Answer: 0.75 W
Explanation:
Given
Current flowing [tex]I=0.5\ A[/tex]
The first flashlight has a single 1.5 V battery
The second flashlight has three 1.5 V batteries connected to provide 4.5 V
Power is the product of voltage and current.
For the first flashlight
[tex]\Rightarrow P=1.5\times 0.5\\\Rightarrow P=0.75\ W[/tex]
Thus, the first flashlight providing 0.75 W of power.
HELP PLZ
Calculate the resistivity of a metal of length 2m at 20°C having the resistance
of 20 ohm and diameter 0.3mm?
Answer:
40*C
Explanation:
I'm a little bit unsure about this question.
Answer:
Option C. 4 Hz
Explanation:
To know the correct answer to the question given above, it is important we know the definition of frequency.
Frequency can simply be defined as the number of complete oscillations or circles made in one second.
Considering the diagram given above, the wave passes through the medium over a period of one second.
Thus, we can obtain the frequency by simply counting the numbers of complete circles made during the period.
From the diagram given above,
The number of circles = 4
Thus,
The frequency is 4 Hz
The diagram below shows a person swinging a hammer.
Which position has the least amount of kinetic energy?
A-1
B-2
C-3
D-4
How long does it take for a 3.5 kW electric water heater to heat 40 kg of water? from 20 ° C to 75 ° C? The specific heat capacity of water is 4190 J/kgK PLEASE HURRY ITS A TEST
Answer:
2633.7 s
Explanation:
From the question,
Heat lost by the water heater = Heat gained by the water
Applying,
P = cm(t₂-t₁)/t.................. Equation 1
Where P = power of the heat, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature, t = time
make t the subject of the equation
t = cm(t₂-t₁)/P.............. Equation 2
From the question,
Given: c = 4190 J/kgK, P = 3.5 kW = 3500 W, m = 40 kg, t₁ = 20°C, t₂ = 75°C
Substitute these values into equation 2
t = 4190×40(75-20)/3500
t = 9218000/3500
t = 2633.7 s
How does rainwater contribute to aquatic succession
Answer:
Rain water carries sediment and then these accumulate on the bottom of ponds, lakes and wetlands. This accumulation build up over time and eventually, the water disappears (because they sink into the ground) and the area once covered with water becomes land.
The illustration shows ?
A)open circuit
B)broken circuit
C)series circuit
D)parallel circuit
Answer:
D)parallel circuit
Explanation:
the components are placed parallel from eachother
PHYSICS HELP !! 30 points please answer correctly !! questions attached below
Which of the following means that an
image is real?
A. +do
B. +di
c. -di
D.-do
Answer:
B
Explanation:
Two identical circular, wire loops 35.0 cm in diameter each carry a current of 2.80 A in the same direction. These loops are parallel to each other and are 24.0 cm apart. Line ab is normal to the plane of the loops and passes through their centers. A proton is fired at 2600 m/s perpendicular to line ab from a point midway between the centers of the loops.
Find the magnitude of the magnetic force these loops exert on the proton just after it is fired.
Answer:
The answer is "[tex]4659.2 \times 10^{-24} \ N[/tex]"
Explanation:
The magnetic field at ehe mid point of the coils is,
[tex]\to B=\frac{\mu_0 i R^2}{(R^2+x^2)^{\frac{3}{2}}}\\\\[/tex]
Here, i is the current through the loop, R is the radius of the loop and x is the distance of the midpoint from the loop.
[tex]\to B=\frac{(4\pi\times 10^{-7})(2.80\ A) (\frac{0.35}{2})^2}{( (\frac{0.35}{2})^2+ (\frac{0.24}{2})^2)^{\frac{3}{2}}}\\\\[/tex]
[tex]=\frac{(12.56 \times 10^{-7})(2.80\ A) \times 0.030625}{( 0.030625+ 0.0144)^{\frac{3}{2}}}\\\\=\frac{ 1.07702 \times 10^{-7} }{0.0095538976}\\\\=112.730955 \times 10^{-7}\\\\=1.12\times 10^{-5}\ \ T\\[/tex]
Calculating the force experienced through the protons:
[tex]F=qvB=(1.6 \times 10^{-19}) (2600)(1.12 \times 10^{-5})= 4659.2 \times 10^{-24}\ N[/tex]
A chemist measures the flow of charged ions through a circuit. Which of these would increase the current? Select all that apply.
What is Hooke's law? what is meant by elastic limit?
please answer me
Answer:
Hooke's law describes the elastic properties of materials only in the range in which the force and displacement are proportional. Hooke's law states that the applied force F equals a constant k times the displacement or change in length x, or F = kx. the maximum extent to which a solid may be stretched without permanent alteration of size or shape, is called elastic limit
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Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A
Answer:
[tex](a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]
[tex](b)\ F_{No} = 771.125N[/tex]
Explanation:
Given
[tex]d_D = 6000ft[/tex] ---- Altitude of container in Denver
[tex]A = 0.0155m^2[/tex] -- Surface Area of the container lid
[tex]P_D = 79000Pa[/tex] --- Air pressure in Denver
[tex]P_{No} = 100250Pa[/tex] --- Air pressure in New Orleans
See comment for complete question
Solving (a): The expression for [tex]F_{No[/tex]
Force is calculated as:
[tex]F = \triangle P * A[/tex]
The force in New Orleans is:
[tex]F_{No} = \triangle P * A[/tex]
Since the inside pressure is half the pressure at sea level, then:
[tex]\triangle P = P_{No} - \frac{P_{area}}{2}[/tex]
Where
[tex]P_{area} = 101000Pa[/tex] --- Standard Pressure
Recall that:
[tex]F_{No} = \triangle P * A[/tex]
This gives:
[tex]F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]
Solving (b): The value of [tex]F_{No[/tex]
In (a), we have:
[tex]F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]
Where
[tex]A = 0.0155m^2[/tex]
[tex]P_{No} = 100250Pa[/tex]
[tex]P_{area} = 101000Pa[/tex]
So, we have:
[tex]F_{No} = [100250 - \frac{101000}{2}] * 0.0155[/tex]
[tex]F_{No} = [100250 - 50500] * 0.0155[/tex]
[tex]F_{No} = 49750* 0.0155[/tex]
[tex]F_{No} = 771.125N[/tex]
A sphere has a surface uniformly charged with 3.30 C. At what distance from its center is the potential 4.50 MV
Answer:
r = 6.6 x 10³ m = 6600 m
Explanation:
The potential at a distance from a charged sphere can be given as follows:
[tex]V = \frac{kq}{r}\\\\r = \frac{kq}{V}[/tex]
where,
r = distance = ?
k = Colomb Constant = 9 x 10⁹ Nm²/C²
q = charge on sphere = 3.3 C
V = potential = 4.5 MV = 4.5 x 10⁶ V
Therefore,
[tex]r = \frac{(9\ x\ 10^9\ Nm^2/C^2)(3.3\ C)}{4.5\ x\ 10^6\ V}[/tex]
r = 6.6 x 10³ m = 6600 m
Two parallel slits are illuminated by light composed of two wavelengths, one of which is 657 nm. On a viewing screen, the light whose wavelength is known produces its third dark fringe at the same place where the light whose wavelength is unknown produces its fourth-order bright fringe. The fringes are counted relative to the central or zeroth-order bright fringe. What is the unknown wavelength
Answer:
λ = 5.75 10⁻⁷ mm
Explanation:
This is a slit interference exercise, we analyze each wavelength separately
λ = 657 nm indicate that the third dark pattern
a sin θ = (m + ½) lam
a sin θ = (3 + ½) 657 10⁻⁹
a sin θ = 2299.5 10⁻⁹ nm
for the other wavelength in the same place we have m = 4 bright
a sin θ = m lam
we substitute
2299.5 10⁻⁹ = 4 λ
λ = [tex]\frac{2299.5 \ 10^9 }{ 4}[/tex]
λ = 5.75 10⁻⁷ mm
g a mass of 1.3 kg is pushed horizontally against a massless spring with a spring constant of 58 n/m until the spring compresses 19.5 cm if the mass is then released what is the kinetic energy of the mass when it is no longer in contact with the spring ignore friction
Answer: [tex]1.102\ J[/tex]
Explanation:
Given
Mass [tex]m=1.3\ kg[/tex]
Spring constant [tex]k=58\ N/m[/tex]
Compression in the spring [tex]x=19.5\ cm\ or\ 0.195\ m[/tex]
When the mass leaves the spring, the elastic potential energy of spring is being converted into kinetic energy of mass i.e.
[tex]\Rightarrow \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}\cdot 58\cdot (0.195)^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}mv^2=1.102\ J[/tex]
The kinetic energy of the mass is 1.102 J.
9. Cellular respiration occurs in what types of cells?
Answer:
Cellular respiration takes place in the cells of all organisms. It occurs in autotrophs such as plants as well as heterotrophs such as animals. Cellular respiration begins in the cytoplasm of cells. It is completed in mitochondria
Explanation:
Cellular respiration takes place in the cells of all organisms. It happening in autotrophs such as plantas as well as heterotrophs such as animals. Cellular respiration starts in the cytoplasm of cells.
It is finished in mitochondria.
David is driving a steady 30.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.10 m/s2 at the instant when David passes. Part A How far does Tina drive before passing David
Answer:
Explanation:
Let after time t , Tina catches up David .
Distance travelled by them are equal ,
Distance travelled by Tina
s = ut + 1/2 a t²
= .5 x 2.10 t²
= 1.05 t²
Distance travelled by David
= 30 t ( because of uniform velocity )
1.05 t² = 30t
t = 28.57 s
Distance travelled by Tina
= 1/2 a t²
= .5 x 2.10 x 28.57²
= 857 m approx.
Answer: [tex]857\ m[/tex]
Explanation:
Given
Speed of David car [tex]v=30\ m/s[/tex]
Tina begins to accelerate [tex]2.1\ m/s^2[/tex] after David pass the tina
Suppose it took t time for tina to catch David
Distance traveled by David in t time
[tex]\Rightarrow s_d=30\times t[/tex]
Using the equation of motion to get the distance of Tina is
[tex]s_t=ut+\dfrac{1}{2}at^2\\\\s_t=0+\dfrac{1}{2}\times 2.1t^2[/tex]
now, [tex]s_d=s_t[/tex]
[tex]30t=\dfrac{2.1}{2}t^2\\\\\Rightarrow 2.1t^2-60t=0\\\Rightarrow t(2.1t-60)=0\\\Rightarrow t=0,28.57\ s[/tex]
Neglecting [tex]t=0[/tex]
Distance traveled by tina in [tex]28.57\ s[/tex] is
[tex]s_t=\dfrac{1}{2}\times 2.1\times (28.57)^2\\\\s_t=857.057\approx 857\ m[/tex]
Help me with this please
Answer:
check out of phase
Explanation:
this is my answer
Write the differential equation that governs the motion of the damped mass-spring system, and find the solution that satisfies the initial conditions specified. Units are mks;
????
is the damping coefficient, with units of kg/sec.
m
This question is incomplete, the complete question is;
Write the differential equation that governs the motion of the damped mass-spring system, and find the solution that satisfies the initial conditions specified. Units are mks; γ is the damping coefficient, with units of kg/sec
m = 0.2, γ = 1.6 and k = 4
Initial displacement is 1 and initial velocity is -2
x" + _____ x' ____x = 0
x(t) =
Answer:
the solution that satisfies the initial conditions specified is;
x(t) = [tex]c_1e^{-4t}cos(2t)[/tex] + [tex]c_2e^{-4t}sin(2t)[/tex]
Explanation:
Given the data in the question ;
m = 0.2, γ = 1.6, k = 4
x(0) = 1, x'(0) = -2
Now, the differential equation that governs the motions of spring mass system is;
mx" + γx' + kx = 0
so we substitute
0.2x" + 1.6x' + 4x = 0
divide through by 0.2
x" + 8x' + 20x = 0
hence, characteristics equation will be;
m² + 8m + 20 = 0
we find m using; x = [ -b±√(b² - 4ac) ] / 2a
m = [ -8 ± √((8)² - 4(1 × 20 )) ] / 2(1)
m = [ -8 ± √( 64 - 80 ) ] / 2
m = [ -8 ± √-16 ) ] / 2
m = ( -8 ± 4i ) / 2
m = -4 ± 2i
Hence, the general solution of the differential equation is;
x(t) = [tex]c_1e^{-4t}cos(2t)[/tex] + [tex]c_2e^{-4t}sin(2t)[/tex]
From the initial conditions;
c₁ = 1, c₂ = 1
the solution that satisfies the initial conditions specified is;
x(t) = [tex]c_1e^{-4t}cos(2t)[/tex] + [tex]c_2e^{-4t}sin(2t)[/tex]
is it possible to have rainbows during the fine sunny day? Explain your answer.
Answer:
As long as there is rain, a rainbow is possible. Rain is possible on a sunny day, and is known as a sunshower.The rainbow can be observed in a sunny day if the water droplets are present in air and the sun rays pass through it reaches the eye of the observer. In this situation, the observer can see a rainbow.
or
If you happened to look up at the sky this past weekend, you might have noticed a rare and beautiful sight: iridescent rainbow clouds, but not a drop of rain in sight. This phenomenon is known, fittingly, as cloud iridescence or irisation. The effect is not unlike seeing a rainbow painted on the clouds.
A system has two possible energy states, E0 and E1 (with E1 > E0). If the difference between these energy states grows (E0 and E1 get further apart while the temperature stays constant), then the ratio of the number of particles in state E0 over the number of particles in state E1 will:__________
Answer:
The number of particles in state E0 over the number of particles in state E1 will reduce
Explanation:
E0 represents the ground level state when all the particles have same energy level.
E1 represents excited state in which only a few particle reaches
E0 and E1 get further apart means that the energy difference between the two level increases.
Thus, the number of particles in state E0 over the number of particles in state E1 will reduce.
You purchased 1.9 kg of apples from Wollaston. You noticed that they used a spring scale with the smallest division of 2.1 g to weigh them. What is the relative error in this weight measurement as a percentage
The relative error in this weight measurement as percentage is
[tex]0.1105\%[/tex]What is relative error?is the ratio of the absolute error of a measurement to the measurement being taken. In other words, this type of error is relative to the size of the item being measured.
Therefore,
[tex]\% relative error = \frac{smallest division}{mass of apples}*100\\\\\% relative error = \frac{2.1g}{1900}*100\\\\\% relative error = 0.1105\%[/tex]
For more information on relative error, visit
https://brainly.com/question/64815
Answer:
The relative error in this weight measurement is approximately 0.1105 %.
Explanation:
We know that,
Relative Error is the ratio of the absolute error of a measurement to the actual measurement . It can be mathematically represented as,
Relative Error = (Measured Value - Actual Value) / Actual Value * 100
In this case, the actual weight is 1.9 kg, and the smallest division of the spring scale is 2.1 g.
Actual Weight = 1.9 * 1000 = 1900 (As 1 kg = 1000g)
We know that the absolute error will be equal to the smallest division hence,
Measured Value - Actual Value = 2.1 g
By replacing this in the general formula we get,
Relative Error = (Measured Value - Actual Value) / Actual Value * 100
= [tex]\frac{2.1}{1900} *100[/tex] %
= [tex]\frac{2.1}{19}[/tex] %
≈ 0.1105 %
Therefore, the relative error in this weight measurement is approximately 0.1105 %.
For more information on relative error, visit
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Light travels at 300,000,000 m/s. This is an example
Answer:
ook soooooo
Explanation:
what happen to the total energy of a moving object if all the applied forces are conserved
Answer:
As objects move around over time, the energy associated with them—e.g., kinetic, gravitational potential, heat—might change forms, but if energy is conserved, then the total will remain the same. Conservation of energy applies only to isolated systems.
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Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard nearly 1000 away in the ocean. What is the wavelength of such a sound in seawater, where the speed of sound is 1531 m/s
Answer:
the wavelength of the sound in seawater is 90.1 m.
Explanation:
Given;
frequency of the sound, f = 17 Hz
speed of the sound in seawater, v = 1531 m/s
The wavelength of the wave is calculated as follows;
v = fλ
λ = v / f
where;
λ is the wavelength of the sound
λ = 1531 / 17
λ = 90.1 m
Therefore, the wavelength of the sound in seawater is 90.1 m.
The springs of a 1500 kg car compress 5.00 mm when its 68 kg driver gets into the driver's seat. Part A If the car goes over a bump, what will be the frequency of oscillations
Answer:
the frequency of the oscillation is 1.5 Hz
Explanation:
Given;
mass of the spring, m = 1500 kg
extention of the spring, x = 5 mm = 5 x 10⁻³ m
mass of the driver = 68 kg
The weight of the driver is calculated as;
F = mg
F = 68 x 9.8 = 666.4 N
The spring constant, k, is calculated as;
k = F/m
k = (666.4 N) / (5 x 10⁻³ m)
k = 133,280 N/m
The angular speed of the spring is calculated;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{133280}{1500} } = 9.426 \ rad/s[/tex]
The frequency of the oscillation is calculated as;
ω = 2πf
f = ω / 2π
f = (9.426) / (2π)
f = 1.5 Hz
Therefore, the frequency of the oscillation is 1.5 Hz
A certain organ pipe, open at both ends, produces a fundamental frequency of 288 Hz in air. Part A If the pipe is filled with helium at the same temperature, what fundamental frequency tHe will it produce
Answer:
773.25 Hz
Explanation:
Concept : In an open organ pipe in fundamental mode of vibration
wave length of wave λ = 2L
where L is length of the pipe
frequency = velocity of sound / λ
Given values: fundamental frequency = 288 Hz
fluid is air. velocity of sound = 340 m/s
⇒ 288 = 340/2L
⇒L = 59.02 cm
The point to be noted is if the pipe is filled with helium initially at the same temperature, there would be change in the sound velocity .Then, frequency of note produced will also be changed .
We know that velocity of sound is inversely proportional to square root of molar mass of gas
velocity of sound in air / velocity of sound in helium = Square root of (Molar mass of Helium/ molar mass of air)
[tex]\frac{V_a}{V_{He}} = \sqrt{\frac{4}{28.8} } \\\frac{340}{V_{He}} =0.3725\\V_{He} =912.5 m/s[/tex]
Now, frequency = velocity of sound / λ
= 912.75 / (2 x 0.5902)
= 773.25 Hz
7) The coefficient of performance (COP) of a refrigerator is defined as the ratio of A) the heat removed from the inside to the heat expelled (dismissed) to the outside. B) the heat expelled (dismissed) to the outside to the heat removed from the inside. C) the heat removed from the inside to the work done to remove the heat. D) the heat expelled (dismissed) to the outside to the work done to remove the heat
Answer:
C) the heat removed from the inside to the work done to remove the heat.
Explanation:
Refrigerator is a heat engine working in reverse direction . Heat from cold source is taken out , some work is done to remove them and total heat and work energy is thrown into outside surrounding .
If q heat is taken out and W is work done to get this heat out .
coefficient of performance ( COP ) = q / W .
Hence C ) is the right choice .
Help
An object’s mass 300 kg is observed to accelerate at the rate of 4 m/s^2. Calculate the force required to produce this acceleration. *
1. 1000 N
2 .75 N
3. 1200 N
4. 1250 N
An object is placed 50cm in front of a concave mirror of radius 60cm. How far from the mirror is the image?
Answer:To explian the exact calculation of this pruduct we must not try!