Answer:
P = 360 Watts
Explanation:
Given that,
The weight of a student, [tex]F=5.4\times 10^2\ N[/tex]
It takes 15 seconds to run up a hill.
The top of the hill is 10 meters vertically above her starting point.
We need to find the power develop during her run. We know that te power developed is given by :
[tex]P=\dfrac{W}{t}\\\\P=\dfrac{mgh}{t}\\\\P=\dfrac{5.4\times 10^2\times 10}{15}\\\\P=360\ W[/tex]
So, the power develop during her run is 360 W.
a car traveling at 30m/s slows down to a stop 10s. what is the acceleration?
Answer:
20 m/s. have a great day
Answer:
since v decreased by 20m/s in 5 sec, a = -4 m/s^2
assuming the 3 seconds started at t=0,
s = 30t - 2t^2
s(3) = 30(3) - 2(9) = 72m
You have been handed an unknown battery. Using your multimeter, you determine that when a 4.30 resistor is connected across the battery's terminals, the current in the battery is 0.500 A. When this resistor is replaced by an 9.4 resistor, the current drops to 0.250 A. From those data, find the emf and internal resistance of your battery.
Answer:
Ri = 0.8 Ω
V= 2.55 V
Explanation:
Since the internal resistance of the battery is connected in series with the resistor connected across the battery's terminals, applying Ohm's Law, we can write the following equation, when R₁=4.30 Ω, and I₁=0.500A:[tex]I_{1} = \frac{V}{R_{i} + 4.3 \Omega } = 0.500 A (1)[/tex]
We can apply exactly this same expression, when R₂ = 9.4Ω, and I₂ = 0.250A:[tex]I_{2} = \frac{V}{R_{i} + 9.4 \Omega} } = 0.250 A (2)[/tex]
Now, we can divide (1) by (2) as follows:[tex]\frac{R_{i} + 9.4 \Omega}{R_{i} + 4.3 \Omega} = \frac{I_{1} }{I_{2}} = 2 (3)[/tex]
Solving for Ri, we get Ri = 0.8 ΩReplacing this value in (1) we can solve for V, as follows:V = I₁ * (Ri + R₁) = 0.500 A * (0.8Ω + 4.3Ω) = 2.55 V⇒ V = 2.55 V (emf of the battery)
To understand the cause of constructive and destructive interference for the double-slit experiment, and to explain how the interference pattern depends on the parameters of the emitted waves.For this tutorial, use the PhET simulation Wave Interference. This simulation allows you to send waves through a variety of barriers and look at the resulting interference patterns.Start the simulation. You will see three possible selections: Waves, Interference, and Slits. To change between simulations at any point, select the desired simulation on the toolbar located at the bottom of the screen. In these simulations you can choose between water waves, sound waves, or light. You can adjust the slit width and slit separation using slider bars, and you can put a barrier containing one or two slits in front of the source of the waves. There are also several measuring tools at the upper-right hand corner of the screen, including a detector that produces plots showing the wave amplitude vs. time for the location of the two sensors on the detector, which can be dragged to any location.Feel free to experiment with all of the simulations to get a feel for how they work. When you are done, and before starting Part A, set the simulation to Waves, and select the Reset icon.Select Light for the type of wave, adjust the wavelength so that the light is red, and increase the amplitude of the light to the max. Then, select the start button at the source location to begin producing the waves.Light is a form of electromagnetic wave, containing oscillating electric and magnetic fields. The wave amplitude detector mentioned above shows how the electric field oscillates in time at the location of the probe. The amplitude of the wave at the location of the probe is equal to the maximum electric field measured.How does the amplitude of the wave depend on the distance from the source?a. The amplitude decreases with distance.b. The amplitude increases with distance.c. The amplitude is constant.
Answer:
The maximum intensity, if it changes with distance, the correct answer is a
Explanation:
After reading your extensive writing on the phenomenon of interference, I see that your question is not very related to it
In the interference experiments the amplitude of the radiation is given by the sum of the intensity of a point of the two waves coming from the slits,
I =[tex]I_{max }[/tex] cos² ([tex]\frac{\pi \ d \ sin \theta}{\lambda}[/tex]) sinx / x
x = [tex]\frac{\pi \ a \ sin \theta }{ \lambda }[/tex]
where d is the separation between the slits and a the width of each slit.
Therefore, we can see that the intensity of the radiation is redistributed in space in lines approximately the same intensity, second factor is due to the diffraction of each slit, this is generally small for the central part of the interference pattern
therefore the interference pattern is made up of lines of equal intensity.
The maximum intensity, if it changes with distance, in general for energy conservation the intensity must decrease with distance, therefore the correct answer is a
An Olympic high jumper, with a mass of 82 kg, has a
maximum gravitational potential energy of 1970 J. How
high was the jump?
Answer:
I don't really know
Explanation:
I really wanted to help you, but then I realized i didnt know how to
What type of reaction is occurring when you activate an instant cold pack?
A. Endothermic, because energy is absorbed
B. Exothermic, because energy is released
O C. Endothermic, because energy is released
D. Exothermic, because energy is absorbed
Answer:
Endothermic, because energy is absorbed (A)
Explanation:
The reaction that take place in the instant cold pack causes the surroundings around it, including the bag that contains it. When the bag loses heat to the cold pack, the cold pack absorbs the heat, thereby causing the environment (the bag) to be cold.
A bird lands on a bird feeder which is connected to a spring. The mass of the bird is exactly the same as the mass of the bird feeder. How does the added mass affect the period of oscillation of the bird feeder?
Answer:
The added mass will mean a longer period of oscillation.
Explanation:
The period of oscillation here is given by the formula;
T = 2π√(m/k)
Where m is mass and k is spring constant
From the equation of oscillation period above, it's obvious that when we increase the mass, the oscillation period will also increase.
Thus, the added mass will mean a longer period of oscillation.
An electron has a mass of 9.1x10-31 kg. What is its momentum if it is travelling at a speed of 3.5x106 m/s?
Answer:
3.19*10^-24
Explanation:
the equation to find momentum is p=mv so you just multiply the mass times velocity
An electron has a mass of 9.1x10⁻³¹ kilograms. if it is traveling at a speed of 3.5x10⁶ meters/seconds then its momentum would be 3.185 ×10⁻²⁴ kgm/s.
What is momentum?It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.
As given in the problem an electron has a mass of 9.1x10⁻³¹ kilograms. if it is traveling at a speed of 3.5x10⁶ meters/seconds then its momentum would be
the momentum of the electrons = mass of the electron×velocity of the electron
= 9.1x10⁻³¹ × 3.5x10⁶
= 3.185 ×10⁻²⁴ kgm/s
Thus, the momentum of the electron would be 3.185 ×10⁻²⁴ kgm/s.
To learn more about momentum from here, refer to the link;
brainly.com/question/17662202
#SPJ2
1. A plane starts from rest and aceelerates in a
straight line along the ground before take-off. It
moves 600 m in 12 s. Calculate the distance
moved during the twelfth second.
s=600 m
t=12 s
s=0.5*a*t² (initial speed V0=0)
a=(2*s)/t²
a=(2*600)/12²
a≈8.33 m/s²
L= s(t2=12s)-s(t1=11s) -> (distance during the twelfth second)
L=0.5*a*(t2²-t1²)
L=0.5*((2*s)/t²)*(t2²-t1²)
L=0.5*((2*600)/12²)*(12²-11²)
L ≈ 95.83 m
A(n) 17.4 g bullet is shot into a(n) 5506 g
wooden block standing on a frictionless sur-
face. The block, with the bullet in it, acquires
a speed of 1.61 m/s.
Calculate the speed of the bullet before
striking the block..
Answer in units of m/s.
The initial speed of the bullet = v₁= 511.07 m/s
Further explanationGiven
17.4 g bullet
5506 g wooden
The velocity of the block+bullet :1.61 m/s
Required
The initial speed
Solution
Momentum
m₁v₁+m₂v₂=m₁v₁'+m₂v₂'
v₂=0 ⇒block at rest
v₁'=v₂'=1.61 m/s
the equation becomes :
m₁v₁=(m₁+m₂)v'
17.4v₁=(17.4+5506)1.61
v₁= 511.07 m/s
Two cellists, one seated directly behind the other in an orchestra, play the same note for the conductor who is directly in front of them. Because of the separation between the cellists, destructive interference occurs at the conductor. This separation is the smallest that produces destructive interference. Would this separation increase, decrease, or remain the same if the cellists produced a note with a higher frequency
Answer:
Please see below as the answer is self-explanatory.
Explanation:
In order to have a destructive interference, the path difference between the sources of the sound, must be equal to an odd multiple of the semi-wavelength, as follows: ⇒ d = d₂ - d₁ = n*(λ/2) The minimum posible value for this distance, is when n=1, as it can be seen here: dmin = λ/2 In any traveling wave, there exists a fixed relationship between the wave speed, the frequency and the wavelength: v = λ*f Therefore, assuming that the speed of sound keeps constant, if the frequency is increased, in order to keep the right side of the expression above balanced, λ must be decreased.As the smallest separation that produces destructive interference is directly proportional to the wavelength, this means that this separation will decrease if the cellists produced a note with a higher frequency.Two stationary positive point charges, charge 1 of magnitude 3.95 nC and charge 2 of magnitude 1.80 nC, are separated by a distance of 39.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.What is the speed of the electron when it is 10.0 cm from charge 1?
Answer:
the speed of the electron from charge q1 is 7.17×10⁶ m/s
Explanation:
Given the data in the question;
the potential at the center of the two charges will be;
V = k( q1/(d/2) + q2/(d/2)
so we substitute
V = (9×10⁹)( (3.95×10⁻⁹/(0.39/2) + 1.80×10⁻⁹/(0.39/2)
V = 265.4 V
the potential at a distance of 10 cm from the charges will be
V = k( q1/(d1) + q2/(d2)
(d1 = 10cm = 0.1m and d2 = 39cm - 10cm = 29cm = 0.29m )
V' = (9×10⁹)( (3.95×10⁻⁹/0.1 + 1.80×10⁻⁹/0.29
V' = 411.4 V
Now, from the conservation of energy the speed of the electron from charge q1 will be;
E = ( V' - V) qe
1/2mv² = ( V' - V) qe
v² = [( V' - V) qe] / 1/2m
v =√ ([( V' - V) qe] / 1/2m)
v =√ ([2( V' - V) qe] / m)
we substitute
v =√ (2[( 411.4 - 265.4) 1.6×10⁻¹⁹] / 9.1×10³¹)
v = 7.17×10⁶ m/s
Therefore, the speed of the electron from charge q1 is 7.17×10⁶ m/s
A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 18.8 N .
(a) Draw two free-body diagrams: one for each block.
(b) What is the acceleration of either block?
(c) Find m.
(d) How does the tension compare to the weight of the hanging block?
Answer and Explanation:
(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope ([tex]F_{t}[/tex]), vertical gravitational force ([tex]F_{g}[/tex]) and vertical normal force ([tex]F_{n}[/tex]), due to the surface. Since there is no vertical movement, [tex]F_{g}[/tex] and [tex]F_{n}[/tex] cancels it out. So, for this block, net force is horizontal due to the rope [tex]F_{t}[/tex].
The block of mass m is hanging from the pulley, so there is the force of the rope ([tex]F_{t}[/tex]) and the gravitational force ([tex]F_{g}[/tex]). Both are vertical, because there is no surface "holding" block m.
(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:
[tex]F_{r}=m.a[/tex]
[tex]a=\frac{F_{r}}{m}[/tex]
[tex]a=\frac{18.8}{3.6}[/tex]
a = 5.22
The acceleration of either block is 5.22 m/s².
(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².
Suppose positive referential is going up. To determine mass:
[tex]F_{r}=m.a[/tex]
[tex]F_{t}-F_{g}=m.a[/tex]
[tex]F_{t}-m.g=m.a[/tex]
[tex]18.8-9.8m=5.22m[/tex]
[tex]15.02m=18.8[/tex]
m = 1.25
Block m has 1.25 kg.
(d) Gravitational force is also called weight. So, as described above: [tex]F_{g}=m.g[/tex].
The weight for the hanging block is
[tex]F_{g}=1.25*9.8[/tex]
[tex]F_{g}=[/tex] 12.25 N
Comparing tension and weight:
[tex]\frac{12.25}{18.8}[/tex] ≈ 0.65
We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.
Are you better off in an accident where the hood is
deformed or where the car is not damaged at all but
bounces backward?
Answer:
you would be better off if the car bounced backwards
Explanation:
because if the hood was dismembered than you have a high chance of very bad injury but if it is just bounced back you would have less chance of getting hurt if properly sitting and seat belted.
What is the total amount of kinetic and potential energy in a system ?
Answer:
Its the sum of the potential energy and the kinetic energy
I’m not sure how to solve this
Answer:
Option 10. 169.118 J/KgºC
Explanation:
From the question given above, the following data were obtained:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1.61 KJ
Mass of metal bar = 476 g
Specific heat capacity (C) of metal bar =?
Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:
1 kJ = 1000 J
Therefore,
1.61 KJ = 1.61 KJ × 1000 J / 1 kJ
1.61 KJ = 1610 J
Next, we shall convert 476 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
476 g = 476 g × 1 Kg / 1000 g
476 g = 0.476 Kg
Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1610 J
Mass of metal bar = 0.476 Kg
Specific heat capacity (C) of metal bar =?
Q = MCΔT
1610 = 0.476 × C × 20
1610 = 9.52 × C
Divide both side by 9.52
C = 1610 / 9.52
C = 169.118 J/KgºC
Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC
4. Taylor Swift pushes Kanye West,
headfirst fearless, off a high dive with and
initial speed of 6m/s. If the platform is 20m
high, what is Kanye's impact velocity with
the water? (velocity and angle)
If vec A and vec B are vectors and vec B = -vec A , which of the following is true ?
a) The magnitude of is equal to the negative of the vec B magnitude of vec A
b) Ā and vec B are perpendicular .
c) The direction angle of is equal to the direction angle of vec B A plus 180 degrees
d ) vec A + vec B =2 vec A .
Answer:
If A + B = C for the vector equation then substituting for B gives
A + (-A) = C = 0
The only thing that can be said about "C" is that it has zero magnitude.
I =p/4pir^2 solving for r
Answer: [tex]r = \sqrt{\frac{p}{4pil}}[/tex]
Explanation:
[tex]l = \frac{p}{4pir^2} \\4pir^2l=p\\r^2 = \frac{p}{4pil} \\r = \sqrt{\frac{p}{4pil}}[/tex]
What is the motion and arrangement of molecules in a liquid
Answer:
particles
Explanation:
in liquids, particle are close together
When an object falls, its:
A. PE increases and KE decreases.
B. PE does not change.
C. PE and KE both increase.
D. PE decreases and kE increases
Answer:
Option D. is correct.
Explanation:
The object's mechanical energy refers to the sum of the potential and kinetic energies of the object. When an object falls, its potential energy (PE) decreases, and its kinetic energy (KE) increases. The increase in kinetic energy is exactly equal to the decrease in potential energy.
Option D. is correct.
___is found in fruits and honey. *
1.Maltose
2.Sucrose
3.Fructose
4.Galactose
Answer:
3. Fructose
Explanation:
Fructose is a sugar found naturally in fruits, fruit juices, some vegetables and honey.
it is number 3 (Fructose)
I need helpp! This is urgent tomorrow is the duedate
Answer:
The Force that resists the motion of one object moving relative to another is Friction. Kinetic Friction Resists the motion of Moving objects, while Static friction keeps the objects from moving.
A 72-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.2 m/s in 0.80 s. The elevator travels with this constant speed for 5.0 s, undergoes a uniform negative acceleration for 1.5 s, and then comes to rest. What does the spring scale register During the first 0.80s of the elevator’s ascent?
Answer:
Explanation:
During the first .8 s , the elevator is under acceleration . It starts from initial velocity u = 0 , final velocity v = 1.2 m /s , time = .8 s
v = u + at
1.2 = 0 + .8 a
a = 1.2 / .8
= 1.5 m /s²
During the acceleration in upward direction , let reaction force of ground on man be R .
Net force on man = R - mg
Applying Newton's 2 nd law
R - mg = ma
R = m ( g + a )
= 72 ( 9.8 + 1.5 )
= 813.6 N .
This reaction force will be measured by spring scale , so reading of spring scale will be 813.6 N .
calculate the peak voltage of a mains supply of 240Vrms.
An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to volume 1450 cm3. If the expansion is isothermal, what are (a) the final pressure and (b) the work done by the gas
Answer:
a) The final pressure is 1.68 atm.
b) The work done by the gas is 305.3 J.
Explanation:
a) The final pressure of an isothermal expansion is given by:
[tex] T = \frac{PV}{nR} [/tex]
[tex] T_{i} = T_{f} [/tex]
[tex] \frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR} [/tex]
Where:
[tex]P_{i}[/tex]: is the initial pressure = 5.79 atm
[tex]P_{f}[/tex]: is the final pressure =?
[tex]V_{i}[/tex]: is the initial volume = 420 cm³
[tex]V_{f}[/tex]: is the final volume = 1450 cm³
n: is the number of moles of the gas
R: is the gas constant
[tex] P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm [/tex]
Hence, the final pressure is 1.68 atm.
b) The work done by the isothermal expansion is:
[tex] W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J [/tex]
Therefore, the work done by the gas is 305.3 J.
I hope it helps you!
chinese wares are wrapped by piece of paper why
Answer:
chinawares are wrapped by paper while packaging to reduce the chances of the wares breaking when falling
Calculate P3 (in W). W (b) Find the total power (in W) supplied by the source. W Compare the total power with the sum of the powers dissipated by the resistors. (Consider two values to be equal if they are within 3% of each other.) The total power is less than the sum of the powers dissipated by the resistors. The total power is equal to the sum of the powers dissipated by the resistors. The total power is greater than the sum of the powers dissipated by the resistors.
Answer:
the principle of conservation of energy cannot be violated.
the correct one is: The total power is equal to the sum of the powers dissipated by the resistors.
Explanation:
The power in an electric circuit is given by
P == I V
In a circuit with several components (resistors) the power dissipated is the current by the voltage in each resistance, by the principle of conservation of energy the current in each resistance is the same if the circuit is in series and the current is the same if The circuit is in parallel, but cannot be greater than the current supplied by the power source.
Therefore, the power dissipated by the entire circuit is the sum of the power dissipated by each part, since the principle of conservation of energy cannot be violated.
When reviewing the answers, the correct one is: The total power is equal to the sum of the powers dissipated by the resistors.
Basketball player Darrell Griffith is on record as
attaining a standing vertical jump of 1.2 m (4 ft).
(This means that he moved upward by 1.2 m after
his feet left the floor.) Griffith weighed 890 N (200
lb). g=9.8 m/s2
1- What is his speed as he leaves the floor?
2- if the time of the part of the jump before his feet left the floor was 0.300s, what was the magnitude of his average acceleration while he was pushing against the floor?
Explanation:
1.
We use the equation
h = [tex]\frac{gt^2}{2}[/tex], where
h is the height traveled,
g is the acceleration due to gravity and
t is the time taken to reach height h.
We can now calculate t to be
[tex]\sqrt{\frac{2*1.2 m}{9.81 m/s^2} }[/tex]
= 0.495 s
Let v be the initial velocity of the player.
The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.
v = 9.81 m/s^2 x 0.495 s = 4.85 m/s
2.
The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is
4.85 m/s / 0.3 s = 16.2 m/s^2
The bat emits a sound wave with a frequency of 25.0 kHz and a wavelength of 0.0136
metres.
Calculate the speed of this sound wave.
Answer:
The calculation that you then need to do is 25000 x 0.0136 = 340. You must not forget the units of speed, which here are metres per second, or m/s. Your final answer is 340 m/s.
A spring with an unknown spring constant is hung vertically, and a 200 g (0.200 kg) mass is attached to the bottom. If the spring stretches 0.250 m from its resting position to the position at which the hanging mass is in equilibrium, what is the spring constant of this spring?
Answer:
k = 7.84 N/m
Explanation:
We are given;
Mass hanging object; m = 0.2 kg
Extension; Δx = 0.25 m
Now, formula for the force is;
F = k•Δx
Where k is the spring constant
Since we have mass, then F = W = mg = 0.2 × 9.8 = 1.96 N
Thus;
1.96 = k × 0.25
k = 1.96/0.25
k = 7.84 N/m