Answer:
The final volume of R-134a is 0.212m³Explanation:
Using one of the general gas equation to find the final volume of the R-134a.
According to pressure law; The volume of a given mas of gas is directly proportional to its temperature provided that the pressure remains constant.
VαT
V = kT
k = V/T
V1/T1 = V2/T2 = k
Given V1 = 0.14-m³ at T1 = –26.4°C = –26.4° + 273 = 246.6K
V2 = ? at T = 100°C = 100+273 = 373K
On substituting this values for T2;
0.14/246.6 = V2/373
373*0.14 = 246.6V2
V2 = 373*0.14 /246.6
V2 = 0.212m³
The final volume of R-134a is 0.212m³
An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associated with this wave points vertically up. The direction of the electric field vector of this wave is:___________
Answer:
either +z direction or -z direction.
Explanation:
The direction of the electric field, in an electromagnetic wave always is perpendicular to the direction of the magnetic field and the direction of propagation of the wave.
You assume a system of coordinates with the negative x axis as the west direction, and the y axis as the up direction
In this case, the wave is propagating toward the west (- x direction), and the magnetic field vector points up (+ y direction), then, it is mandatory that the electric field vector points either +z direction or -z direction.
250cm3 of fres
er of density 1000kgm-3 is mixed with 100cm3 of sea water of density 1030kgm-3. Calculate the density of the mixture. *
Answer:
1008.57kg/m3
Explanation:
Now the mass of fresh water is 250×1000 /1000000 = 0.25kg
Now the mass of salt water is
100×1030 /1000000 = 0.103kg
Note Density = mass / volume
Mass = volume × density
Note that converting from cm3 to m3 we divide by 1000000
Total mass = 0.25kg +0.103kg= 0.353kg.
Total volume also is (250 +100 )/1000000= 35 × 10^{-5}m3
Hence the density of the mixture= total mass / total volume
0.353kg/35 × 10^{-5}m3=1008.57kg/m3
A point charge is located at the center of a thin spherical conducting shell of inner and outer radii r1 and r2, respectively. A second thin spherical conducting shell of inner and outer radii R1 and R2, respectively, is concentric with the first shell. The flux is as follows for the different regions of this arrangement. Ф -10.3 103 N-m2/C for
0 for r<2 4:
-36.8 x 10נ N-m2/c
0 for r > R2
36.8 x 10נ N-m2/c
Determine the magnitude ond sign of the point chorge ond the charge on the surface of the two shels point charge inner shell outer shel.
Answer:
the magnitude is 7 and sign of the point charge on the surface shell is -13
Explanation:
Which symbol in a chemical equation separates the reactants from the products?
Answer:
the arrow symbol ⇒ in irreversible reactions and doble arrow symbol in reversible reactios⇔
Explanation:
i hope this will help you
A tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12 cm below the top of the tube. Resonance occurs again when the water is 34 cm below the top of the tube. If the air temperature is 23 degrees Celsius, find the frequency of the tuning fork.
Answer:
786 Hz
Explanation:
Recall, the speed of sound is
v = 332 + 0.6t
Where t = 23°
v = 332 + 0.6(23)
v = 332 + 13.8
v = 345.8 m
Also, it is known that distance between two consecutive resonance length is half of the wavelength.
L2 - L1 = λ/2
34 - 12 = λ/2
λ/2 = 22
λ = 44 cm
Finally, remember that also
Frequency = speed/ wavelength
Frequency = 345.8/0.4
Frequency = 786 Hz
Therefore, the frequency of the tuning fork is 786 Hz
Unit conversion
The choices are in units
A,GA,MA,uA,kA,mA,nA,pA. Pick one the units
Answer:
1.234567 kA
Explanation:
The prefix k stands for kilo-, or 10³. The prefix m stands for milli-, or 10⁻³. The sum shown is ...
1.234 kA + 0.000567 kA = 1.234567 kA
The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.
a. What is the electric field strength and direction between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules
70 pointss yall !!! helpp
Answer:
A= The type of plant
B= How tall the plant is
Explanation:
A woman weighs 129 lb. If she is standing on a spring scale in an elevator that is traveling downward, but slowing up, the scale will read:___________.
A) more than 129 lb
B) 129 lb
C) less than 129 lb
D) It is impossible to answer this question without knowing the acceleration of the elevator.
Answer:
C) less than 129 lb.
Explanation:
Let the elevator be slowing up with magnitude of a . That means it is accelerating downwards with magnitude a .
If R be the reaction force
For the elevator is going downwards with acceleration a
mg - R = ma
R = mg - ma
R measures its apparent weight . Spring scale will measure his apparent weight.
So its apparent weight is less than 129 lb .
The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by B. What is the magnitude of the magnetic field at twice the distance from the conductor
Answer:
B/4
Explanation:
The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:
The field at twice the distance is B/4.
n astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a time interval 11.2 s . What are the average acceleration in each interval? Assume that the positive direction is to the right.
Answer:-
-1 m/s^2
Explanation:
The average acceleration is given by dividing the change in velocity by change in time;
[tex]a_f=\frac{v_f-v_i}{t_f-t_i}[/tex]
[tex]=\frac{(0-11.2)}{(11.2-0)}=-1 m/s^2[/tex]
the point to be noted here is if the velocity is to the left we substitute it with a negative sign and if it is to the right we substitute it with a positive sign.
Of one of the planets becomes a black hole , what would the escape speed be?
Answer:
If, instead, that rocket was on a planet with the same mass as Earth but half the diameter, the escape velocity would be 15.8 km/s Any object that is smaller than its Schwarzschild radius is a black hole – in other words, anything with an escape velocity greater than the speed of light is a black hole.
Explanation:
brainlies plssssssssssssssssss!
g it as been suggested that solar powered space ships could get a boost from a laser either on earth or in orbit around earth. the laser would have to be very powerful to give any measurable benefit to the ship. if the laser produces a 0.18-m diameter beam of 490-nm light, what is the minimum angular spread of the beam?
Answer:
The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians (or 3.32 μrad).
Explanation:
The minimum angular spread of a wave is the ratio of its narrowest diameter to its wavelength.
From Rayleigh's formula,
Angular spread = 1.22 (wavelength ÷ diameter)
= 1.22 (λ ÷ D)
Given that:
diameter, D = 0.18 m and wavelength, λ = 490 nm, then;
Angular spread of the laser beam = 1.22 (λ ÷ D)
= 1.22[tex](\frac{490*10^{-9} }{0.18})[/tex]
= 1.22× 2.7222 × [tex]10^{-6}[/tex]
= 3.3211 × [tex]10^{-6}[/tex] rad
The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians.
28 points!! please help
Lebron James and Stephen Curry are playing an intense game of minigolf. The final(18th) hole is 8.2 m away from the tee box (starting location) at an angle of 20◦ east of north. Lebron’s first shot lands 8.6 m away at an angle of 35.2◦ east of north and Steph’s first shot lands 6.1 m away at an angle of 20◦ east of north. Assume that the minigolf course is flat.
(A) Which ball lands closer to the hole?
(B) Each player sunk the ball on the second shot. At what angle did each player hit their ball to reach the hole?
Answer:
A. we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B. Lebron James hits at an angle of 17.48° North -East.
The direction of Stephen is = 20° due to East of North
Explanation:
Let [tex]r ^ {\to[/tex] represent the position vector of the hole;
Also; using the origin as starting point. Let the east direction be along the positive x axis and the North direction be + y axis
Thus:
[tex]r ^ {\to[/tex] = [tex]8.2 \ sin 20^0 \hat i + 8.2 \ cos 20 \hat j[/tex]
[tex]r ^ {\to[/tex] = [tex](2.8046 \hat i + 7.7055 \hat j ) m[/tex]
Let [tex]r_1 ^ \to[/tex] be the position vector for Lebron James's first shot
So;
[tex]r_1 ^ \to[/tex] = [tex](8.6 \ sin \ 35.2 )^0 \hat i + 8.6 \ cos \ ( 35.2)^0 \hat j[/tex]
[tex]r^ \to = (4.9573 \hat i + 7.02745 \hat j) m[/tex]
Let [tex]r_2 ^ \to[/tex] be the position vector for Stephen Curry's shot
[tex]r_2 ^ \to[/tex] [tex]=6.1 \ sin 20^0 \hat i + 6.1 \ cos \ 20 \hat j[/tex]
[tex]r_2 ^ \to[/tex] = [tex](2.0863 \hat i + 5.7321 \hat j )m[/tex]
However;
[tex]r ^ \to - r_1 ^\to = (-2.1527 \hat i + 0.67805 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_1 ^\to| =2.25696 \ m }[/tex]
Also;
[tex]r ^ \to - r_2 ^\to = (0.71013 \hat i - 1.9734 \hat j) m[/tex]
[tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex]
Thus; from above ; we will see that the notion [tex]\mathbf{|r ^ \to - r_2 ^\to| = 2.10006 \ m}[/tex] which denotes Stephen Curry illustrates that Stephen Curry minigolf ball shot is closer
B .
For Lebron James ;
The angle can be determine using the trigonometric function:
[tex]tan \theta = ( \dfrac{0.67805}{-2.1527}) \\ \\ tan \theta = -0.131498 \\ \\ \theta = tan ^{-1} ( -0.31498) \\ \\ \mathbf{\theta = -17.48^0}[/tex]
Thus Lebron James hits at an angle of 17.48° North -East.
For Stephen Curry;
[tex]tan \theta = ( \dfrac{-1.9734}{0.7183}) \\ \\ tan \theta = -2.74732 \\ \\ \theta = tan ^{-1} ( -2.74732) \\ \\ \mathbf{\theta = -70.0^0}[/tex]
The direction of Stephen is = 90° - 70° = 20° due to East of North
Which of these boxes will not accelerate!
30 Newtons
40 Newtons
50 kg
15 Newton
B.
10 kg
30 Newtons
C.
30 Newtons
80 kg
20 Newtons
20 Newtons
20 Newtons
D.
75 kg
Answer:
(possibly) Box D
Explanation:
The one that has balanced forces will not accelerate. The forces are unbalanced in figures A, B, C. We cannot tell about figure D, because the downward force is not shown. If that force is 20 N, box D will not accelerate.
A turntable rotates with a constant 1.85 rad/s2 clockwise angular acceleration. After 4.00 s it has rotated through a clockwise angle of 30.0 rad . Part A What was the angular velocity of the wheel at the beginning of the 4.00 s interval?
Answer: The angular velocity of the wheel at the beginning of the 4.00 s interval is 3.8 rad/s
Explanation: Please see the attachment below
The angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.
The given parameters:
Angular speed of the turn table = 1.85 rad/s²Time of motion, t = 4.0 sAngular displacement, θ = 30.0 radThe angular velocity of the wheel at the beginning of the 4.0 s time is calculated as follows;
[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2[/tex]
where;
[tex]\omega_i[/tex] is the initial angular velocity
[tex]30 = \omega_i (4) \ + \frac{1}{2}(1.85)(4)^2\\\\30 = 4\omega _i + 14.8\\\\4\omega _i = 30 - 14.8\\\\ 4\omega _i = 15.2\\\\\omega _i = \frac{15.2}{4} \\\\\omega _i = 3.8 \ rad/s[/tex]
Thus, the angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.
Learn more about angular velocity here: https://brainly.com/question/540174
Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistivities of these two wires compare?
Explanation:
The relation between resistance and resistivity is given by :
[tex]R=\rho \dfrac{l}{A}[/tex]
[tex]\rho[/tex] is resistivity of material
l is length of wire
A is area of cross section of wire
Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is
Part A The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. 3.0 m, 5.0 m 4.0 m, 5.0 m 2.0 m, 3.0 m 1.0 m
Answer:
The location are [tex]x_1 = 2 \ and \ x_2 = 3[/tex]
Explanation:
From the question we are told that
The potential energy is [tex]U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J[/tex]
The force on the mass can be mathematically evaluated as
[tex]F = - \frac{d U(x)}{d x } = -( 6 x^2 - 30x +36)[/tex]
The negative sign shows that the force is moving in the opposite direction of the potential energy
[tex]F = - 6 x^2 + 30x - 36[/tex]
At critical point
[tex]\frac{d U(x)}{dx} = 0[/tex]
So
[tex]- 6 x^2 + 30x - 36 = 0[/tex]
[tex]- x^2 + 5x - 6 = 0[/tex]
Using quadratic equation formula to solve this we have that
[tex]x_1 = 2 \ and \ x_2 = 3[/tex]
please help! i will be giving 50 points, this is for my psychology class.
Iris has been ahead of her classmates for as long as she has been in school. Lately, her classmates have started making fun of her for being a “teacher’s pet,” and they mock her whenever she raises her hand to answer a question.
Iris is most likely being negatively stereotyped as being __________.
A.
below average
B.
normal
C.
intellectually disabled
D.
gifted
Answer:
D
Explanation:
the other students are making fun of her most likely because they are jealous that she is successing in school. hope this helps :)
Answer:
D
Explanation:
Use the position function s(t) = -16t + v_0t + s_0 for free falling objects. A ball is thrown straight down from the top of a 600-foot building with an initial velocity of -30 feet per second. (a) Determine the position and velocity functions for the ball. (b) Determine the average velocity on the interval [1, 3]. (c) Find the instantaneous velocities when t=1 and t=3. (d) Find the time required for the ball to reach ground level. (e) Find the velocity of the ball at impact.
Answer:
a) v = -30 - 32 t , s (t) = 600 - 30 t -16 t² , b) v = -32 ft / s
c) v (1) = -62 ft / s, v (3) = -126 ft / s , d) t = 7.13 s , e) v = -258.16 ft / s
Explanation:
a) For this exercise they give us the function of the position of the ball
s (t) = s (o) + v_o t - 16 t²
notice that you forgot to write the super index
indicate the initial position of the ball
s (o) = 600 ft
also indicates initial speed
v_o = - 30 ft / s
let's substitute in the equation
s (t) = 600 - 30 t -16 t²
to find the speed we use
v = ds / dt
v = v_o - 32 t
v = -30 - 32 t
b) To find the average speed, look for the speed at the beginning and end of the time interval
t = 1 s
v (1) = -30 -32 1
v (1) = - 62 ft / s
t = 3 s
v (3) = -30 -32 3
v (3) = -126 ft / s
the average speed is
v = (v (3) -v (1)) / (3-1)
v = (-126 +62) / 2
v = -32 ft / s
c) instantaneous speeds, we already calculated them
v (1) = -62 ft / s
v (3) = -126 ft / s
d) the time to reach the ground
in this case s = 0
0 = 600 - 30 t -16 t²
t² + 1,875 t - 37.5 = 0
we solve the quadratic equation
t = [-1,875 ±√ (1,875² + 4 37.5)] / 2
t = [1,875 ± 12.39] / 2
t₁ = 7.13 s
t₂ = negative
Since the time must be positive, the correct answer is t = 7.13 s
e) the speed of the ball on reaching the ground
v = -30 - 32 t
v = -30 - 32 7.13
v = -258.16 ft / s
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
Answer:
[tex]\large \boxed{42\, \mu \text{C}}$[/tex]
Explanation:
The formula for the force exerted between two charges is
[tex]F=k \dfrac{ q_1q_2}{r^2}[/tex]
where k is the Coulomb constant.
The charges are identical, so we can write the formula as
[tex]F=k\dfrac{q^{2}}{r^2}[/tex]
[tex]\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}[/tex]
An object is known to have a coefficient of kinetic friction (µk) of 0.167 and a coefficient of static friction (µk) of 0.42. If the normal force is 200 N, how much frictional force will it encounter while it is moving?
Answer:
Ff = 33.4N
Explanation:
To find the frictional force while the object is moving, you take into account that the friction force depends of the coefficient of kinetic friction.
The frictional force is given by:
[tex]F_f=\mu_kN[/tex] (1)
Ff: frictional force = ?
µk: coefficient of kinetic friction = 0.167
N: normal force of the object = 200N
You replace the values of the parameters in the equation (1):
[tex]F_f=(0.167)(200N)=33.4N[/tex]
The frictional force, while the objects is moving, is 33.4N
A 2-kilogram toy car is traveling forward at 1 meter per second when it is hit in the rear by a 3-kilogram toy truck that was traveling at 3 meters per second just before impact. If the two toys stick together, their speed immediately after the collision is
Answer:
v = 1.4 m/s
Explanation:
This problem is about an inelastic collision. The total momentum before the collision is equal to total momentum after (because of the conservation of momentum law):
[tex]m_1v_1-m_2v_2=(m_1+m_2)v[/tex] (1)
m1: mass of the toy car = 2 kg
m2: mass of the toy truck = 3 kg
v1: speed of the toy car = 1 m/s
v2: speed of the truck car = 3 m/s
v: speed of both car and truck after the collision = ?
In the equation (1) the negative sign of m2v2 is because of the opposite direction of the toy truck respect to the toy car.
You solve the equation (1) for v, and you replace the values of all variables involved:
[tex]v=\frac{m_1v_1-m_2v_2}{m_1+m_2}\\\\v=\frac{(2kg)(1m/s)-(3kg)(3m/s)}{2kg+3kg}=-1.4\frac{m}{s}[/tex]
this velocity is negative, then, the direction of motion of both car and truck is in the direction of the truck
Hence, the speed of both car and truck toys is 1.4 m/s
A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected across the primary coil has a voltage given by the function Δv = (130 V)sin(ωt). What rms voltage (in V) is measured across the secondary coil?
Answer:
The rms voltage (in V) measured across the secondary coil is 459.62 V
Explanation:
Given;
number of turns in the primary coil, Np = 375 turns
number of turns in the secondary coil, Ns = 1875 turns
peak voltage across the primary coil, Ep = 130 V
peak voltage across the secondary coil, Es = ?
[tex]\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V[/tex]
The rms voltage (in V) measured across the secondary coil is calculated as;
[tex]V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V[/tex]
Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V
A block of mass 15.0 kg slides down a ramp inclined at 28.0∘ above the horizontal. As it slides, a kinetic friction force of 30.0 N parallel to the ramp acts on it. If the block slides for 5.50 m along the ramp, find the work done on the block by friction.
Answer:
Work is done by friction = -165 J
Explanation:
Given:
Mass of block (m) = 15 kg
Ramp inclined = 28°
Friction force (f) = 30 N
Distance (d) = 5.5 m
Find:
Work is done by friction.
Computation:
Work is done by friction = -Fd
Work is done by friction = -(30)(5.5)
Work is done by friction = -165 J
Coherent light that contains two wavelengths, 660 nm and 470 nm , passes through two narrow slits with a separation of 0.280 mm and an interference pattern is observed on a screen which is a distance 5.50 m from the slits.
Required:
What is the disatnce on the screen between the first order bright fringe for each wavelength?
Answer:
λ1 = 0.0129m = 1.29cm
λ2 = 0.00923m = 0.92 cm
Explanation:
To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:
[tex]y_m=\frac{m\lambda D}{d}[/tex] (1)
m: order of the bright fringe = 1
λ: wavelength of the light = 660 nm, 470 nm
D: distance from the screen = 5.50 m
d: distance between slits = 0.280mm = 0.280 *10^⁻3 m
ym: height of the m-th fringe
You replace the values of the variables in the equation (1) for each wavelength:
For λ = 660 nm = 660*10^-9 m
[tex]y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm[/tex]
For λ = 470 nm = 470*10^-9 m
[tex]y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm[/tex]
4. A neutrally charged conductor has a negatively charged rod brought close to it, and thus has an induced positive charge on the surface closest to the rod. What can we say about the overall charge on the conductor
Answer:
Overall charge still remains zero on conductor until touched by charged rod.
Explanation:
Here, we want to know what has happened to the overall charge on the conductor.
Since the conductor is neutral, the overall charge on the conductor must remain zero because positive charge is induced on close end to rod then equal and negaitve charge is induced on far end to rod.
Thus, overall charge still remains zero on conductor until touched by charged rod.
Which of the following best describes the current age of the Sun?
A.) It is near the end of its lifespan.
B.) It is about halfway through its lifespan.
C.) It is early in its lifespan.
D.) We do not have a good understanding of the Sun's age.
Answer: Its b, The only problem with this is is there supposed to be a picture?
Explanation: NASA has used there fancy gadgets to figure this out but if there was a picture, this answer could be different.
As your bus rounds a flat curve at constant speed, a package with mass 0.900 kg , suspended from the luggage compartment of the bus by a string 50.0 cm long, is found to hang at rest relative to the bus, with the string making an angle of 30.0 â with the vertical. In this position, the package is 55.0 m from the center of curvature of the curve.
Required:
a. What is the radial acceleration of the bus?
b. What is the radius of the curve?
Answer:
a.[tex]5.66ms^{-2}[/tex]
b.55 m
Explanation:
We are given that
Mass ,m=0.9 kg
Length of string,l=50 cm=[tex]\frac{50}{100}=0.50 m[/tex]
1 m=100 cm
[tex]\theta=30^{\circ}[/tex]
R=55 m
a.Centripetal acceleration
[tex]a_c=gtan\theta[/tex]
[tex]a_c=9.8tan30^{\circ}[/tex]
[tex]a_c=5.66 m/s^2[/tex]
Hence, the radial acceleration of the bus=[tex]5.66m/s^2[/tex]
b. Radius of curve,R=55 m
