Answer:
233.33°F
Explanation:
(385K - 273.15) * 9/5 + 32 = 233.33°F
An airplane propeller is 2.16 m in length (from tip to tip) with mass 100 kg and is rotating at 2900 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. What is its rotational kinetic energy? Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?
a) The rotational kinetic energy of the airplane propeller is 1792152.287 joules.
b) The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute.
How to determine the angular speed of a airplane propeller
Let consider the airplane propeller a rigid body, the rotational kinetic energy of the propeller (K), in joules, is described by the following formula:
K = 0.5 · I · ω² (1)
Where:
I - Moment of inertia of the airplane propeller, in kilogram-square meters.ω - Angular speed, in radians per secondIn addition, the moment of inertia of a slender rod rotating around its center is:
I = 0.0833 · M · L² (2)
Where:
M - Mass of the propeller, in kilogramsL - Length of the propeller, in metersa) If we know that M = 100 kg, L = 2.16 m and ω = 303.687 rad/s, then the rotational kinetic energy of the propeller is:
K = 0.5 · [0.0833 · (100 kg) · (2.16 m)²] · (303.687 rad/s)²
K = 1792152.287 J
The rotational kinetic energy of the airplane propeller is 1792152.287 joules. [tex]\blacksquare[/tex]
b) By (1) and (2) we know that the mass of the propeller is inversely proportional to the square of the angular speed. Therefore, we have the following relationship:
[tex]M_{o}\cdot \omega_{o}^{2} = M_{f}\cdot \omega_{f}^{2}[/tex]
[tex]\omega_{f} = \sqrt{\frac{M_{o}}{M_{f}} }\cdot \omega_{o}[/tex] (3)
If we know that [tex]\omega_{o} = 2900\,\frac{rev}{min}[/tex], [tex]M_{o} = 100\,kg[/tex] and [tex]M_{f} = 75\,kg[/tex], then the angular speed of the airplane propeller is:
[tex]\omega_{f} = \left(2900\,\frac{rev}{min} \right)\cdot \sqrt{\frac{100\,kg}{75\,kg} }[/tex]
[tex]\omega_{f} \approx 3348.631\, \frac{rev}{min}[/tex]
The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute. [tex]\blacksquare[/tex]
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F=(4i+3j)N acts on an object of mass m=2k.g and drags it by moving the object from origion to x=5m. Find the workdone on the object and the angle between the force and the displacement
Answer:
nnnjjdndbsnnshfhhgbfbdbdh
What is measurement
Answer:
Measurements refers to a process which typically involves identifying and determining the dimensions of a physical object.
Explanation:
A scientific method can be defined as a research method that typically involves the use of experimental and mathematical techniques which comprises of a series of steps such as systematic observation, measurement, and analysis to formulate, test and modify a hypothesis.
Measurements refers to a process which typically involves identifying and determining the dimensions of a physical object.
Basically, the dimensions include important parameters such as width, height, length, area, volume, circumference, breadth, etc.
II) One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4-49. ( ) If the buckets are at rest, what is the tension in each cord? ( ) If the two buckets are pulled upward with an acceleration of 1.25 m/s by the upper cord, calculate the tension in each cord
Answer:
Here , mass of bucket ,m = 3.2 Kg
Now , let the tension in upper rope is T1
the tension in the middle rope is T2
a)
For lower bucket, balancing forces in vertical direction
T2 - mg = 0
T2 = mg
T2 = 3.2 *9.8
T2 = 31.36 N
tension in the middle rope is 31.36 N
For the upper bucket , balancing forces in vertical direction
T1 - T2 - mg = 0
T1 = T2 + 3.2 *9.8
T1 = 62.72 N
the tension in the upper rope is 62.72 N
B)
for a = 1.25 m/s^2
Using second law of motion ,for both the buckets
Fnet = ma
T1 - 2mg = 2m*a
T1 = 2*3.2*(9.8 +1.25)
T1 = 70.72 N
the tension in the upper rope is 70.7 N
Now , the lower bucket
Using second law of motion,
T2 - mg = ma
T2 = 3.2 * (9.8 + 1.25)
T2 = 35.36 N
the tension in the lower rope is 35.36 N
what aspect of the US justice system has its roots in Jewish scripture?
The aspect of the US justice system that has its roots in Jewish scripture is:
the idea that all people are subject to the same rules and laws.
It is the doctrine of "equality before the law." Equality before the law means that every individual is equal in the eyes of the law, whether the individual is a lawmaker, a judge, a law enforcement officer, etc. Equality before the law is also known as equality under the law, equality in the eyes of the law, legal equality, or legal egalitarianism. It is a legal principle that treats each independent being equally and subjects each to the same laws of justice and due process.
Answer:
answer is C
the idea that all people are subject to the same rules and laws
Explanation:
hope this helps!
Which technological device makes an energy conversion in the same way that a human ear makes an energy conversion?
a.) a loudspeaker
b.) a headphone
c.) a light bulb
d.) a microphone
I think it's c because of the concept of mechanical energy to electrical energy but I'm not sure
Answer:
I THINK C
Explanation:
BECAUSE A Light Emitting Diode (LED) glows even when a weak electric current passes through it.
What is the de Broglie wavelength of a red blood cell with a mass of 1.00 * 10-11 g that is moving with a speed of 0.400 cm/s? Do we need to be concerned with the wave nature of the blood cells when we describe the flow of blood in the body?
Answer:
The wavelength is "[tex]=16.5675\times 10^{-18} \ m[/tex]".
Explanation:
Given:
Mass,
m = [tex]1\times 10^{-11} \ g[/tex]
Speed,
V = [tex]0.400 \ cm/s[/tex]
or,
= [tex]0.4\times 10^{-2}[/tex]
According to De Broglie,
The wavelength will be:
⇒ [tex]\lambda = \frac{h}{mV}[/tex]
[tex]=\frac{6.627\times 10^{-34}}{1\times 10^{-11}\times 10^{-3}\times 0.4\times 10^{-2}}[/tex]
[tex]=16.5675\times 10^{-18} \ m[/tex]
So, blood cells move these wavelength.
A 40 kg boy is standing on the edge of a stationary 30-kg platform that is free to rotate without friction. The boy tries to walk around the platform in a counterclockwise direction. As he does:
a. the platform doesn't rotate.
b. the platform rotates in a clockwise direction just fast enough so that the boy remains stationary relative to the ground.
c. the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground.
d. both go around with equal angular velocities but in opposite directions
Answer:
the correct one is C
Explanation:
To find the answer, let's propose the solution of the problem
We create a system formed by the child and the platform so that all the forces have been internal and the angular momentum is conserved.
Initial instant. Before starting to walk
L₀ = 0
Final moment. After the child is walking
L_f = I₁ w₁ + m r v₂
where index 1 is used for the platform and index 2 for the child
linear and angular velocity are related
v₂ = w₂ r
angular momentum is conserved
0 = I₁ w₁ + m r (w₂ r)
w₁ = [tex]- \frac{m r^2}{I1} \ w_2[/tex]
the moment of inertia of the platform bringing it closer to a disk or cylinder
I₁ = [tex]\frac{1}{2}[/tex] M r²
sustitute
w₁ = [tex]- \frac{2 m }{M} \ w_2[/tex]
W₁ = - [tex]- \frac{2 40}{30} \ w_2 = - \frac{8}{3} \ w_2[/tex]
from here we can see that the platform and the child rotate in the opposite direction and with different angular speeds
when examining the answers the correct one is C
Answer:
Option C (the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground)Explanation:
relative to the ground the boy moves in a counter clockwise motion , now the boy and the wheel are one system
so by conservation of angular momentum their net sum of angular momentum relative to a point outside the system(say ground) should be zero
so the wheel moves in a clockwise direction , their angular velocity may or may not be same depending on I. so option D is wrong
option B is wrong because relative to ground their angular momentum should be equal and opposite
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The standard unit of brightness is called the candela.
True
False
Answer:
TRUE
Explanation:
A ball of mass m is dropped from a height h above the ground. neglecting air resistance then determine the speed of the ball when it is at a height y above the ground and determine the speed of the ball at y if at the instant of release it already has an initial upward speed vi at the initial altitude h.
Answer:
Explanation:
kinematic equation (g will have a negative value if we assume UP is positive)
v² = u² + 2as
a) v = √(0² + 2(g)(y - h))
b) v = √(vi² + 2(g)(y - h))
3) A lead bullet initially at 30 C just melts upon striking a target. Assuming that all of the initial kinetic energy of the bullet goes into the internal energy of the bullet to raise its temperature and melt it, calculate the speed of the bullet upon impact. (Specific heat of lead is 0.128 kJ/kg K and lead latent heat of fusion is 24.7 kJ/Kg and melting point of lead is 600 K).
Answer:
The speed of bullet is 354.2 m/s
Explanation:
initial temperature, T = 30 degree C
specific heat, c = 128 J/kg K
Latent heat, L = 24.7 x 1000 J/kg
melting point = 600 K = 327 degree C
Let the mass is m and the speed is v.
Kinetic energy = heat used to increase the temperature + Heat used to melt
[tex]\frac{1}{2} mv^2 = m c (T' - T) + m L\\\\0.5 v^2 = 128 \times (327 - 30) + 24.7\times 1000\\\\0.5 v^2 = 38016 + 24700 \\\\0.5 v^2 = 62716\\\\v = 354.2 m/s[/tex]
Question 3 of 10
What has the same value no matter where it is located in the universe?
A. Volume
B. Weight
C. Mass
D. Density
Reset Selection
Answer:
C. Mass
Explanation:
Relative to a stationary observer, a moving clock Group of answer choices can do any of the above. It depends on the relative energy between the observer and the clock. always runs faster than normal. can do any of the above. It depends on the relative velocity between the observer and the clock. always runs slower than normal. keeps its normal time.
Answer:
always runs slower than normal.
Explanation:
The basic concept of theory of relativity was given famous scientist, Albert Einstein. The relativity theory provides the theory of space and time, which are the two aspects of spacetime.
According to the theory of relativity, the laws of physics are same for all the non-accelerating observers.
In the context, according to the theory of relativity, a moving clock relative tot a stationary observer always runs slower than the normal time.
what do we mean by thrust?
Answer:
the answer is push example: she thrust her hand into her pocket
3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic
The index of refraction of the plastic is approximately 1.461
The known values in the question are;
The thickness of the piece of plastic placed on the dot = 1.30 cm
The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm
The unknown values in the question are;
The index of refraction
Strategy;
Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic
[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]
The real depth of the dot below the piece of plastic, d₁ = 1.30 cm
The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised
Therefore;
The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm
[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]
Therefore, n = 1.30/0.89 ≈ 1.461
The refractive index of the plastic block, n ≈ 1.461
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Four identical metallic objects carry the following charges 1.08 6.74 4.61 and 9.41 C The objects are brought simultaneously into contact so that each touches the others Then they are separated a What is the final charge on each object b How many electrons or protons make up the final charge on each object
Answer:
(a) 5.46 C
(b) 3.4125 x 10^19
Explanation:
q' = 1.08 C, q'' = 6.74 C, q''' = 4.61 C, q'''' = 9.41 C
When the charges are in contact to each other.
(a) So, the net charge is
[tex]q = \frac{q' + q'' + q''' + q''''}{4}[/tex]
[tex]q = \frac{1.08+6.74+4.61+9.41}{4}\\\\q = 5.46 C[/tex]
(b) As the charge is positive in nature, so the protons are there. The number of protons is
[tex]n = \frac{q}{e}\\\\n = \frac{5.46}{1.6\times 10^{-19}}\\\\n = 3.4125\times 10^{19}[/tex]
HELP ME ASAP PLSSS!!
Suppose your actual height is 5 feet and 5.2 inches. A tape measure which can be read tothe nearest 1/8 of an inch gives your height as 65 3/8 inches. The laser device at the clinic that givesreadings to the nearest hundredth of an inch says you are 65.31 inches.
Required:
a. Which measuring device is more accurate?
b. Which measuring device is more precise?
Answer:
a) The laser device
b) The tape
Explanation:
First, there is a need to understand what accuracy and precision mean.
Accuracy is the closeness of a measurement to its true (pre-determined) value.
Precision is the closeness of repeated measurements to each other.
Since 1 feet = 12 inches, then, 5 feet and 5.2 inches would be equivalent to 65.2 inches. This value represents the true value of my height.
The tape measured the height as 65 3/8, which is equivalent to 65.375 inches.
The laser device measured the height as 65.31.
Error = true value - measured value
Absolute error from the tape = 65.2 - 65.375
= -0.175 inches
Absolute error from laser device = 65.2 - 65.31
= -0.11
a) The magnitude of error from the tape is more than that of the laser device. Hence, the laser device is said to be more accurate.
b) Even though there were just single readings from both instruments, the tape can be read to the nearest 1/8 of an inch and as such, can give more precisive measurements than the laser device.
A lead ball is dropped into a lake from a diving board 5.20 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 s after it is released. How deep is the lake?
Answer:
35.047 m
Explanation:
The time it takes the lead ball to reach the surface of the water is
s = ut+gt²/2............. Equation 1
Where t = time it takes the lead ball to reach the surface of water, u = initial velocity of the lead ball, g = acceleration due to gravity, s = heigth.
From the question,
Given: s = 5.20 m, u = 0 m/s (dropped from a height)
Constant: g = 9.8 m/s²
5.2 = 0+9.8t²/2
t² = (5.2×2)/9.8
t² = 10.4/9.8
t² = 1.06
t = √(1.06)
t = 1.03 s
Hence, time taken for the lead ball to reach the bottom of the lake is
t' = 4.5-1.03
t' = 3.47 seconds
v² = u²+2gs............... Equation 2
Where v = final velocity of the lead ball
Substitute into equation 2
v² = 0+2(9.8)(5.2)
v² = 101.92
v = √(101.92)
v = 10.1 m/s
Therefore, depth of the lake is
D = vt'
D = 10.1(3.47)
D = 35.047 m
You attach a 2.30 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched by 0.500 m and release it from rest. Assume the weight slides on a horizontal surface with negligible friction. The weight reaches a speed of zero again 0.400 s after release (for the first time after release). What is the maximum speed of the weight (in m/s)
Answer: [tex]3.92\ m/s[/tex]
Explanation:
Given
Mass of the attached object is [tex]m=2.3\ kg[/tex]
Spring is stretched by [tex]A=0.5\ m[/tex]
Speed reaches zero after [tex]t=0.4\ s[/tex]
Speed is zero at the extremities of the S.H.M motion that is
[tex]\Rightarrow \dfrac{T}{2}=0.4\\\\\Rightarrow T=0.8\ s[/tex]
Time period of motion is [tex]0.8\ s[/tex] which can also be given by
[tex]\Rightarrow \omega T=2\pi\\\\\Rightarrow \omega=\dfrac{2\pi }{T}\\\\\Rightarrow \omega =\dfrac{2\pi }{0.8}\\\\\Rightarrow \omega=\dfrac{5\pi }{2}[/tex]
Maximum speed for S.H.M. is [tex]v_{max}=A\omega[/tex]
[tex]\Rightarrow v_{max}=0.5\times 2.5\pi\\\Rightarrow v_{max}=3.92\ m/s[/tex]
When a charged particle moves at an angle of 26.1 with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90o) with respect to this field will this particle, moving at the same speed?
Answer:
The angle is 153.9 degree.
Explanation:
Let the magnetic field is B and the charge is q. Angle = 26.1 degree
The force is F.
Let the angle is A'.
Now equate the magnetic forces
[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]
A bicycle tire with a volume of 0.00210 m^3 is filled to its recommended absolute pressure of 495 kPa on a cold winter day when the tire's temperature is -14°C. The cyclist then brings his bicycle into a hot laundry room at 32°C.
a. If the tire warms up while its volume remains constant, will the pressure increase be greater than, less than, or equal to the manufacturer's stated 10% overpressure limit?
b. Find the absolute pressure in the tire when it warms to 32 degrees Celcius at constant volume.
(A) The pressure will be greater than 10% overpressure limit.
(B) The final pressure will be "582.915 kPa".
Given:
Volume,
[tex]V = 0.0021 \ m^3[/tex]Initial pressure,
[tex]P_o= 495 \ kPa[/tex]Initial temperature,
[tex]T_o = -14^{\circ} C[/tex][tex]= 259 \ K[/tex]
Final temperature,
[tex]T = 32^{\circ} C[/tex](B)
Number of moles,
→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]
then,
The final absolute pressure,
→ [tex]P = \frac{nRT}{V}[/tex]
[tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]
[tex]=(\frac{T}{T_o} )P_o[/tex]
[tex]= (\frac{305}{259} )\times 495[/tex]
[tex]= 582.915 \ kPa[/tex]
Thus the above approach is correct.
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At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder arrive 7.5 seconds later. How many miles away was the lighting strike? (Assume the light takes essentially no time to reach you).
Answer:
1.6031 miles
Explanation:
Given the following data;
Speed = 344 m/s
Time = 7.5 seconds
To find how many miles away was the lighting strike;
Mathematically, the distance travelled by an object is calculated by using the formula;
Distance = speed * time
Distance = 344 * 7.5
Distance = 2580 meters
Next, we would have to convert the value of the distance travelled in meters to miles;
Conversion:
1609.344 metres = 1 mile
2580 meters = X mile
Cross-multiplying, we have;
X * 1609.344 = 2580
X = 2580/1609.344
X = 1.6031 miles
Se lanza un cohete en un ángulo de 53° sobre la horizontal con una rapidez inicial de 100 m/s. El cohete se mueve por
3.00 s a lo largo de su línea inicial de movimiento con una aceleración de 30.0 m/s2
. En este momento, sus motores fallan,
y el cohete procede a moverse como un proyectil. Determine: (a) la altitud máxima que alcanza el cohete, (b) su tiempo
total de vuelo y (c) su alcance horizontal
Answer:
Explanation:
v = u + at
v₃ = 100 +30.0(3.00) = 190 m/s
s = vt + ½at²
y₃ = (100sin53)(3.00) + ½(30sin53)(3.00²) = 347.4 m
x₃ = (100cos53)(3.00) + ½(30cos53)(3.00²) = 261.8 m
a) v² = u² + uas
s = (v² - u²) / 2a
ymax = 347.4 + (0² - (190sin53)²) / (2(-9.80)) = 1,522 m
b) t₁ = 3.00 s
t₂ = (190sin53) / 9.80 = 15.5 s
t₃ = √(2(1522) / 9.80) = 17.6 s
t = 3.00 + 15.5 + 17.6 = 36.1 s
c) xmax = 261.8 + (190cos53)( 15.5 + 17.6) = 4,047 m
Consider a box with two gases separated by an impermeable membrane. The membrane can move back and forth, but the gases cannot penetrate the membrane. The left side is filled with gas A and the right side is filled with gas B. We will assume that equipartition applies to both gases, but gas A has an excluded volume due to large molecules so its entropy has a different formula.
SA=NAkln(VA+ bNA)+f(UA,NA)
SB=NBkln(VB)+f(UB,NB)
Required:
If NA= 1 moles, NB = 2 moles, the total volume of the box is 1 m3, and b= 4 × 10-4 m3/mole, then find the equilibrium value of VA by maximizing the total entropy.
Answer:
The answer is "[tex]0.3336\ m^3[/tex]"
Explanation:
Using the Promideal gas law:
[tex]P_A=P_B\\\\P_A(V_A-\eta_A b)= \eta_A RT......(1)\\\\P_B V_B=\eta_B \bar{R}T........(2)\\\\From (1) \zeta (2)\\\\[/tex]
[tex]\frac{\eta_A}{V_A-\eta_A b}=\frac{\eta B}{V B}\\\\ \frac{V A- \eta_A b}{V B}=\frac{\eta A}{\eta B }\\\\ \frac{V A-b}{V B}=\frac{1}{2}\\\\V A+V B=1\\\\V B =1- V A\\\\\frac{V A-b}{1-V A}=\frac{1}{2}\\\\2V A-2b=1-V A\\\\3 V A=1+2b\\\\V A=\frac{1+2b}{3}\\\\[/tex]
[tex]=\frac{1+2(4\times 10^{-4})}{3}\\\\=0.3336\ m^3[/tex]
The equilibrium value of Va is 0.3336 m³.
Ideal gas lawThe equilibrium value of Va is determine by applying ideal gas law as shown below;
Pressure of gas A = Pressure of gas B
Pa = Pb
Pa(Va - nab) = naRT----(1)
PbVb = nbRT -----(2)
Solve equation (1) and (2)
[tex]\frac{P_b}{RT} = \frac{n_b}{V_b} \\\\\frac{P_b}{P_a(V_a- n_ab)/n_a} = \frac{n_b}{V_b}\\\\\frac{n_a}{V_a - n_ab} = \frac{n_b}{V_b} \\\\\frac{V_a - n_ab}{V_b} = \frac{n_a}{n_b} \\\\\frac{V_a - b}{V_b} = \frac{1}{2}[/tex]
Va + Vb = 1
Vb = 1 - Va
[tex]\frac{V_a - b}{1 - V_a} = \frac{1}{2}[/tex]
2Va - 2b = 1 - Va
3Va = 1 + 2b
[tex]V_ a = \frac{1 + 2b}{3} \\\\V_a = \frac{1 + (2 \times 4\times 10^{-4})}{3} \\\\V_a = 0.3336 \ m^3[/tex]
Thus, the equilibrium value of Va is 0.3336 m³.
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Action and Reaction are equal in magnitude and opposite in direction.Then,Why do they not balance each other?
Answer:
Newton's third law of motion states that every action has an equal and opposite reaction. This indicates that forces always act in pairs. Reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel each other out.
What type of potential energy is a 9 volt battery an example of?
Gravitational potential energy
Elastic potential energy
Electrical potential energy
chemical potential energy
Answer:
chemical potential energy
Explanation:
A 9v battery comes in different formats, such that the most common one is the carbon-zinc and alkaline chemistry, so these are alkaline batteries (there are also rechargeable or lithium batteries, these also depend on chemical interactions).
These batteries "draw" the energy from chemical interactions of the materials inside of it, so the type of potential energy that is stored in a battery is actually chemical (regardless of the fact that the energy can be transformed into electrical energy later) the "potential" refers to how the energy is stored.
Then the correct option is chemical potential energy
Answer:
Chemical Potential Energy
Explanation:
Hope this helps!!
Have a blessed day/night!! <33
widely accepted scientific principles do not change. true or false
Answer:
False
Explanation:
As technology advances and new evidence is found which either contradicts or supports accepted scientific principles, the principles are susceptible to change.
train starts from rest and accelerates at 1m/ s²
for 10 seconds how far does it move
Answer:
s=50m
Explanation:
you can use the formula
s=ut+1/2at²
s=0t+1/2(1)10²
=1/2(100)
=50
I hope this helps
A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r < R) from the center of the sphere the electric field has a value E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R the magnitude of the electric field at a radius r would be equal to:__________
Answer:
Hence the answer is E inside [tex]= KQr_{1} /R^{3}[/tex].
Explanation:
E inside [tex]= KQr_{1} /R^{3}[/tex]
so if r1 will be the same then
E [tex]\begin{bmatrix}Blank Equation\end{bmatrix}[/tex] proportional to 1/R3
so if R become 2R
E becomes 1/8 of the initial electric field.
Answer:
The electric field is E/8.
Explanation:
The electric field due to a solid sphere of uniform charge density inside it is given by
[tex]E =\frac{\rho r}{3}[/tex]
where, [tex]\rho[/tex] is the volume charge density and r is the distance from the center.
For case I:
[tex]\rho = \frac{Q}{\frac{4}{3}\pi R^3}[/tex]
So, electric field at a distance r is
[tex]E = \frac { 3 Q r}{3\times 4\pi R^3}\\\\E = \frac{Q r}{4\pi R^3}[/tex]
Case II:
[tex]\rho = \frac{Q}{\frac{4}{3}\pi 8R^3}[/tex]
So, the electric field at a distance r is
[tex]E' = \frac { 3 Q r}{3\times 32\pi R^3}\\\\E' = \frac{Q r}{8\times 4\pi R^3}\\\\E' = \frac{E}{8}[/tex]
