Answer:b
Explanation:
The current in a conductor is 2.5A .explain the meaning of this statement
Answer:
In 1 second the amount of charge flowing through the conductor is 2.5 Q.
Explanation:
How many loops are in this circuit?
I see six (6) loops.
I attached a drawing to show where I get six loops from.
Increased air pressure on the surface of hot water tends to
A) prevent boiling.
B) promote boiling.
C) neither of these
A 165-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,015 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable?
Answer:
22.1 years
Explanation:
Since the current in the wire is I = nevA where n = electron density = 8.50 × 10²⁸ electrons/cm³ × 10⁶ cm³/m³= 8.50 × 10³⁴ electrons/m³, e = electron charge = 1.602 × 10⁻¹⁹ C, v = drift velocity of electrons and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.00 cm = 2 × 10⁻² m
Making v subject of the formula, we have
v = I/neA
So, v = I/neπd²/4
v = 4I/neπd²
Since I = 1,015 A, substituting the values of the other variables into the equation, we have
v = 4I/neπd²
v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π ×(2 × 10⁻² m)²]
v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π × 4 × 10⁻⁴ m²]
v = (1,015 A)/[42.779 × 10¹¹ electronsC/m]
v = 23.73 × 10⁻¹¹ m/s
v = 2.373 × 10⁻¹⁰ m/s
Since distance d = speed, v × time, t
d = vt
So, the time it takes one electron to travel the full length of the cable is t = d/v
Since d = distance moved by free charge = length of transmission line = 165 km = 165 × 10³ m and v = drift velocity of charge = 2.373 × 10⁻¹⁰ m/s
t = 165 × 10³ m/2.373 × 10⁻¹⁰ m/s
t = 69.54 × 10⁷ s
t = 6.954 × 10⁸ s
Since we have 365 × 24 hr/day × 60 min/hr × 60 s/min = 31536000 s in a year = 3.1536 × 10⁷ s
So, 6.954 × 10⁸ s = 6.954 × 10⁸ s × 1yr/3.1536 × 10⁷ s = 2.21 × 10 yrs = 22.1 years
It will take one electron 22.1 years to travel the full length of the cable
A box that is sliding across the floor experiences a net force of 10.0 N. If the box has a mass of 1.50 kg, what is the resulting acceleration of the box g
Answer:
a = 6.67 m/s²
Explanation:
F = 10.0 N
m = 1.50 kg
a = ?
F = ma
10.0 = (1.50)a
6.67 = a
Four toy racecars are racing along a circular race track. The cars start at the 3-o'clock position and travel CCW along the track. Car A is constantly 2 feet from the center of the race track and travels at a constant speed. The angle Car A sweeps out increases at a constant rate of 1 radian per second.
Required:
How many radians θ does car A sweep out in t seconds?
Answer:
in t seconds, Car A sweep out t radian { i.e θ = t radian }
Explanation:
Given the data in the question;
4 toy racecars are racing along a circular race track.
They all start at 3 o'clock position and moved CCW
Car A is constantly 2 feet from the center of the race track and moves at a constant speed
so maximum distance from the center = 2 ft
The angle Car A sweeps out increases at a constant rate of 1 radian per second.
Rate of change of angle = dθ/dt = 1
Now,
since dθ/dt = 1
Hence θ = t + C
where C is the constant of integration
so at t = 0, θ = 0, the value of C will be 0.
Hence, θ = t radian
Therefore, in t seconds, Car A sweep out t radian { i.e θ = t radian }
The Cleveland City Cable Railway had a 14-foot-diameter pulley to drive the cable. In order to keep the cable cars moving at a linear velocity of 14 miles per hour, how fast would the pulley need to turn (in revolutions per minute)
Answer:
13.94 rpm
Explanation:
Given that,
The diameter of the pulley, d = 14 foot
Radius, r = 7 foot
The linear velocity of the pulley, v = 14 mph = 20.53 ft/s
We need to find the angular velocity in rpm.
We know that, the relation between the linear velocity and the angular velocity is as follows :
[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{20.53}{14}\\\\\omega=1.46\ rad/s[/tex]
or
[tex]\omega=13.94\ rpm[/tex]
So, the angular velocity of the pulley is 13.94 rpm.
An electron has an initial speed of 8.06 x10^6 m/s in a uniform 5.60 x 10^5 N/C strength electic field.The field accelerates the electron in the direction opposite to its initial velocity.
(a) What is the direction of the electric field?
i. opposite
ii. direction to the electron's initial velocity
iii. same direction as the electron's initial velocity
iv. not enough information to decide
(b) How far does the electron travel before coming to rest? m
(c) How long does it take the electron to come to rest? s
(d) What is the electron's speed when it returns to its starting point?
Answer:
Explanation:
a)
The force on electron acts opposite to the velocity , and direction of force on electron is always opposite to direction of electric field .
Hence direction of electric field must be in the same in which electrons travels.
Hence option iii is correct.
b )
deceleration a = force / mass
= qE / m
= 1.6 x 10⁻¹⁶ x 5.6 x 10⁵ / 9.1 x 10⁻³¹
= .98 x 10²⁰ m /s²
v² = u² - 2 a s
0 = (8.06 x 10⁶ )² - 2 x .98 x 10²⁰ s
s = 64.96 x 10¹² / 1.96 x 10²⁰
= 33.14 x 10⁻⁸ m
c ) time required
= 8.06 x 10⁶ / .98 x 10²⁰
= 8.22 x 10⁻¹² s .
d ) Its speed will be same as that in the beginning ie 8.06 x 10⁶ m/s .
Answer:
(a) Option (i)
(b) 6.6 x 10^-4 m
(c) 8.2 x 10^-11 s
Explanation:
initial velocity, u = 8 .06 x 10^6 m/s
Electric field, E = 5.6 x 10^5 N/C
(a) The direction of field is opposite.
Option (i).
(b) Let the distance is s.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times \frac{qE}{m}\times s\\\\8.06\times 10^6\times 8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} s\\\\s = 6.6\times 10^{-4} m[/tex]
(c) Let the time is t.
Use first equation of motion.
[tex]v = u + a t \\\\0 = u - \times \frac{qE}{m}\times t\\\\8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} t\\\\t = 8.2\times 10^{-11} s[/tex]
convert 2.4 milimetres into metre
Answer: 2.4 millimeters = 0.0024 meters
Explanation: A millimeter is 1/1000 of a meter. By diving 2.4 by 1000, you get 0.0024.
As you move farther away from a source emitting a pure tone, the ___________ of the sound you hear decreases.
Answer:
frequency
Explanation:
The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.
So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is
[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]
where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.
George Frederick Charles Searle
Answer:
George Frederick Charles Searle FRS was a British physicist and teacher. He also raced competitively as a cyclist while at the University of Cambridge. WikipediaExplanation:
GIVE BRAINLISTAn inductor of inductance 0.02H and capacitor of capatance 2uF are connected in series to an a.c. source of frequency 200 Hz- Calculate the Impedance in the circuit . TC
Explanation:
Given:
L = 0.02 H
C = [tex]2\:\mu \text{F}[/tex]
f = 200 Hz
The general form of the impedance Z is given by
[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]
Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to
[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]
[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]
A seesaw made of a plank of mass 10.0 kg and length 3.00 m is balanced on a fulcrum 1.00 m from one end of the plank. A 20.0-kg mass rests on the end of the plank nearest the fulcrum. What mass must be on the other end if the plank remains balanced?
Answer:
7.5 kg
Explanation:
We are given that
[tex]m_1=10 kg[/tex]
Length of plank, l=3 m
Distance of fulcrum from one end of the plank=1 m
[tex]m_2=20 kg[/tex]
We have to find the mass must be on the other end if the plank remains balanced.
Let m be the mass must be on the other end if the plank remains balanced.
In balance condition
[tex]20\times 1=10\times (1.5-1)+m\times (1.5+0.5)[/tex]
[tex]20=10(0.5)+2m[/tex]
[tex]20=5+2m[/tex]
[tex]2m=20-5=15[/tex]
[tex]\implies m=\frac{15}{2}[/tex]
[tex]m=7.5 kg[/tex]
Hence, mass 7.5 kg must be on the other end if the plank remains balanced.
Answer:
The mass at the other end is 7.5 kg.
Explanation:
Let the mass is m.
Take the moments about the fulcrum.
20 x 1 = 10 x 0.5 + m x 2
20 = 5 + 2 m
2 m = 15
m = 7.5 kg
PLZ help asap :-/
............................
Explanation:
[16][tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]
Here,
[tex]\rm { R_1} [/tex] = 2Ω[tex]\rm { R_2} [/tex] = 2Ω[tex]\rm { R_3} [/tex] = 2Ω[tex]\rm { R_4} [/tex] = 2ΩWe have to find the equivalent resistance of the circuit.
Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,
[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]
[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]
[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]
Reciprocating both sides,
[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]
Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]
[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]
Henceforth, Option A is correct.
_________________________________[17][tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]
Here, we have to find the amount of flow of current in the circuit. By using ohm's law,
[tex] \longrightarrow [/tex] V = IR
[tex] \longrightarrow [/tex] 3 = I × 3.33
[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I
[tex] \longrightarrow [/tex] 0.90 Ampere = I
Henceforth, Option B is correct.
____________________________[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]
Topic: Chapter 10: Projectory or trajectile?
Projectile range analysis:
A projectile is launched from the ground at 10 m/s, at
an angle of 15° above the horizontal and lands 5.1 m away.
What other angle could the projectile be launched at, with the same velocity,
and land 5.1 m away?
90°
75°
45
50°
30°
Answer:
The other angle is 75⁰
Explanation:
Given;
velocity of the projectile, v = 10 m/s
range of the projectile, R = 5.1 m
angle of projection, 15⁰
The range of a projectile is given as;
[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]
To find another angle of projection to give the same range;
[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]
Check:
sin(2θ) = sin(2 x 75) = sin(150) = 0.5
sin(2θ) = sin(2 x 15) = sin(30) = 0.5
Water is falling on the blades of a turbine at a rate of 100 kg/s from a certain spring. If the height of spring be 100m, then the power transferred to the turbine will be: a) 100 KW b) 10 KW c) 1 KW d) 100 W
Answer:
Natae Si Jordan Kaya Sya Napaihe
Explanation:
haha
What is the energy equivalent of an object with a mass of 2.5 kg? 5.5 × 108 J 7.5 × 108 J 3.6 × 1016 J 2.25 × 1017 J
Answer:
E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules
Answer:
The answer is D. 2.25 × 1017 J
Explanation:
got it right on edge 2021
A
cook
holds a 3.2 kg carton of milk at arm's length.
75.9
w
25,5 cm
What force FB must be exerted by the bi-
ceps muscle? The acceleration of gravity is
9.8 m/s2. (Ignore the weight of the forearm.)
Answer in units of N.
Answer:
Explanation:
From the given information:
From the rotational axis, the distance of the force of gravity is:
d_g = 25+5.0 cm
d_g = 30.0 cm
d_g = 30.0 × 10⁻² m
However, the relative distance of FB cos 75.9° from the axis is computed as:
d_B = 5.0 cm
d_B = 5.0 × 10⁻² m
The net torque rotational equilibrium = zero (0)
i.e.
[tex]\tau_g -\tau_B = 0 \\ \\ F_gd_g -F_gcos 75.9^0 d_B = 0 \\ \\ F_B = \dfrac{F_g d_g}{F_g cos 65.6} \\ \\ F_B = \dfrac{(3.2)(9.8)(30*10^{-2})}{(5.0*10^{-2} * cos 75.9)} \\ \\ \mathbf{F_B = 772.4 N}[/tex]
= 772.4 N
Thus, the force exerted = 1772.4 N
What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)
Answer: hello your question is incomplete below is the missing part
A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.
answer:
- q
Explanation:
Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero
given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q
A uniformly charged thin rod of length L and positive charge Q lies along the x-axis with its left end at the origin as shown in Figure 1.
a. Set up a correct integral expression for the potential at point A,which lies a distance H above the right end of the rod. Point A has coordinates (L, H). You need to give appropriate limits of integration and expressions for r and dq
b. Set up a correct integral expression for the potential at point B on the x-axis, a distance D from the left end of the rod with the appropriate limits of integration.You need to give appropriate limits of integration and expressions for r and dq.
Answer:
b)
Explanation:
2. g A spring extends by 20 cm when a force of 2 N is applied. What is the value of the spring constant in N/m
10N/m
Explanation:
f=kx
k=f/x
k=20N/0.2m
k=10N/m
11. An object moves in circular path with constant speed
a. Is the object's velocity constant? Explain.
b. Is its acceleration constant? Explain.
Answer:
B. Is its acceleration constant
Explanation:
Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. ... An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction.
A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18 m/s in the x-direction, 24 m/s in the y-direction, and 72 m/s in the z-direction. The segment has length 0.50 m and is parallel to the z-axis as it moves.
Required:
a. Find the motional emf induced between the ends of the segment.
b. What would the motional emf be if the wire segment was parallel to the y-axis?
Answer:
Explanation:
From the information given:
The motional emf can be computed by using the formula:
[tex]E = L^{\to}*(V^\to*\beta^{\to})[/tex]
[tex]E = L^{\to}*((x+y+z)*\beta^{\to})[/tex]
[tex]E = 0.50*((18\hat i+24 \hat j +72 \hat k )*0.0800)[/tex]
[tex]E = 0.50*((18*0.800)\hat k +0j+(72*0.080) \hat -i ))[/tex]
[tex]E = 0.50*((18*0.800)[/tex]
E = 0.72 volts
According to the question, suppose the wire segment was parallel, there will no be any emf induced since the magnetic field is present along the y-axis.
As such, for any motional emf should be induced, the magnetic field, length, and velocity are required to be perpendicular to one another .
Then the motional emf will be:
[tex]E = 0.50 \hat j *((18*0.800)\hat k -(72*0.080) \hat i ))[/tex]
E = 0 (zero)
A system is acted on by its surroundings in such a way that it receives 50 J of heat while simultaneously doing 20 J of work. What is its net change in internal energy
Answer:
30J
Explanation:
Given data
The total quantity of heat recieved= 50J
Quantity of heat used to do work= 20J
Hence the net change is
ΔU= Total Heat - Net work
ΔU= 50-20
ΔU= 30J
Hence the change in the internal energy is 30J
A 9.0 V battery is connected across two resistors in series. If the resistors have resistances of and what is the voltage drop across the resistor?
Select one:
A. 4.6 V B. 9.4 V C. 8.6 V D. 4.4 V
Answer:
the answer to the question is known as D
TIME REMAINING
45:13
A framed picture hangs from two cords attached to the ceiling.
A picture of a picture frame hanging by two cables at the center of the frame at the same length and angle from the vertical.
Which shows the correct free body diagram of the hanging picture?
A free body diagram with two force vectors, the first pointing downward labeled F Subscript g Baseline, the second pointing upward labeled F Subscript N Baseline.
A free body diagram with three force vectors, the first pointing south labeled F Subscript p Baseline, the second pointing northeast labeled F Subscript T Baseline, and the third pointing northwest labeled F Subscript N.
A free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T.
A free body diagram with two force vectors, the first pointing downward labeled F Subscript p Baseline, the second pointing upward labeled F Subscript T Baseline.
Answer:The answer is C
Explanation:
David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s2 at the instant when David passes. How far does Tina drive before passing David?
1
An astronaut weighs 202 lb. What is his weight in newtons?
Answer:
978.6084 Newton
Explanation:
Given the following data;
Weight = 220 lbTo find the weight in Newtown;
Conversion:
1 lb = 4.448220 N
220 lb = 220 * 4.448220 = 978.6084 Newton
220 lb = 978.6084 Newton
Therefore, the weight of the astronaut in Newton is 978.6084.
Weight can be defined as the force acting on a body or an object as a result of gravity.
Mathematically, the weight of an object is given by the formula;
Weight = mg
Where;
m is the mass of the object.g is the acceleration due to gravity.Note:
lb is the symbol for pounds.N is the symbol for Newton.A body of mass 4kg is moving with a velocity of 108km/h . find the kenetic energy of the body.
Answer:
KE = 2800 J
Explanation:
Usually a velocity is expressed as m/s. Then the energy units are joules.
[tex]\frac{108 km}{hr} * \frac{1000m}{1 km} * \frac{1 hour}{3600 seconds} =\frac{108*1000 m}{3600sec}[/tex]
v = 30 m / sec
KE = 1/2 * 4 * (30)^2
KE =2800 kg m^2/sec^2
KE = 2800 Joules
Part B
What is the approximate amount of thrust you need to apply to the lander to keep its velocity roughly constant? Explain why, using Newton's first
law of motion.
Answer:
Force is zero.
Explanation:
According to the Newton's second law, when an object is moving with an acceleration the force acting on the object is directly proportional to the rate of change of momentum of the object.
F = m a
if the object is moving with uniform velocity, the acceleration is zero, and thus, the force is also zero.
Answer: Near the moon’s surface, a thrust over 11,250 N but under 13,500 N would make it travel at a constant vertical velocity.
Explanation: .Newton’s first law of motion states that an object in motion continues to move in a straight line at a constant velocity unless acted upon by an unbalanced force. In accordance with this law, the lunar lander moves in a downward direction toward the surface of the moon under the influence of force due to gravity. A thrust somewhere between 11,250 and 13,500 balances this gravitational force out.