Resistors of 10 and 30 ohms are connected in series to a 120-V source. What is the current l owing in the 30 ohm resistor
Answer:
the current through 30 ohms resistor is 30 A.
Explanation:
Given;
resistor, R₁ = 10 ohms
resistor, R₂ = 30 ohms
voltage of the source = 120 V
The effective resistance, Rt = 30 ohms + 10 ohms = 40 ohms
The current through 30 ohms resistor is the same as the current through 10 ohms resistor because they are connected in series.
Appy ohms law to determine the currecnt through 30 ohms resistor;
I = V / Rt
I = 120 / 40
I = 3 A
Therefore, the current through 30 ohms resistor is 30 A.
A toy helicopter takes off and moves 6 m up and then 1 m back down. What
is the displacement of the helicopter?
A. 7 m
5 m up
C. 6 m
D. 1 m down
the answer is 5 m up AKA "B"
An element with 7 valence electrons will most likely-
be found in group 17 and be highly reactive
be found in group 7 and be highly reactive
be found in group 17 and be inert
be found in group 17 and be somewhat reactive
Answer: be found in group 17 and be highly reactive
Explanation:
Elements are distributed in groups and periods in a periodic table.
Elements that belong to same groups will show similar chemical properties because they have same number of valence electrons.
Flourine, chlorine, bromine and iodine are elements which belong to Group 17. All of them contain 7 valence electrons each and need one electron to complete their octet.
The chemical reactivity of elements is governed by the valence electrons present in the element and thus all of them are highly reactive.
The period of a wave is equal to the time it takes for one wavelength to pass by a fixed point. You stand on a pier watching water waves and see 10.9 waves pass by in a time of 28 seconds.What is the period of the water waves? Round to nearest .01 and do not include units in your answer.
Answer:
T = 2.57 s
Explanation:
The period of a wave is equal to the time it takes for one wavelength to pass by a fixed point.
No of waves observed = 10.9
Time taken, t = 28 s
We need to find the period of the water waves. Number of waves per unit time is called frequency. Let it is f. So,
[tex]f=\dfrac{n}{t}\\\\f=\dfrac{10.9}{28}\\\\f=0.389\ Hz[/tex]
If T is period of the wave. It is equal to the reciprocal of frequency.
[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.389}\\\\T=2.57\ s[/tex]
So, the period of the water waves is 2.57 seconds.
A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught at its original position. What is the initial velocity of the ball? Consider upwards to be the positive direction.
Answer:
The initial velocity of the softball is 14.711 meters per second.
Explanation:
This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.
From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:
[tex]y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex] (Eq. 1)
Where:
[tex]y_{o}[/tex] - Initial height of the softball, measured in meters.
[tex]y[/tex] - Final height of the softball, measured in meters.
[tex]v_{o}[/tex] - Initial velocity of the softball, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]y = y_{o}[/tex], [tex]t = 3.56\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the initial velocity of the softball is:
[tex]v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0[/tex]
[tex]3\cdot v_{o} -44.132\,m= 0[/tex]
[tex]v_{o} = 14.711\,\frac{m}{s}[/tex]
The initial velocity of the softball is 14.711 meters per second.
61) A bicycle wheel of radius 0.36 m and mass 3.2 kg is set spinning at 4.00 rev/s. A very light bolt is attached to extend the axle in length, and a string is attached to the axle at a distance of 0.10 m from the wheel. Initially the axle of the spinning wheel is horizontal, and the wheel is suspended only from the string. We can ignore the mass of the axle and spokes. At what rate will the wheel process about the vertical
Answer:
The rate the wheel will process about the vertical is 2.86 RPM
Explanation:
Given;
radius of the bicycle wheel, R = 0.36 m
mass of the wheel, m = 3.2 kg
angular velocity, ω = 4 rev/s
The rate at which the wheel will process about the vertical is given by;
Ф = τ/L
Where;
τ is the torque
L is the angular momentum
τ = Fr
τ = mgr = 3.2 x 9.8 x 0.1 = 3.126 N.m
L = Iω = MR²ω
L = 3.2 x (0.36)² x (4 x 2π)
L = 10.4244 kg.m²/s
Ф = τ/L
Ф = (3.126) / (10.4244)
Ф = 0.29987 rad/s
Ф = 0.29987 rad/s x (60 / 2π)
Ф = 2.86 RPM
Therefore, the rate the wheel will process about the vertical is 2.86 RPM
In an RL series circuit, an inductor of 4.74 H and a resistor of 9.33 Ω are connected to a 26.4 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Answer in units of J.
Answer:
The energy is [tex]U = 18.98 \ J [/tex]
Explanation:
From the question we are told that
The inductor is [tex]L = 4.74 \ H[/tex]
The resistance of the resistor is [tex]R = 9.33 \ \Omega[/tex]
The voltage of the battery is [tex]V = 26.4 \ V[/tex]
Generally the current flowing in the circuit is mathematically represented as
[tex]I = \frac{V}{R}[/tex]
=> [tex]I = \frac{26.4}{9.33 }[/tex]
=> [tex]I = 2.83 \ A[/tex]
Generally the corresponding energy stored in the circuit is
[tex]U = \frac{1}{2} * L * I^2[/tex]
[tex]U = \frac{1}{2} * 4.74 * 2.83 ^2[/tex]
[tex]U = 18.98 \ J [/tex]
A truck is speeding up as it travels on an interstate. The truck's momentum (in kg · m/s) is proportional to the truck's speed (in m/s).At some moment the truck's momentum is 50600 kg · m/s and the truck's speed is 23 m/s. At this moment, the truck's momentum (in kg · m/s) is how many times as large as the truck's speed (in m/s)?IncorrectThis means that as the truck travels, the truck's momentum (in kg · m/s) is always Incorrect times as large as the truck's speed (in m/s).If the truck is traveling at 16 m/s, what is the truck's momentum?
Answer:
At the moment, the truck's momentum (in kg · m/s) is 2200 times as large as the truck's speed (in m/s).
This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).
If the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.
Explanation:
From the question,
The truck's momentum (in kg · m/s) is proportional to the truck's speed (in m/s).
Let the truck's momentum be P and the truck's speed be v,
Then we can write that
P∝v
Then,
P = kv
Where k is the proportionality constant
From the question,
At some moment the truck's momentum is 50600 kg · m/s and the truck's speed is 23 m/s,
To determine how many times the truck's speed is as large as the truck's momentum at this moment, we will divide the truck's momentum by the speed, that is
50600 ÷ 23 = 2200
Hence, at the moment, the truck's momentum (in kg · m/s) is 2200 times as large as the truck's speed (in m/s).
Since, dividing the truck's momentum by the truck's speed gives the proportionality constant k (that is, P/v = k), then
This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).
From
P = kv
Then, k = P/v
At a moment, P = 50600 kg · m/s and v = 23 m/s
∴ k = 50600 kg · m/s ÷ 23 m/s = 2200 kg
k = 2200 kg
To determine the truck's momentum if the truck is traveling at 16 m/s
From
P = kv
k = 2200 kg
v = 16 m/s
∴ P = 2200 kg × 16 m/s
P = 35200 kg · m/s
Hence, if the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.
The smallest living organism discovered so far is called a mycoplasm. Its mass is estimated as
1.0 × 10–16 g. Express this mass in a. petagrams. b. femtograms. c. attograms.
Answer:
10-16= -4 so, 1.0x-4 is -4. So the answer is -4.
Explanation:
Water flows from one reservoir to another a height, 41 m below. A turbine (η=0.77) generates power from this flow. 1 m3/s passes through the turbine. If 12 m head loss occurs between the two reservoirs, determine the actual (i.e. useable) power generated by the turbine (in kW).
Complete Question
A diagram representing this question is shown on the first uploaded image
Answer:
The value is [tex]P = 294594.3 \ W[/tex]
Explanation:
From the question we are told that
The height is [tex]h = 41 \ m[/tex]
The efficiency of the turbine is [tex]\eta = 0.77[/tex]
The flow rate is [tex]\r V = 1 m^3 / s[/tex]
The head loss is g = 2 m
Generally the head gain by the turbine is mathematically represented as
[tex]H = h - d[/tex]
=> [tex]H = 41 - 2[/tex]
=> [tex]H = 39 \ m [/tex]
Generally the actual power generated by the turbine is mathematically represented as
[tex]P = \eta *\gamma * \r V * H[/tex]
Here [tex]\gamma[/tex] is the specific density of water with value
[tex]\gamma = 9810 N/m^3 [/tex]
So
[tex]P =0.77 *9810 * 1 * 39[/tex]
[tex]P = 294594.3 \ W[/tex]
Predict how the total pressure varies during the gas-phase reaction 2 ICl(g)+H2 (g)→I 2 (g)+2 HCl(g) in a constant-volume container.
Answer:
Explanation:
First thing we should note, is remember Avogadro's law, which states that "equal no of moles of gas occupy equal volume". And thus, the number of moles on both the sides of above are the same, this means that the number of particles that are colliding with wall due to which pressure will be present. Also, if the number of moles are the same, which they are, then the pressure will also be the same.
The Initial number of moles is(see first attachment), without ammonia.
At a point in the reaction, α is the amount or number of N2 that has reacted. Therefore, the total number of gas moles is get by is(see second attachment)
The first two terms represents a change in number of moles of reactant while the last one stands for the number of ammonia produced.
The forces exerted on an object are shown. (3 points)
A box has an arrow pointing up labeled F and an arrow pointing down labeled 3 N.
If the net force on the object along the vertical plane is zero, which statement is correct?
F equals 3 N and the object moves up.
F equals 3 N and the object remains stationary.
F equals 0 N and the object moves down.
F equals 0 N and the object remains stationary.
Answer:
F equals 3 N and the object remains stationary. (second option in the list)
Explanation:
For sure to cancel acting forces, F must be 3N pointing up. But with regards to the object stationary or not, the question is tricky. We could have a ZERO net force applied, and the object moving at constant speed, which could still verify Newton's Laws. But considering the first answer option that refers to vertical motion upward where the object could be gaining potential energy, the most accurate response is that the force F has to be 3 N pointing up to make the object in equilibrium, and no motion in the vertical axis.
Answer: F equals 3 N and the object remains stationary.
Explanation:
g A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 1.97 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1680 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 4.34 V/m, (b) in the negative z direction and has a magnitude of 4.34 V/m, and (c) in the positive x direction and has a magnitude of 4.34 V/m
Answer:
a) 1.22*10^-18 N in the positive z direction
b) 1.65*10^-19 N in the negative z direction
c) (6.94*10^-19 N) in the positive x direction + (5.30*10^-19 N) in the positive z direction
Explanation:
See attachment for calculations
(a) The electromagnetic force on the proton is 1.224 × 10⁻¹⁸ [tex]\hat k[/tex] N
(b) The force is 1.65 × 10⁻¹⁹ [tex](-\hat k)[/tex] N
(c) The force is (6.94 [tex]\hat i[/tex] + 5.29 [tex]\hat k[/tex]) × 10⁻¹⁹ N
Electromagnetic force on the proton:Given a proton moving in the positive y-direction with a speed of :
v = 1680 m/s [tex]\hat j[/tex]
The magnetic field is in the negative x-direction with magnitude:
B = 1.97 mT [tex](-\hat i)[/tex]
(a) Electric field applied in positive z-direction :
E = 4.34 V/m [tex]\hat k[/tex]
The net force on the proton is iven by:
F = q (E + v×B)
where q is the charge on proton, given by:
q = 1.6×10⁻¹⁹ C
So,
F = 1.6×10⁻¹⁹( 4.34 [tex]\hat k[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )
F = 1.6×10⁻¹⁹ ( 4.34 [tex]\hat k[/tex] + 3.309 [tex]\hat k[/tex])
F = 1.224 × 10⁻¹⁸ [tex]\hat k[/tex] N
(b) Electric field applied in negative z-direction :
E = 4.34 V/m [tex](-\hat k)[/tex]
The net force on the proton is iven by:
F = q (E + v×B)
where q is the charge on proton, given by:
q = 1.6×10⁻¹⁹ C
So,
F = 1.6×10⁻¹⁹( 4.34 [tex](-\hat k)[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )
F = 1.6×10⁻¹⁹ ( 4.34 [tex](-\hat k)[/tex] + 3.309 [tex]\hat k[/tex])
F = 1.65 × 10⁻¹⁹ [tex](-\hat k)[/tex] N
(c) Electric field applied in positive x-direction :
E = 4.34 V/m [tex]\hat i[/tex]
The net force on the proton is iven by:
F = q (E + v×B)
where q is the charge on proton, given by:
q = 1.6×10⁻¹⁹ C
So,
F = 1.6×10⁻¹⁹( 4.34 [tex]\hat i[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )
F = 1.6×10⁻¹⁹ ( 4.34 [tex]\hat i[/tex] + 3.309 [tex]\hat k[/tex])
F = (6.94 [tex]\hat i[/tex] + 5.29 [tex]\hat k[/tex]) × 10⁻¹⁹ N
Learn more about electromagnetic force:
https://brainly.com/question/807785?referrer=searchResults
Time it takes stone to fall from the height of 80 m is approximately equal to *
A. 1 s
B. 2 s
C. 4 s
D. 8 s
Answer:
D
Explanation:
Answer:
c.4s
Explanation:
How would you measure and compare the size of the two balls ?
Answer:
you would use a measuring tape and compare your answers
A wire loop with 3030 turns is formed into a square with sides of length ss . The loop is in the presence of a 1.20 T1.20 T uniform magnetic field B⃗ B→ that points in the negative yy direction. The plane of the loop is tilted off the x-axisx-axis by θ=15∘θ=15∘ . If i=1.10 Ai=1.10 A of current flows through the loop and the loop experiences a torque of magnitude 0.0256 N⋅m0.0256 N⋅m , what are the lengths of the sides ss of the square loop, in centimeters?
Answer:
2.59 cm
Explanation:
The torque τ on a current carrying loop of wire is given by τ = NiABsinθ where N = number of turns of loop, i = current in loop, A = area of loop and B = magnetic field.
Now, given that τ = 0.0256 Nm, i = 1.10 A, B = 1.20 T,N = 30 and since the loop is tilted 15° off the x-axis and the magnetic field points in the negative y- direction, the angle between the normal to the loop and the magnetic field is thus 90° - 15° = 75°. So, θ = 75°.
We now find the area of the loop A from
τ = NiABsinθ
A = τ/NiBsinθ
substituting the values of the variables, we have
A = 0.0256 Nm/30 × 1.10 A × 1.20 T × sin75°
A = 0.0256 Nm/38.25
A = 6.69 × 10⁻⁴ m²
Since the loop is a square, with length of side L, its area A = L² and
L = √A
= √(6.69 × 10⁻⁴ m²)
= 2.59 × 10⁻² m
converting to cm, we have
L = 2.59 × 10⁻² m × 100 cm/m
L = 2.59 cm
So, the lengths of sides of the loop is 2.59 cm
A Labrador retriever runs 50 m in 7.2 s to retrieve a toy bird. The dog then runs half way
back in 3.85 s. Determine the average speed and velocity of the dog
Answer:
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
Explanation:
From Physics we must remember the definitions of average speed and average velocity, both measured in meters per second. Velocity is a vectorial quantity, that is, it has both magnitude and direction, whereas speed is an scalar quantity, which is a quantity that is represented solely by its magnitude. We assume that dog moves at constant speed.
For the case of the dog, we get that average speed and average velocity of the animal are, respectively:
Average velocity:
[tex]\vec v_{avg} = \frac{1}{\Delta t}\cdot (\vec r_{B}-\vec r_{A})[/tex] (Eq. 1)
Where:
[tex]\Delta t[/tex] - Travelling time of the dog, measured in seconds.
[tex]\vec r_{A}[/tex] - Initial vector position of the dog, measured in meters.
[tex]\vec r_{B}[/tex] - Final vector position of the dog, measured in meters.
Average speed:
[tex]v_{avg} = \frac{1}{\Delta t} \cdot (s_{A}+s_{B})[/tex] (Eq. 2)
Where [tex]s_{A}[/tex] and [tex]s_{B}[/tex] are the travelled distances of each stage, measured in meters.
If we know that [tex]\Delta t = 11.05\,s[/tex], [tex]\vec r_{A} = 0\,\hat{i}\,\,\,[m][/tex] and [tex]\vec r_{B} = 25\,\hat{i}\,\,\,[m][/tex], [tex]s_{A} = 50\,m[/tex] and [tex]s_{B} = 25\,m[/tex], average velocity and average speed are, respectively:
[tex]\vec v_{avg} = \frac{1}{11.05\,s}\cdot (25\,\hat{i})\,\,\,[m][/tex]
[tex]\vec v_{avg} = 2.262\,\hat{i}\,\,\,\left[\frac{m}{s} \right][/tex]
[tex]v_{avg} = \frac{75\,m}{11.05\,s}[/tex]
[tex]v_{avg} = 6.787\,\frac{m}{s}[/tex]
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
What is the ratio of the displacement amplitudes of two sound waves given that they are both5.0 kHz but have a 3.0 dB intensity level difference?
Answer:
The ratio of the displacement amplitudes of two sound waves is 1.16.
Explanation:
Given that,
Frequency = 5.0 kHz
Intensity level difference = 3.0 dB
We know that,
The sound intensity is inversely proportional to the square of distance.
[tex]I\propto\dfrac{1}{r^2}[/tex]
The sound intensity for first wave is
[tex]\beta_{1}=10\log\dfrac{I_{1}}{I_{0}}[/tex]...(I)
The sound intensity for second wave is
[tex]\beta_{2}=10\log\dfrac{I_{2}}{I_{0}}[/tex]...(II)
We need to calculate the ratio of intensity
From equation (I) and (II)
[tex]\beta_{2}-\beta_{1}=10\log\dfrac{I_{2}}{I_{0}}-10\log\dfrac{I_{1}}{I_{0}}[/tex]
[tex]\Delta \beta=10\log(\dfrac{I_{2}}{I_{1}})[/tex]
Put the value into the formula
[tex]3.0=10\log(\dfrac{I_{2}}{I_{1}})[/tex]
[tex]\dfrac{I_{2}}{I_{1}}=e^{\dfrac{3.0}{10}}[/tex]
[tex]\dfrac{I_{2}}{I_{1}}=1.34[/tex]
We need to calculate the ratio of the displacement
Using formula of displacement
[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{\dfrac{I_{2}}{I_{1}}}[/tex]
Put the value into the formula
[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{1.34}[/tex]
[tex]\dfrac{r_{1}}{r_{2}}=1.16[/tex]
Hence, The ratio of the displacement amplitudes of two sound waves is 1.16.
1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal. The coefficient of friction between the crate and the floor is 0.12.
a) Calculate the net force and acceleration of the crate.
b) If the crate was moving at 1.0 m/s when it was pulled, what would be its velocity after pulling it for 5.0 s?
Answer:
(a) The net force is 80.394 N
The acceleration of the crate is 0.804 m/s²
(b) the final velocity of the crate is 5.02 m/s
Explanation:
Given;
mass of the crate, m = 100 kg
applied force, F = 250 N
angle of inclination, θ = 45°
coefficient of friction, μ = 0.12
Applied force in y-direction, [tex]F_y = Fsin \theta = 250sin45 = 176.78 \ N[/tex]
Applied force in x-direction, [tex]F_x = Fcos \theta = 250cos45 = 176.78 \ N[/tex]
The normal force is calculated as;
N + Fy -W = 0
N = W - Fy
N = (100 x 9.8) - 176.78
N = 980 - 176.78 = 803.22 N
The frictional force is given by;
Fk = μN
Fk = 0.12 x 803.22
Fk = 96.386 N
(a) The net force is given by;
[tex]F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N[/tex]
Apply Newton's second law of motion;
F = ma
[tex]a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2[/tex]
(b) the velocity of the crate after 5.0 s
[tex]F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s[/tex]
I need help with this answer
Choose the best description of a magnet
O A. Something with magnetic domains
O B. Something that attracts iron
O C. Iron, cobalt, and nickel
O D. Something that becomes magnetic with the application of a
current
Answer:
Explanation:
Something that attracts iron...duuhhh
At a distance of 10.0 m from a loudspeaker, the sound intensity level is measured to be 70 dB. At what distance from the source will the intensity be 40 dB?
Answer:
At a distance of 100 m from the source the intensity will be 40 dB.
Explanation:
Sound intensity is the acoustic power transferred by a sound wave per unit area normal to the direction of propagation.
The sound intensity depends on the power of the sound source, where the higher the power the greater the intensity, the distance to the sound source, the greater the distance being the lower the intensity, and the nature of the transmission medium.
The conversion between intensity and decibels corresponds to:
[tex]L=10*log\frac{I}{I0}[/tex]
where I0 = 10⁻¹² W/m² and corresponds to a level of 0 decibels therefore.
In this case, you can apply the following relationship between two intensities and distance, considering that the intensity of the sound level decreases with distance:
[tex]L1 - L2=10*log\frac{I1}{I0} - 10*log\frac{I2}{I0}[/tex]
[tex]L1 - L2=10*(log\frac{I1}{I0} - *log\frac{I2}{I0})[/tex]
[tex]L1 - L2=10*[log(\frac{I1}{I0}\frac{I0}{I2})][/tex]
[tex]L1 - L2=10*[log(\frac{I1}{I2})][/tex]
Being L1= 70 dB and L2= 40 dB
[tex]70 dB - 40 dB=10*[log(\frac{I1}{I2})][/tex]
[tex]30=10*[log(\frac{I1}{I2})][/tex]
[tex]\frac{30}{10} =log(\frac{I1}{I2})[/tex]
[tex]3=log(\frac{I1}{I2})[/tex]
[tex]10^{3} =\frac{I1}{I2}[/tex]
[tex]1,000=\frac{I1}{I2}[/tex]
The intensity is inversely proportional to the square of the distance to the source. The relationship between the intensities I1 and I2 at distances d1 and d2 respectively is:
[tex]\frac{I1}{I2} =\frac{d2^{2} }{d1^{2} }[/tex]
Then:
[tex]1,000=\frac{d2^{2} }{d1^{2} }[/tex]
Being d1= 10 m
[tex]1,000=\frac{d2^{2} }{10^{2} }[/tex]
[tex]1,000=\frac{d2^{2} }{100}[/tex]
1,000*100= d2²
10,000= d2²
√10,000= d2
100 m= d2
At a distance of 100 m from the source the intensity will be 40 dB.
Why does the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery?
a. All the current is used up in the bulb, so the connecting wires don't matter.
b. Very little energy is dissipated in the thick connecting wires.
c. Electric field in the connecting wires is zero, so emf = E_bulb * L_bulb.
d. Current in the connecting wires is smaller than current in the bulb.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
Answer:
Options B & E are correct
Explanation:
Looking at all the options, B & E are the correct ones.
Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.
Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.
Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.
Option C is not correct because although the Electric field along the wire is not zero, it is very small.
Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.
The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is :
b. Very little energy is dissipated in the thick connecting wires.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
"Energy"The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is very little energy is dissipated in the thick connecting wires and the electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
The thicker the wire per unit length, the lesser resistance it'll posses and the lesser the vitality that will be scattered by the wire and in return more vitality will be disseminated by the bulb.
The resistance of the copper wires is low sufficient to guarantee that there's not much drop in voltage over the copper wires. Hence, there will not be any noticeable contrasts within the voltage over the bulb.
Thus, the correct answer is B and E.
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A spring has natural length 16 cm. A force of 3 N is required to holdthe spring compressed compressed to 11 cm. Find the amount ofwork instretching the spring from 17 cm to 19 cm.
Answer:
W = 0.012 J
Explanation:
For this exercise let's use Hooke's law to find the spring constant
F = K Δx
K = F / Δx
K = 3 / (0.16 - 0.11)
K = 60 N / m
Work is defined by
W = F. x = F x cos θ
in this case the force and the displacement go in the same direction therefore the angle is zero and the cosine is equal to 1
W = ∫ F dx
W = k ∫ x dx
we integrate
W = k x² / 2
W = ½ k x²
let's calculate
W = ½ 60 (0.19 -0.17)²
W = 0.012 J
A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sphere about an axis through its center
Answer:
2I/7
Explanation:
Formula for moment of inertia through the centre of mass of a solid sphere is given as; I_sc = (2/5) mR²
Now, from parallel axis theorem, moment of inertia of solid sphere from tangent is given as;
I = (2/5) m R² + m R²
I = (7/5) mR²
Thus,mR² = 5I/7
Putting 5I/7 for mR² in first equation, we have;
I_sc = (2/5) × 5I/7
I_sc = 2I/7
In the figure below the two blocks are connected by a string of negligible mass passing over a frictionless pulley. m1 = 10.0 kg and m2 = 3.80 kg and the angle of the incline is = 37.0°. Assume that the incline is smooth. (Assume the +x direction is down the incline of the plane.)
For what value of m1 (in kg) will the system be in equilibrium?
Answer:
For equilibrium, the mass m1 must be about 2.29 kg
Explanation:
The forces acting on m1 are : the weight (m1 * g) and the tension (T1) of the string.
The forces acting on m2 are:
1) along the ramp: the component of m2's weight (m2 g * sin(37)) and the tension (T2) of the string. Which by the way, must be equal in magnitude to T1 since the string is inextensible.
2) perpendicular to the ramp; the component of m2's weight (m2 g * cos(37)) and the normal from the contact with the ramp. These two compensate each other.
We therefore want the net force on each block to be zero for the system to be in equilibrium. This means:
T1 = m1 g
T2 = T1 = m2 g sin(37)
Then we have that if m2 = 3.8 kg, then:
m1 g = m2 g sin(37)
cancelling "g" on both sides:
m1 = m2 * sin(37) = 3.8 * sin(37) = 2.28689 kg
which may be rounded to about 2.29 kg.
WHAT IS ACCURACY, PRECISION, AND REPRODUCIBILITY? AND WHY ARE THEY SO NECESSARY IN CONDUCTING/DESIGNING EXPERIMENTS? 30 POINTS AND WILL MARK BRAINLIEST
Answer:
Explanation:
Accuracy can be said to mean the degree to which the particular result of a measurement, or calculation, and even possibly specification agrees or is the same with respect to the correct value or an established standard. Succinctly put, it's is how close a value is to the actual value it ought to be.
Precision on the other hand, is a change in a measurement, or calculation, and even as far as specification, much especially as represented by the number of digits that has been established. In other words, it is the proximity of two or more measurements with respect to one another.
Reproducibility occurs when a measurement(for example) is made by another person, or a different instrument is used. Yet, the same values are obtained.
They are very important in design because they account for very important part of an experiment. Neglecting these quantities means exposing an instrument to unknown danger to the factory and even the personnels.
Also, neglect in taking note of accuracy, precision and reproducibility can lead to poor data processing and even human errors.
Now, vote brainliest, will you? :)
A cannon fires a shell straight upward; 1.6 s after it is launched, the shell is moving upward with a speed of 19 m/s. Assuming air resistance is negligible, find the speed (magnitude of velocity) of the shell at launch and 5.5 s after the launch.
Answer:
-19.259m/s
Explanation:
Given;
Final velocity = 19m/s
time t = 1.6s
u is the initial velocity
g is the acceleration due to gravity = 9.81m/s²
Using the equation of motion to first get the initial velocity of the shell:
v = u-gt
19 = u - (9.81)(1.6)
19 = u - 15.696
u = 19+15.696
u = 34.696m/s
The initial velocity of the shell is 34.696m/s
Next is to find the speed of the shell 5.5s after the launch
Using the equation of motion:
v = u-gt
v = 34.696-9.81(5.5)
v = 34.696 - 53.955
v = -19.259m/s
The negative value of the velocity shows that the velocity is travelling in the downward direction
Which is a belief held by sociologists who work from a social-conflict
perspective?
O A. The best approach for a study is from a micro-level orientation.
O B. Personal background has little impact on how individuals react
with one another.
C. Some social patterns are helpful, while others are harmful.
D. Data are irrelevant to the study of sociology.
SUBMIT
Answer:
C. Some social patterns are helpful, while others are harmful.
Explanation:
Hope this was helpful, Have an amazing,spooky Halloween!!
A small glass bead has been charged to 20 nC. What is the magnitude of acceleration in m/s^2 of an electron that is 1.0 cm from the center of the bead? (mass of an electron= 9.1x10^-31 kg)
Answer:
The acceleration is 3.16x10¹⁷ m/s².
Explanation:
First, we need to find the magnitude of the Coulombs force (F):
[tex] |F| = \frac{Kq_{1}q_{2}}{d^{2}} [/tex]
Where:
K is the Coulomb constant = 9x10⁹ Nm²/C²
q₁ is the charge = 20x10⁻⁹ C
q₂ is the electron's charge = -1.6x10⁻¹⁹ C
d is the distance = 1.0 cm = 1.0x10⁻² m
[tex]|F| = \frac{Kq_{1}q_{2}}{d^{2}} = \frac{9\cdot 10^{9}Nm^{2}/C^{2}*20 \cdot 10^{-9} C*(-1.6\cdot 10^{-19} C)}{(0.01 m)^{2}} = 2.88 \cdot 10^{-13} N[/tex]
Now, we can find the acceleration:
[tex] a = \frac{F}{m} = \frac{2.88 \cdot 10^{-13} N}{9.1 \cdot 10^{-31} kg} = 3.16 \cdot 10^{17} m/s^{2} [/tex]
Therefore, the acceleration is 3.16x10¹⁷ m/s².
I hope it helps you!
