Answer: 28
Explanation:
Given
Circuit breaker current is [tex]I=20\ A[/tex]
Power of the light bulb is [tex]P=100\ W[/tex]
Voltage of the DC-circuit is [tex]V=120\ V[/tex]
If the resistance are connected in parallel, they must have same voltage i.e. 120 V
So, Resistance is given by
[tex]\Rightarrow R=\dfrac{V^2}{P}\\\\\Rightarrow R=\dfrac{120^2}{100}\\\\\Rightarrow R=144\ \Omega[/tex]
For the 20 A current and 120 V battery, net resistance is
[tex]\Rightarrow R_{net}=\dfrac{120}{20}\\\\\Rightarrow R_{net}=6\ \Omega[/tex]
Suppose there are n resistance in the circuit connected in parallel.
[tex]\Rightarrow \dfrac{144}{n}=R_{net}\\\\\Rightarrow n=\dfrac{144}{6}\\\\\Rightarrow n=28.8\approx 28\ \text{for current to be less than 20A}[/tex]
Thus, there can maximum of 28 bulbs.
A 40-kg crate is being lowered with a downward acceleration is 2.0 m/s2 by means of a rope. (a) What is the magnitude of the force exerted by the rope on the crate
Answer:
F = 312 N
Explanation:
Given that,
The mass of a crate, m = 40 kg
Acceleration of the crate, a = 2 m/s²
As the carte is falling downward, the net force exerted by the rope on the carte is given by :
F = m(g-a)
Put all the values,
F = 40(9.8-2)
F = 312 N
Hence, the required force exerted by the rope on the crate is equal to 312 N.
5N
5 N
19 N
19 N
Pls help look at the pic
Answer:
b. is the correct answer ....
magnetism/ magnetic field ana magnetic forces
Answer:
Magnetism is a physical phenomenon that manifests itself in a force acting between magnets or other magnetized or magnetisable objects, and a force acting on moving electric charges, such as in current-carrying cables. The force action takes place by means of a magnetic field, which is generated by the objects themselves or otherwise. There are natural and artificial magnets. All magnets have two poles called the north pole and the south pole. The north pole of one magnet repels the north pole of another magnet and attracts the south pole of another magnet; the same with south poles.
Which of the following statements about the electromagnetic spectrum is true?
A. It moves slower than the speed of light
B. It's consisting of waves of varying lengths
C. the slowest is wavelengths are orange and red
D. Scientist can only detect waves of visible light
Answer:
B. its consist of waves of varying lengtu
The decibel level of the sound of a subway train was measured at 92 dB. Find the intensity in watts per square meter (W/m2). (Give your answer in scientific notation, correct to one decimal place.)
Answer:
I = 1.58 x 10⁻³ watt/m²
Explanation:
Here, we will use the following formula:
[tex]\beta = 10\ log_{10}(\frac{I}{I_o})[/tex]
where,
β = decibel level = 92 dB
I = Intenisty of sound in watt/m² = ?
I₀ = reference intensity = 10⁻¹² watt/m²
Therefore,
[tex]92\ dB =10\ log_{10}(\frac{I}{10^{-12}\ watt/m^2} )\\\\[/tex]
[tex]10^{9.2} = \frac{I}{10^{-12}}\ watt/m^2\\\\I = (1.58\ x\ 10^9)(10^{-12}\ watt/m^2)[/tex]
I = 1.58 x 10⁻³ watt/m²
Please help! ❤️
I’ll make you the Brainlyest, I can’t get this one wrong.
Which of these is a source of thermal energy inside earth
There's no multiple answers that you added if that's what you meant but it possibly could be Magma or radioactive decay of particles from the earths core if those two are any of the options
What kind of model is shown below?
о
A. Experimental model
O B. Computer model
O C. Mathematical model
O D. Physical model
Answer:
.....where's the model-
consider a circular loop of wire carrying a counterclockwise current as shown. Indicate the direction of the magnetic field at points both inside and outside of the loop.
Answer:
in this case around the loop the field points downwards on the outside and upwards on the inside
Explanation:
To find the direction of the magnetic field in a wire you must use the right hand rule.
The thumb points in the direction of the current flow and the other created fingers point in the direction of the magnetic field.
Therefore in this case around the loop the field points downwards on the outside and upwards on the inside
According to Newton's second law, how are mass and acceleration related?
A. They are directly proportional to each other
B. They are inversely proportional to each other
Answer:
B. They are inversely proportional to each other
[tex] \frac{momentum}{time} = force \\ \\ \frac{mass \times velocity}{time} = force \\ \\ \frac{mass \times velocity}{time} = mass \times acceleration[/tex]
The accepted speed of sound at atmospheric pressure and 0 *C is 331.5 m/s. The speed of sound increases 0.607 m/s for every *C. Calculate the speed of sound at the temperature of your room and compare your measured value to the accepted value.
Complete Question
The accepted speed of sound at atmospheric pressure and 0 *C is 331.5 m/s. The speed of sound increases 0.607 m/s for every *C. Calculate the speed of sound at the temperature of your room(70F) and compare your measured value to the accepted value.
Answer:
[tex]V_{Tc}=344.314m/s[/tex]
Explanation:
From the question we are told that:
Speed of sound at Temperature [tex]0 \textdegree[/tex] [tex]V_0=331.5m/s[/tex]
Rate of Speed increase [tex]\triangle V_{infty}=0.607[/tex]
Generally the equation for Temperature in Celsius is mathematically given by
[tex]Tc=\frac{100}{180}(T_f-32)[/tex]
[tex]Tc=0.56*38[/tex]
[tex]Tc=21.11 textdegree C[/tex]
Therefore speed at Tc
[tex]V_{Tc}=V_0+(Tc)( V_{infty})[/tex]
[tex]V_{Tc}=331.5+(21.11)(0.607)[/tex]
[tex]V_{Tc}=344.314m/s[/tex]
One of the wavelengths of light emitted by hydrogen atoms under normal laboratory conditions is at ?0 = 656.3nm in the red portion of the electromagnetic spectrum. In the light emitted from a distant galaxy this same spectral line is observed to be Doppler-shifted to ? = 953.3nm , in the infrared portion of the spectrum.
How fast are the emitting atoms moving relative to the earth?
Answer:
1.07 × 10⁸ m/s
Explanation:
Using the relativistic Doppler shift formula which can be expressed as:
[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]
here;
[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v
[tex]\lambda_s =[/tex] observed wavelength from the source's frame.
Given that:
[tex]\lambda _o[/tex] = 656.3 nm
[tex]\lambda_s =[/tex] 953.3 nm
We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.
From the above equation, let's make (v/c) the subject of the formula: we have:
[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]
[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]
[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]
[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]
[tex]\dfrac{v}{c} =0.357[/tex]
v = 0.357 c
To m/s:
1c = 299792458 m/s
∴
0.357c = (299 792 458 × 0.357) m/s
= 107025907.5 m/s
= 1.07 × 10⁸ m/s
one of the lady spartans was falling to the ground after dunking the winning basket. At the end of her fall, she was falling 4 m/s. If she was 60kg, how much potential energy did she have at the top of her jump?
Answer:
15 is the correct answer I.t.
what is the magnitude of an electric field (in 106 n/c) that balances the weight of a plastic sphere of mass 2.1 g that has been charged to 3.0 nc
Answer:
[tex]E=6.86\times 10^6\ N/C[/tex]
Explanation:
Given that,
Mass of the sphere, m = 2.1 g = 0.0021 kg
Charge, q = 3 nC
We need to find the magnitude of the electric field that balanced the weight of sphere. Let it is E. So,
qE = mg
[tex]E=\dfrac{mg}{q}[/tex]
Put all the values,
[tex]E=\dfrac{0.0021\times 9.8}{3\times 10^{-9}}\\\\E=6.86\times 10^6\ N/C[/tex]
So, the magnitude of the elecric field is [tex]6.86\times 10^6\ N/C[/tex].
A disk of Radius R with a uniform distibution of mass"m" rotater about an axis perpendicular to its place at the rim with angular speed "w" the moment of Inertia of the disc about an axis through the contre MR² What is the KE of the disk?
Answer:
[tex]\frac{1}{2}mR^2\omega^2[/tex]
Explanation:
The rotational kinetic energy of an object is given by [tex]KE_r=\frac{1}{2}I\omega^2[/tex], where [tex]I[/tex] is the object's moment of inertia/rotational inertia and [tex]\omega[/tex] is the object's angular speed.
What we're given:
The object's moment of inertia: [tex]I=MR^2[/tex] The object's radius, mass, and angular speed: [tex]R, m, \omega[/tex], respectivelySince no numerical value is given for any of these, it is implied the desired answer will be an equation in terms of the variables given.
Substituting [tex]I=MR^2[/tex]:
[tex]KE_r=\boxed{\frac{1}{2}mR^2\omega^2}[/tex]
Compared to its weight on Earth, a 5kg object on the moon will weigh
The same amount
Less
More
Answer:
Less
Explanation:
Weight is a force measurement. The object's mass is 5kg not its weight. To find its weight you have to take the mass of an object and multiply it by the acceleration of gravity. The acceleration of gravity is greater on earth than on the moon so therefore the object will weigh less on the moon.
Which image shows an example of potential energy?
Answer:
D
Explanation:
Potential energy involves the change of an object's position, which in this case a rocket is increasing its vertical displacement from the ground.
When a rocket is increasing its vertical displacement from the ground, it exhibits both potential and kinetic energy. Therefore option D is correct.
At the initial stage, when the rocket is on the ground and not moving, it possesses potential energy. This potential energy is in the form of stored energy due to its elevated position above the ground.
As the rocket launches and gains altitude, it continues to accumulate potential energy because it is moving higher against the force of gravity.
Simultaneously, as the rocket moves upward, it also gains kinetic energy. Kinetic energy is the energy associated with the rocket's motion.
The faster the rocket moves, the greater its kinetic energy becomes. As the rocket ascends, its speed increases, resulting in an increase in kinetic energy.
Therefore, in the context of a rocket increasing its vertical displacement from the ground, both potential energy (due to its height) and kinetic energy (due to its motion) are present.
Know more about potential energy:
https://brainly.com/question/24284560
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If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
This question is incomplete, the complete question is;
In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration?
b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
Answer:
a) the angular acceleration is 80 rad/s²
b) the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
Explanation:
Given the data in the question;
from the graph below;
Angular Velocity at time 0s [tex]w_o[/tex] = 12 rad/s
Angular Velocity at time 0.15s [tex]w_f[/tex] = 24 rad/s
a) What is the angular acceleration;
Angular acceleration ∝ = ( [tex]w_f[/tex] - [tex]w_o[/tex] ) / dt
we substitute
Angular acceleration ∝ = ( 24 - 12 ) / 0.15
Angular acceleration ∝ = 12 / 0.15
Angular acceleration ∝ = 80 rad/s²
Therefore, the angular acceleration is 80 rad/s²
b)
If the ball is 0.60 m from her shoulder, i.e s = 0.6 m
the tangential acceleration of the ball will be;
a = ∝ × s
we substitute
a = 80 × 0.6
a = 48 m/s²
a = ( 48 / 9.8 )g
a = 4.9 g
Therefore, the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
How do a parachutes work??4-5 sentences plsss help rn
Answer:
Explanation:
A parachute works by forcing air into the front of it and creating a structured 'wing' under which the canopy pilot can fly. Parachutes are controlled by pulling down on steering lines that change the shape of the wing, cause it to turn. The main forces acting on a parachute are gravity and drag. When you first release the parachute, the force of gravity pulls it downward, and the parachute speeds toward the ground. The faster the parachute falls, though, the more drag it creates.
A 3.50 kg basket of cookies sits on a 2.00 m high shelf. What is the gravitational potential energy of the basket?
pls help
Answer:
68.6 J
Explanation:
Applying,
P.E = mgh............... Equation 1
Where P.E = Potential Energy of the basket, m = mass of the basket, g = acceleration due to gravity of the basket, h = height of the basket
From the question,
Given: m = 3.5 kg, h = 2.00 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
P.E = 3.5×2×9.8
P.E = 68.6 J
Hence the potential energy of the basket is 68.6 J
Isotopes of the same element always have the same
(2 points)
O atomic mass number
O A-number
O Z-number
O neutrinos
Answer:
Z-number
Explanation:
The Z number is the number of protons in an atom, and this does not change when an isotope is created. I got it right on the test.
You want to produce a magnetic field of magnitude 5.50 x 10¹ T at a distance of 0.0 6 m from a long, straight wire's center. (a) What current is required to produce this field? (b) With the current found in part (a), how strong is the magnetic field 8.00 cm from the wire's center?
Answer:
(a) I = 1650000 A
(b) 4.125 T
Explanation:
Magnetic field, B = 5.5 T
distance, r = 0.06 m
(a) Let the current is I.
The magnetic field due to a long wire is given by
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\5.5= 10^{-7}\times \frac{2\times I}{0.06}\\I =1650000 A[/tex]
(b) Let the magnetic field is B' at distance r = 0.08 m.
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\B = 10^{-7}\times \frac{2\times 1650000}{0.08}\\B'= 4.125 T[/tex]
Is a measurement is precuse it must also be accurate
An illustration with two positive spheres 0.1m apart. The one on the left is labeled q Subscript 1 baseline = 6 microcoulombs and the sphere on the right is labeled q Subscript 2 baseline = 2 microcoulombs.
Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.
Recall that k = 8.99 × 109 N•meters squared per Coulomb squared.
What is the force applied between q1 and q2?
In which direction does particle q2 want to go?
Answer:
F = 10.78 N
Hence q₂ will move away from the charge q₁ towards right side.
Explanation:
The force between two charged particles can be found by using Colomb's Law:
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
where,
F = Force = ?
k = Colomb Constant = 8.99 x 10⁹ N.m²/C²
q₁ = charge on first particle = 6 μC = 6 x 10⁻⁶ C
q₂ = charge on second particle = 2 μC = 2 x 10⁻⁶ C
r = distance between particles = 0.1 m
Therefore,
[tex]F = \frac{(8.99\ x\ 10^9\ N.m^2/C^2)(6\ x\ 10^{-6}\ C)(2\ x\ 10^{-6}\ C)}{(0.1\ m)^2}[/tex]
F = 10.78 N
Since both particles have a positive charge. Therefore this force will be the force of repulsion.
Hence q₂ will move away from the charge q₁ towards right side.
Answer:
Explanation:
E2020
A parallel plate vacuum capacitor has 8.40 J of energy stored. The separation between plates is 2.30 mm. If the separation is decreased to 1.15 mm what is the energy stored if (a) the charge Q on the plates is held constant, and (b) the voltage V across the plates is held constant
(a) 4.20 J
(b) 16.74 J
Explanation:For a parallel plate vacuum capacitor with area A and whose plates are separated by by a distance of d, its capacitance C is given by;
C = A∈₀ / d --------------------(i)
Where;
∈₀ = constant called permittivity of vacuum.
The energy U stored in such capacitor is given by;
U = [tex]\frac{1}{2}[/tex]CV² ----------------------(ii)
or
U = [tex]\frac{1}{2}[/tex](Q²/C) -------------------(**)
Where;
V = potential difference or voltage across the plates.
Q = charge on the plates.
(a) If the charge is held constant
Combine equations (i) and (**) to give;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d) -----------------------(iii)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iii)
8.40 = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / 0.023)
8.40 = [tex]\frac{1}{2}[/tex]Q² x (0.023 / A∈₀)
Multiply through by 2
2 x 8.40 = Q² x (0.023 / A∈₀)
16.80 = Q² x (0.023 / A∈₀)
Divide through by 0.023
16.80 / 0.023 = Q² x (0.023 / A∈₀) / 0.023
730.4 = Q² / (A∈₀)
Make Q² subject of the formula
Q² = 730.4(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
Q = constant [this means that Q² still remains 730.4(A∈₀) ]
The energy stored is found by substituting these values of d and Q² into equation (iii) as follows;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀)) / (A∈₀ / 0.0115)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀))(0.0115 / A∈₀)
U = [tex]\frac{1}{2}[/tex](730.4)(0.0115)
U = 4.20J
Therefore, the energy stored if the charge Q on the plates is held constant is 4.20 J
(b) If the voltage is held constant
Combine equations (i) and (ii) to give;
U = [tex]\frac{1}{2}[/tex](A∈₀ / d)V² -----------------------(iv)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iv)
8.40 = [tex]\frac{1}{2}[/tex](A∈₀ / 0.023)V²
Multiply through by 2 x 0.023
2 x 0.023 x 8.40 = (A∈₀)V²
2 x 0.023 x 8.40 = (A∈₀)V²
0.385 = (A∈₀)V²
Make V² subject of the formula
V² = 0.385/(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
V = constant [this means that V² still remains 0.385/(A∈₀) ]
The energy stored is found by substituting these values of d and V² into equation (iv) as follows;
U = [tex]\frac{1}{2}[/tex](A∈₀ / 0.0115)[0.385/(A∈₀)]
U = [tex]\frac{1}{2}[/tex](0.385/0.0115)
U = 16.74
Therefore, the energy stored if the voltage V across the plates is held constant is 16.74 J
What do interplanetary space missions study?
the moon
stars in other galaxies
planets in the solar system
the sun
Answer:
C. Planets in the solar systemExplanation:
The one above is incorrect, and I know this is late. Even if it doesn't help you I hope it helps people in the future! YES I AM TALKING ABOUT YOU FUTURE PEOPLE!! I know this is the answer because I have taken 5.11 Quiz: Uncrewed Spacecraft in K12. There will only be the questions and correct answers below.
1. Which planetary body was Spirit designed to explore?
Mars.
2. What is the name of the most distant manmade object in space? (Credit: shathaadnan64/lak521)
Voyager 1.
3. Which group was designed to study Saturn? (Credit: Brainly User/snowballandtigoya1xa
Voyager 1, Huygens, and Cassini.
4. Why are scientists interested in exploring Mars?
Possible evidence of life.
5. What do interplanetary space missions study?
Planets in the solar system.
Have an amazing day!!
What is sieving? Give an example where this method is used. (2)
Answer:
sieving is when you separate particles of different sizes.
Explanation:
separating sand mixtures
separating chaffs from local garri
A force of 350 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 20 centimeters to 50 centimeters
Answer:
52.5 J
Explanation:
Applying,
Hook's law,
F = ke............... Equation 1
Where F = Force, k = spring constant, e = extension.
make k the subject of the equation
k = F/e............ Equation 2
From the question,
Given: F = 350 Newtons, e = 30 cm = 0.3 m
Substitute these values into equation 2
k = 350/0.3 N/m
Also,
W = 1/2(ke²).................. Equation 3
Where W = work done in stretching the spring.
Also given: e = (50-20) cm = 30 cm = 0.3 m, k = 350/0.3 N/m
Substitute these values into equation 3
W = 1/2(350/0.3)(0.3²)
W = 350×0.3/2
W = 52.5 J
In the early 1900s, it was proposed that the law of conservation of mass should be simultaneously considered with the law of conservation of energy to explain particular phenomena. Thus, a theory of conservation of mass-energy was proposed. Which of the following reasons could provide evidence to support the proposed theory?
A. After charged particles travel a complete loop around a circuit, the electric potential energy of the charged particles does not change, but the number of available charged particles that can move through the circuit is reduced. This is because charged particles are used in order for circuit elements to operate correctly.
B. After a photon of light is absorbed by certain metals, electrons are found to be ejected from the metals. This is because the energy contained in the massless photon is used to eject an electron with mass out of the metal.
C. After particles of a hot gas collide with other particles in the gas, the initial combined mass of all particles of the gas immediately before the collisions occur is not equal to the final combined mass of all particles immediately after the collisions. This is because some of the particles in the gas are destroyed in the collisions.
D. After the decay of certain unstable nuclei, the initial mechanical energy of an unstable nucleus is not equal to the final mechanical energy of the resultant particles immediately after the decay process. This is because some of the available mechanical energy is converted into a particle that was originally not accounted for.
Answer:
B. After a photon of light is absorbed by certain metals, electrons are found to be ejected from the metals. This is because the energy contained in the massless photon is used to eject an electron with mass out of the metal.
Explanation:
Before, in the early days, it was proposed to form a combined theory by joining the theory of conservation of mass and the theory of conservation of energy and form a combined theory of conservation of mass-energy. It was done to explain a particular theory of [tex]$\text{photoelectric effect}$[/tex].
The [tex]$\text{photoelectric effect}$[/tex] is the emission of the electrons form the surface of a metal when light energy strikes on it. Here, in this phenomenon, both mass and energy is conserved.
When the light strikes a metal surface, electrons gets ejected from the surface. The energy of the photon is used to eject the electron form the metal surface.
The loudness of a sound is the wave's _______
Answer:
amplitude
Explanation:
The loudness of a musical sound is a measure of the sound wave's ?
is amplitude explanation:- The loudness of a sound depends upon the amplitude.Loudness of a sound depends on the amplitude of the vibration producing that sound. Greater is the amplitude of vibration, louder is the sound produced by it. if you find this answer helpful please rate positive thank you so much.