image which is going to be seen as 38.8 +30.1 = 68.9cm behind the right Mirror .
Hence the shortest distances are 8.7cm and 68.9cm
3. The number of reflections is infinite for both cases.
John heats 1 kg of soup from 25 °C to 70 °C for 15 minutes by a heater. How long does the same heater take to heat 1.5 kg of the same kind of soup from 20 °C to 80 °C? The energy output per unit time by the heater is constant.
Answer:
30 minutes
Explanation:
Energy per time is constant, so:
E₁ / t₁ = E₂ / t₂
m₁C₁ΔT₁ / t₁ = m₂C₂ΔT₂ / t₂
(1 kg) C (70°C − 25°C) / 15 min = (1.5 kg) C (80°C − 20°C) / t
(1 kg) (45°C) / 15 min = (1.5 kg) (60°C) / t
3/min = 90 / t
t = 30 min
A ball is projected upward at time t= 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to:____________A. 9.0 sB. 9.4 sC. 9.7 sD. 8.7 sE. 10 s
Answer:
B. 9.4 s
Explanation:
In order to calculate the total time taken by the ball to hit the ground, we first analyze the upward motion. We will use subscript 1 for upward motion. Now, using 1st equation of motion:
Vf₁ = Vi₁ + gt₁
where,
Vf, = Final Velocity in upward motion = 0 m/s (ball stops at highest point)
Vi = Initial Velocity in upward motion = 36.2 m/s
g = - 9.8 m/s² (negative due to upward motion)
t₁ = Time taken in upward motion = ?
Therefore,
0 m/s = 36.2 m/s + (-9.8 m/s²)(t₁)
t₁ = (36.2 m/s)/(9.8 m/s²)
t₁ = 3.7 s
Now, using 2nd equation of motion:
h₁ = (Vi₁)(t₁) + (0.5)(g)(t₁)²
where,
h₁ = distance from top of building to highest point ball reaches = ?
Therefore,
h₁ = (36.2 m/s)(3.7 s) + (0.5)(-9.8 m/s²)(3.7 s)²
h₁ = 133.58 - 66.86 m
h₁ = 66.72 m
No, considering downward motion and using subscript 2, for it.
Using 2nd equation of motion:
h₂ = (Vi₂)(t₂) + (0.5)(g)(t₂)²
where,
h₂ = height of the highest point from ground = h₁ + height of building
h₂ = 66.72 m + 90 m = 156.72 m
Vi₂ = Initial Speed during downward motion = 0 m/s (ball stops for a moment at highest point)
t₂ = Time Taken in downward motion = ?
g = 9.8 m/s²
Therefore,
156.72 m = (0 m/s)(t₂) + (0.5)(9.8 m/s²)(t₂)²
t₂² = (156.72 m)/(4.9 m/s²)
t₂ = √31.98 s²
t₂ = 5.7 s
Now, the total time taken by ball to reach the ground is"
Total Time = T = t₁ + t₂
T = 3.7 s + 5.7 s
T = 9.4 s
Therefore, the correct answer is:
B. 9.4 s
The inhabitants of a small island export a cloth made from a plant that grows only on their island. A clothier from New York, believing that he can save money by "cutting out the middleman," decides to travel to the island and buy the cloth himself. Ignorant of the local custom where strangers are offered outrageous prices initially, the clothier accepts (much to everyone's surprise) the initial price of 400 tepizes/m^2. The price of this cloth in New York is 120 dollars/yard^2. If the clothing maker bought 500 m^2 of this fabric, how much money did he lose? Use 1tepiz= 0.625dollar and 0.9144m = 1yard.
Answer:
Explanation:
purchase price = 400 tepizes / m²
1 tepiz = .625 dollar
purchase price in terms of dollar = 400 x .625 dollar / m²
= 250 dollar / m²
.9144 m = 1 yard
1 m = 1.0936 yard
1m² = 1.196 yard²
price in terms of dollar / yards²
= 250 / 1.196 dollar / yard²
= 209 dollar / yard²
Price of cloth in New York = 120 dollar / yard²
loss = 209 - 120 = 89 dollar / yard²
500 m² = 500 x 1.196 yard²
= 598 yard²
net loss in purchasing 500 m² cloth
= 598 x 89
= 53222 dollar .
A sphere of diameter 6.0cm is moulded into a thin uniform wire of diameter 0.2mm. Calculate the length of the wire in metres (Take π = 22/7) *
Answer:
2025m
Explanation:
Since all materials of the sphere is made to a cylindrical wire, it implies the volume of the sphere material is same as that of the cylinder. This is expressed mathematically thus.
Volume of Sphere= volume of cylinder
4/3 ×π×R^3= π× r2× L
4/3 ×R^3= r^2×L
Hence
L = 3/4 × R^3/ r^2
But R = 6.0/2 = 3.0cm{ Diameter is twice raduis}
r= 0.2/2 = 0.1mm=>0.01cm{ Diameter is twice raduis and unit converted by dividing by 10 since 10mm = 1cm}
Substituting R and r into the expression for L, we have :
L = 3/4 × 3^3/ 0.01^2= 0.75 ×27/0.0001 = 202500cm
202500/100= 2025m{ we divide by 100 because 100cm=1m}
PLEASE HELP !
Complete the following sentence. Choose the right answer from the given ones. The internal energy of the body can be changed A / B / C. A. only when the body is warmed or cooled B. when work is done on the body or heat flow C. only when the body does work
B
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Potential difference is measured in which units?
volts
amps
currents
watts
Answer:
Potential difference is measured in volts
Explanation:
The standard metric unit on electric potential difference is the volt, abbreviated V and named in honor of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb.
Answer:
Your answer is A.) volts
Explanation:
A surveyor measures the distance across a river that flows straight north by the following method. Starting directly across from a tree on the opposite bank, the surveyor walks distance, D = 130 m along the river to establish a baseline. She then sights across to the tree and reads that the angle from the baseline to the tree is an angle θ = 25°. How wide is the river?
Answer:
The width of the river is [tex]z = 60.62 \ m[/tex]
Explanation:
From the question we are told that
The distance of the base line is D = 130 m
The angle is [tex]\theta = 25^o[/tex]
A diagram illustration the question is shown on the first uploaded image
Applying Trigonometric Rules for Right-angled Triangle,
[tex]tan 25 = \frac{z}{130}[/tex]
Now making z the subject
[tex]z = 130 * tan (25)[/tex]
[tex]z = 60.62 \ m[/tex]
two blocks with masses 2 kg and 4 kg are pushed from rest by the same amount of fore for a distance of 100 m on a frictionless floor. the final kinetic energy of the 2 kg block after the 100 m distance is
Answer:
the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J
Explanation:
mass of block A = 2 kg
mass of block B = 4 kg
distance the blocks were pushed = 100 m
NB: Blocks were pushed the same distance at the same equal time period. And the ground is without friction.
Work done in moving the 2 kg mass along the 100 m distance is,
work = force x distance moved
force exerted by the 2 kg mass = 2 x 9.81 m/s^2(acceleration due to gravity)
force = 19.62 N
therefore,
work done = 19.62 x 100 = 1962 Joules of work.
According to energy conservation principles, the kinetic energy impacted of the 2 kg mass through this distance will be equal to the work done in moving the 2 kg mass through this distance.
Therefore, the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J
Jackson heads east at 25 km/h for 20 minutes before heading south at 45 km/h for 20 minutes. Hunter heads south at 45 km/h for 10 minutes before heading east at 40 km/h for 30 minutes. Find average velocity (magnitude and direction) of each person
Answer:
The average velocity of Jackson is 18.056 m/s South
The average velocity of Hunter is 10.65 m/s East
Explanation:
initial velocity of Jackson, u = 25 km/h east = 6.944 m/s east
time for this motion, [tex]t_i[/tex] = 20 minutes = 1200 seconds
⇒initial displacement of Jackson, [tex]x_i[/tex] = (6.944 m/s) x (1200 s) = 8332.8 m
Final velocity of Jackson, v = 45 km/h South = 12.5 m/s South
time at Jackson's final position, [tex]t_f[/tex] = 20 minutes + [tex]t_i[/tex] = 20 minutes + 20 minutes
time at Jackson's final position, [tex]t_f[/tex] = 40 minutes = 2400 s
⇒Final displacement of Jackson,[tex]x_f[/tex] = (12.5 m/s) x (2400 s) = 30,000m
Average velocity of Jackson;
[tex]= \frac{x_f-x_i}{t_f-t_i} \\\\= \frac{30,000-8332.8}{2400-1200} \\\\= 18.056 \ m/s \ South[/tex]
initial velocity of Hunter, u = 45 km/h South = 12.5 m/s South
time for this motion, [tex]t_i[/tex] = 10 minutes = 600 seconds
⇒initial displacement of Hunter, [tex]x_i[/tex] = (12.5 m/s) x (600 s) = 7500 m
Final velocity of Hunter, v = 40 km/h east = 11.11 m/s east
time at Hunter's final position, [tex]t_f[/tex] = 30 minutes + [tex]t_i[/tex] = 30 minutes + 10 minutes
time at Hunter's final position, [tex]t_f[/tex] = 40 minutes = 2400 s
⇒Final displacement of Hunter,[tex]x_f[/tex] = (11.11 m/s) x (2400 s) = 26,664m
Average velocity of Hunter;
[tex]= \frac{x_f-x_i}{t_f-t_o} \\\\= \frac{26,664-7500}{2400-600} \\\\= 10.65 \ m/s \ east[/tex]
An LC circuit has a 6.00 mH inductor. The current has its maximum value of 0.570 A at t =0s. A short time later the capacitor reaches its maximum potential difference of 66.0 V. What is the value of the capacitance?
Answer:
C = 44.75 x 10⁻⁸ F
Explanation:
Assuming no loss of energy between capacitor and inductor
energy in inductor initially = 1/2 Li₀² where L is inductance and i₀ is peak current .
= .5 x 6 x 10⁻³ x .57²
= .97 x 10⁻³ J .
This energy is transferred to capacitor .
energy of capacitor = 1/2 CV²
= .5 x C x 66²
= 2178 C
2178C = .97 x 10⁻³
C = 44.75 x 10⁻⁸ F .
The magnetic energy stored in the inductor is transformed into electrical energy stored in the capacitor. The value of capacitance for the given circuit is 44.75 x 10⁻⁸ F
Finding the capacitance:According to the law of conservation of energy, the magnetic energy stored in the inductor will be gradually lost and this energy will be stored in the capacitor as electrical energy.
Initially, the energy in the inductor is:
E = 1/2 Li₀²
where L is inductance
and i₀ is peak current.
E = 0.5 × 6 × 10⁻³ × (0.57)²
E = 0.97 × 10⁻³J
This energy is transformed into electrical energy stored in the capacitor.
So the capacitor energy is:
E = 1/2 CV²
where C is the capacitance
E = 0.5 × C × 66²
E = 2178 C
0.97 x 10⁻³ = 2178 C
C = 44.75 x 10⁻⁸ F
Learn more about capacitance:
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A dimension is a physical nature of a quantity.
(i) give two (2) limitations of dimensional analysis..
(ii) if velocity (v), time (T) and force (F) were chosen as basic quantities, find the dimensions of mass?
Answer:
i) A dimension is the physical nature of a quantity. The two limitations of dimensional analysis is as following:
Dimesnional analysis is unable to derive relation when a physical quantity depends on more than three factors with dimensions. It is unable to derive a formula that contain exponential function, trigonometric function, and logarithmic function.ii) Given:
Velocity = v
Time = t
Force = F
Force = mass x acceleration
= mass x velocity/time
So, mass= (force x time) / velocity
[mass] = Ftv^-1
Hence, dimesnion of mass is Ftv^-1.
A sample of silver (with work function Φ=4.52 eV ) is exposed to an ultraviolet light source (????=200 nm), which results in the ejection of photoelectrons. What changes will be observed if:
1. The silver is replaced with copper (Φ= 5.10 eV)?
a. more energetic photoelectrons (on average)
b. no photoelectrons are emitted more photoelectrons ejected
c. less energetic photoelectrons (on average)
d. fewer photoelectrons ejected
2. A second (identical) light source also shines on the metal?
a. fewer photoelectrons ejected
b. no photoelectrons are emitted more
c. energetic photoelectrons (on average)
d. less energetic photoelectrons (on average)
e. more photoelectrons ejected
3. The ultraviolet source is replaced with an X-ray source that emits the same number of photons per unit time as the original ultraviolet source?
a. no photoelectrons are emitted
b. less energetic photoelectrons (on average)
c. fewer photoelectrons ejected
d. more energetic photoelectrons (on average)
e. more photoelectrons ejected
Answer:
1. c
2. e
3. d
Explanation:
1.
From Einstein's Photoelectric Equation, we know that:
Energy given up by photon = Work Function + K.E of Electron
hc/λ = φ + K.E
where,
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light source = 200 nm = 2 x 10⁻⁷ m
φ = (5.1 eV)(1.6 x 10⁻¹⁹ J/eV) = 8.16 x 10⁻¹⁹ J
Therefore,
(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2 x 10⁻⁷ m) - 8.16 x 10⁻¹⁹ = K.E
K.E = (9.939 - 8.16) x 10⁻¹⁹ J
K.E = 1.778 x 10⁻¹⁹ J
The positive answer shows that electrons will be emitted. Since it is clear from the equation the the K.E of electron decreases with the increase in work function. Therefore:
c. less energetic photo-electrons (on average)
2.
The increase in light sources means an increase in the intensity of light. The no. of photons are increased, due to increase of intensity. Thus, more photons hit the metal and they eject greater no. of electrons. Therefore,
e. more photo-electrons ejected
3.
X-rays have smaller wavelength and greater energy than ultraviolet rays. Thus, the photons with greater energy will strike the metal and as a result, electrons with higher energy will be ejected.
d. more energetic photo-electrons (on average)
A 330-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,110 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? (Use 3.156 107 for the number of seconds in a year.)
Answer:
t = 402 years
Explanation:
To find the number of year that electrons take in crossing the complete transmission line, you first calculate the drift speed of the electrons. Then, you use the following formula for the current in a wire:
[tex]I=nqv_dA[/tex] (1)
n: number of mobile charge carrier per volume = 8.50*10^28 e/m^3
q: charge of the electron = 1.6*10^-19 C
vd: drift velocity of electron in the metal = ?
A: cross sectional area of the wire = π r^2 = π (0.02m/2)^2 = 3.1415*10^-4 m^2
I: current in the wire = 1110 A
You solve the equation (1) for vd:
[tex]v_d=\frac{I}{nqA}=\frac{110A}{(8.50*10^{28}m^{-3})(1.6*10^{-19}C)(3.1415*10^{-4}m^2)}\\\\v_d=2.59*10^{-4}m/s[/tex]
Next, you calculate the time by using the information about the length of the line transmission:
[tex]x=v_dt\\\\x=330km=330000m\\\\t=\frac{x}{v_d}=\frac{330000m}{2.59*10^{-4}m/s}=1,270,184,865s\\\\1,270,184,865s*\frac{1\ year}{3,156,107}=402.45\ years[/tex]
hence, the electrons will take aproximately 402 years in crossing the line of transmission
When Marcel finds the distance L from the previous part, it turns out to be greater than Lend, the distance from the pivot to the end of the seesaw. Hence, even with Jacques at the very end of the seesaw, the twins Gilles and Jean exert more torque than Jacques does. Marcel now elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is at height h above the pivot. (Figure 3)With what force in the rightward direction, Fx, should Marcel push? If your expression would give a negative result (using actual values) that just means the force should be toward the left.Express your answer in terms of W, Lend, w, L2, L3, and h.
Answer:
Fx = - (1/h)( wL2 + wL3 - wLend )
Explanation:
Assuming The twins Gilles and Jean has a weight ( w ) each
The torque that would balance the equation would be = wL2 + wL3 -------- 1
THEREFORE the ccw torques are = wLend + Fh ----------- 2
hence equation 2 equals equation 1
= wLend + Fh = wL2 + wL3 --------- 3
equation 3 can as well be represented as
F = ( 1/h) ( wL2 + wL3 - wLend )---------- 4
From equation 4 it can be seen that F is on the left hand side therefore the value of Fx is negative
therefore equation 4 is represented as
Fx = - (1/h)( wL2 + wL3 - wLend )
An aluminium pot whose thermal conductivity is 237 W/m.K has a flat, circular bottom
with diameter 15 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in
the pot through its bottom at a rate of 1400 W. If the inner surface of the bottom of the pot
is at 105 °C, determine the temperature at the outer surface of the bottom of the pot
Answer:
T₁ = 378.33 k = 105.33°C
Explanation:
From Fourier's Law of heat conduction, we know that:
Q = - KAΔT/t
where,
Q = Heat Transfer Rate = 1400 W
K = Thermal Conductivity of Material (Aluminum) = 237 W/m.k
A =Surface Area through which heat transfer is taking place=circular bottom
A = π(radius)² = π(0.15 m)² = 0.0707 m²
ΔT = Difference in Temperature of both sides of surface = T₂ - T₁
T₁ = Temperature of outer surface = ?
T₂ = Temperature of inner surface = 105°C + 273 = 378 k
ΔT = 388 k - T₁
t = thickness of the surface (Bottom of Pot) = 0.4 cm = 0.004 m
Therefore,
1400 W = - (237 W/m.k)(0.0707 m²)(378 k - T₁)/0.004 m
(1400 W)/(4188.14 W/k) = - (378 k - T₁)
T₁ = 0.33 k + 378 k
T₁ = 378.33 k = 105.33°C
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed
Complete Question
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?
A It should stay the same
B It should be quadrupled.
C It should be quintupled
D It should be doubled.
E It should be tripled
Answer:
Option D is the correct option
Explanation:
Generally electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon_o}[/tex]
Where [tex]\sigma[/tex] is the charge per unit area (Charge density )
From the question we are told that [tex]\sigma[/tex] is doubled hence the
[tex]E = \frac{2 \sigma }{\epsilon_o}[/tex]
Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density
A 1100 kg car pushes a 1800 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 N.A) What is the magnitude of the force of the car on the truck?B) What is the magnitude of the force of the truck on the car?
Answer:The answer is 3000 N.
Force (F) is the multiplication of mass (m) and acceleration (a).
F = m · a
It is given:
mc = 1000 kg
mt = 2000 kg
total force: F = 4500 N
total mass: m = mc + mt
Let's calculate acceleration which is common:
a = F/m = F/(mc + mt) = 4500/(1000 + 2000) = 4500/3000 = 1.5 m/s²
Now, when we know acceleration, let's calculate force on the truck:
Ft = mt · a = 2000 · 1.5 = 3000 N
Explanation:
A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.
Answer:
4.93 m
Explanation:
According to the question, the computation of the height is shown below:
But before that first we need to find out the speed which is shown below:
As we know that
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]Speed = \frac{5}{0.504}[/tex]
= 9.92 m/s
Now
[tex]v^2 - u^2 = 2\times g\times h[/tex]
[tex]9.92^2 = 2\times 9.98 \times h[/tex]
98.4064 = 19.96 × height
So, the height is 4.93 m
We simply applied the above formulas so that the height i.e H could arrive
Zinc is added to a breaker containing hydrochloric acid and the beaker gets warm what type os reaction is this
Answer:
Exothermic
Explanation:
Depending on the unit you are in, the answer may vary.
This is an exothermic reaction because it produces heat (the beaker gets warm).