Answer:
Angular velocity (w) = 8.86 rad/s
Explanation:
Angular velocity (w) = [tex]\sqrt{} 4ghi/R^{2}[/tex]
g= 9.81 m/s
R= 0.5
hi (initial depth) = 0.5m
Hence= [tex]\sqrt4* 9.81* 0.5/0.5^{2}[/tex] = 8.86 rad/s
heat travel through vacuum by
a. conduction. b.convention
c. radiation. d. both a&b
Answer:
Option C
Explanation:
C. Radiation.....
Answer:
heat travel through vacuum by radiation
Three resistors 4ohms 6ohms8ohms are connected in series and are connected to cell of EMF 60volt and negilible resistance calculate the current in circuit , potential difference in resistors ,total terminal potential difference in circuit and lost volt
Explanation:
Given that,
Three resistors 4ohms 6ohms 8ohms are connected in series and are connected to cell of EMF 60volt.
The equivalent resistance in series combination is given by :
R = R₁ + R₂ +R₃
Put all the values,
R = 4 + 6 + 8
R = 18 ohms
Let I is the current in circuit. So,
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{60}{18}\\\\I=3.33\ A[/tex]
Potential difference in 4 ohms,
[tex]V_1=IR_1\\\\V_1=3.33\times 4\\\\=13.32\ V[/tex]
Potential difference in 6 ohms,
[tex]V_2=IR_2\\\\V_2=3.33\times 6\\\\=19.98\ V[/tex]
Potential difference in 8 ohms,
[tex]V_3=IR_3\\\\V_3=3.33\times 8\\\\=26.64\ V[/tex]
Terminal potential difference in circuit is :
V = IR
Put all the values,
V = 3.33 × 18
V = 59.94 volts
Hence, this is the required solution.
g A car with a mass of 1,120.0 kg accelerates from 0 to 80.0 km/h in 10.0 s. The driver applies the brakes when the car is moving at 80.0 km/h, and the car comes to rest after traveling 44.0 m. What is the net force (in N) on the car during its deceleration
Answer:
8761.5 N
Explanation: Given that a car with a mass of 1,120.0 kg accelerates from 0 to 80.0 km/h in 10.0 s. The driver applies the brakes when the car is moving at 80.0 km/h, and the car comes to rest after traveling 44.0 m. What is the net force (in N) on the car during its deceleration
Mass m = 1120kg
First change km/h to m/s
(80 × 1000) ÷ 3600
Velocity = 22.2 m/s
Acceleration = ( 22.2 - 0 ) ÷ 10
Acceleration = 2.2 m/s^2
Force F1 = mass × acceleration
Substitutes all the parameters into the formula
F1 = 1120 × 2.2
F1 = 2489 N
For the car to come to rest, it will be decelerating.
Using 3rd law of motion
V^2 = U^2 + 2as
V = 0 since it is coming to rest.
Substitute all the parameters into the formula
0 = 22.2^2 - 2 × a × 44
88a = 492.84
a = 492.84/88
a = 5.6 m/s^2
The force F2 = 1120 × 5.6
F2 = 6272.5N
Net force = 6272.5 + 2489
Net force = 8761.5 N
Therefore, the net force is 8761.5 N
A tennis player swings at a ball at a constant speed, taking 0.50 s to rotate her arms and racquet from horizontal to vertical. What acceleration is felt by a small bug at the tip of her racquet if it is 1.3 m from her shoulder
Answer:
the acceleration of the small bug is 12.83 m/s²
Explanation:
Given;
time of motion, t = 0.5 s
radius of the circular path created by his arm, r = 1.3 m
if he rotates his arm from horizontal to vertical, the angular displacement = 90⁰
The centripetal acceleration of the ball is calculated as;
[tex]a_c = \omega^2 r\\\\a_c = (\frac{\theta}{t} )^2 r\\\\[/tex]
[tex]a_c = (\frac{90}{360} \times\frac{ 2\pi }{t} )^2r\\\\a_c = (\frac{\pi}{2t} )^2 r\\\\a_c = \frac{\pi^2r}{4t^2} = \frac{\pi^2 \times1.3 }{4\times 0.5^2} = 12.83 \ m/s^2[/tex]
Therefore, the acceleration of the small bug is 12.83 m/s²
(A) State the relation between acceleration and momentum (10 marks).
(B) Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively. They are moving towards each other in opposite directions on a smooth horizontal table when they collide directly. Immediately before the collision, the speed of A is 8 m/s and the speed of B is 2 m/s. Immediately after the collision, the direction of motion of A is unchanged and the speed of B is twice the speed of A. Find : a. The speed of A immediately after the collision (10 marks), b. The magnitude of the impulse exerted on B in the collision (10 marks).
Answer:
momentum is directly proportional to acceleration
A magnetic field of magnitude 0.550 T is directed parallel to the plane of a circular loop of radius 43.0 cm. A current of 5.80 mA is maintained in the loop. What is the magnetic moment of the loop? (Enter the magnitude.)
Answer:
[tex]\mu = 3.36\times 10^{-3}\ A-m^2[/tex]
Explanation:
Given that,
The magnitude of magnetic field, B = 0.55 T
The radus of the loop, r = 43 cm = 0.43 m
The current in the loop, I = 5.8 mA = 0.0058 A
We need to find the magnetic moment of the loop. It is given by the relation as follows :
[tex]\mu = AI\\\\\mu=\pi r^2\times I[/tex]
Put all the values,
[tex]\mu=\pi \times (0.43)^2\times 0.0058\\\\=3.36\times 10^{-3}\ A-m^2[/tex]
So, the magnetic moment of the loop is equal to[tex]3.36\times 10^{-3}\ A-m^2[/tex].
Hệ số giản nở nhiệt là gì?
Answer:
Hâ ÇÒ ÑÁ ÚIT LÃ
Explanation:
MÃÑJTÎWJWJChoose the smallest item from the list below.
1 glass of water
1 droplet of water
1 atom of oxygen
1 molecule of water
Answer
One molecule of water
While running, a person dissipates about 0.6 J of mechanical energy per step per kilogram of body mass. If a 60-kg person runs with a power of 70 Watts during a race, how fast is the person running
Complete question is: While running, a person dissipates about 0.60 J of mechanical energy per step per kilogram of body mass. If a 60-kg person develops a power of 70 W during a race, how fast is the person running? (Assume a running step is 1.5 m long).
Answer: The person running at a speed of 2.91 m/s.
Explanation:
Given: Mass of runner = 60 kg
Runner dissipates = 0.6 J/kg per step
Average power = 70 W
1 step = 1.50 m
Energy dissipated by the runner is as follows.
[tex]\Delta E_{step} = 0.60 \times 60\\= 36 J[/tex]
Formula used to calculate the value of one step 'S' is as follows.
[tex]\frac{S}{\Delta t} = \frac{P_{avg}}{\Delta E_{step}} = \frac{70}{36}\\= 1.94\\S = 1.94 \Delta t[/tex]
It is known that average velocity is equal to the total distance divided by time interval.
So, total distance for the given situation is as follows.
[tex]d = S \times 1.5[/tex]
Hence, speed of the person is calculated as follows.
[tex]v_{avg} = \frac{d}{\Delta t}\\= \frac{S \times 1.5}{\Delta t}\\= \frac{1.94 \Delta t \times 1.5}{\Delta t}\\= 2.91 m/s[/tex]
Thus, we can conclude that the person running at a speed of 2.91 m/s.
You need to produce a set of cylindrical copper wire 3.5 m long that will have a
resistance of 0.125 Ω each. What will be the mass of each of these wires?
(ρ = 1.72X10-8 Ωm, density of copper = 8.9X103 kg/m3)
Solution :
We know, resistance is given by :
[tex]R = \dfrac{\rho l}{A}[/tex]
[tex]A = \dfrac{\rho l }{R}\\\\A = \dfrac{1.72\times 10^{-8} \times 3.5 }{0.125}\\\\A = 4.816 \times 10^{-7} \ m^2[/tex]
Now, we know mass of wire is given by :
[tex]Mass = Density \times Volume\\\\\M = 8.9 \times 10^3 \times 4.816 \times 10^{-7} \times 3.5 kg\\\\M = 0.01500\ kg\\\\M = 15.00\ gram[/tex]
Hence, this is the required solution.
Given:
Length of wire, l = 3.5 mResistance, R = 0.125 ΩThe resistance will be:
→ [tex]R = \frac{\rho l}{A}[/tex]
or,
→ [tex]A = \frac{\rho l}{R}[/tex]
By substituting the values, we get
[tex]= \frac{1.72\times 10^{-8}\times 3.5}{0.125}[/tex]
[tex]= 4.816\times 10^{-7} \ m^2[/tex]
hence,
The mass will be:
→ [tex]Mass = Density\times Volume[/tex]
[tex]= 8.9\times 10^3\times 4.816\times 10^{-7}\times 3.5[/tex]
[tex]= 0.01500 \ kg[/tex]
[tex]= 15.00 \ g[/tex]
Thus the above answer is right.
Learn more about mass here:
https://brainly.com/question/17108656
At what frequency would an inductor of inductance 0.8H have a reactance of 12000^?
Explanation:
The inductive reactance [tex]X_L[/tex] is given by
[tex]X_L = \omega L = 2 \pi fL[/tex]
Solving for f, we get
[tex]f = \dfrac{X_L}{2 \pi L} = \dfrac{12000\:\text{ ohms}}{2\pi (0.8\:H)}[/tex]
[tex]\:\:\:\:\:\:\:= 2387.3\:\text{Hz}[/tex]
An object of mass m 1 moving with speed v collides with another object of mass m 2 at rest and stick to it. Find the impulse to the second object.
Answer:
(m1+m2)vo=m1v+m2×0
⟹vo=m1+m2m1v
The impulse imparted to second object is equal to change is momentum is -
J=m2vo=m1+m2m2m1v
Calculate how fast the ball would be moving at the instant it leaves the projectile launcher of the spring is compressed by 3.75 cm. Use a value of k = 500 N/m for the spring constant, 10 g for the mass of the ball, and 75 g for the effective mass of the ball holder. Show your work.
Answer:
V = 8.34m/s
Explanation:
Given that
1/2ke^2 = 1/2mv^2 ......1
Where e = 3.75cm = (3.75/100)m
e = 0.0375m
K = 500 N/m
m = 10g = 10/1000
= 0.01kg
Substitute the values into equation 1
0.5×500×(0.0375)^2 = 0.5×0.01×v^2
250×0.001395 = 0.005v^2
0.348 = 0.005v^2
v^2 = 0.348/0.005
v^2 = 69.6
V = √69.6
V = 8.34m/s
The ball launches at the speed of V = 8.34m/s
If v = 5.00 meters/second and makes an angle of 60° with the negative direction of the y–axis, calculate all possible values of vx.
Question 24 of 33 Which of the following is an example of uniform circular motion? A. A car speeding up as it goes around a curve O B. A car slowing down as it goes around a curve 23 C. A car maintaining constant speed as it goes around a curve D. A car traveling along a straight road
Answer:
Pluto revolves around sun with constant speed in a circular orbit, so it is an example of uniform circular motion.
3. How can a generator that otherwise produces AC
current be modified to produce DC current?
Un gas a una temperatura de 38°C, tienen un volumen de 21 Lts Litros. ¿Qué volumen tendrá si la temperatura sube a 67°C?
English Translation :
A gas at a temperature of 38 °C, have a volume of 21 Lts Liters. What volume will it have if the temperature rises to 67°C?
Solution :
Since, there is no information about pressure, let us assume it is constant.
So, by ideal gas equation at constant pressure :
[tex]V_1 T_2 = V_2 T_1[/tex]
Putting given volume and temperature ( in Kelvin ) in above equation, we get :
[tex]21 \times ( 67 + 273 ) = V_2 \times ( 38 + 273 )\\\\V_2 = \dfrac{21 \times (67+273) }{(38+273)}\\\\V_2 = 22.96 \ L[/tex]
Hence, this is the required solution.
A singly charged ion (q=−1.6×10−19) makes 7.0 rev in a 45 mT magnetic field in 1.29 ms. The mass of the ion in kg is
Answer:
[tex]m=1.47\times 10^{-24}\ Kg[/tex]
Explanation:
Given that,
Charge, [tex]q=1.6\times 10^{-19}\ C[/tex]
Revolution = 7 rev
magnetic field, B = 45 mT
Time, t = 1.29 ms
We need to find the mass of the ion. Let m be the mass. The formula for the mass in terms of time period is given by :
[tex]m=\dfrac{qBT}{2\pi}\\\\m=\dfrac{1.6\times 10^{-19}\times 45\times 10^{-3}\times 1.29\times 10^{-3}}{2\pi}\\\\m=1.47\times 10^{-24}\ Kg[/tex]
So, the mass of the ion is equal to [tex]1.47\times 10^{-24}\ Kg[/tex].
What are two things that are different about this refugee crisis compared to other
refugee events in the past?
Answer:
The current humanitarian crisis is unprecedented with an appalling and unacceptable human cost. The number of
refugees is unparalleled in recent times. The diversity of nationalities, motives for migration and individual profiles
also creates a huge challenge for asylum systems and welcoming communities in main European destination
countries. Moreover, given the complexity of its main driving forces, there is unfortunately little hope that the
situation will improve significantly in the near future.
This issue of Migration Policy Debates looks at the most recent developments in the humanitarian migration crisis
and what makes this crisis different from previous ones.
_______________________________________________________
An electric motor consumes 8.40 kJ of electrical energy in 1.00 min. Part A If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2900 rpmrpm
Answer:
The torque is 0.31 Nm.
Explanation:
Electrical energy, E = 8400 J
time, t = 1 min
Angular speed, w = 2900 rpm = 303.53 rad/s
efficiency = 2/3 of input power
The toque is given by
[tex]P =\tau w\\\\\frac{2}{3}\times \frac{E}{t}=\tau w\\\\\frac{2}{3}\times \frac{8400}{60}=\tau \times 303.53\\\\\tau =0.31 Nm[/tex]
Q2 A source of frequency 500 Hz emits waves of
wavelength 0.2m. How long does it take the waves to
travel 400m?
Answer:
4 secs
Explanation:
The first step is to calculate the velocity
V= frequency × wavelength
= 500× 0.2
= 100
Therefore the time can be calculated as follows
= distance/velocity
= 400/100
= 4 secs
At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what distance from the line will the field strength to be 2000 N/C?
a. D/√2.
b. √2D.
c. D/2.
d. 2D.
e. D/4.
Answer:
The correct option is (a).
Explanation:
We know that, the E is inversely proportional to the distance as follows :
[tex]E=\dfrac{k}{d^2}[/tex]
We can write it as follows :
[tex]\dfrac{E_1}{E_2}=(\dfrac{d_2}{d_1})^2[/tex]
Put all the values,
[tex]\dfrac{1000}{2000}=(\dfrac{d_2}{d})^2\\\\\sqrt{\dfrac{1000}{2000}}=(\dfrac{d_2}{D})\\\\0.7071=\dfrac{d_2}{d}\\\\d_1=0.7071D\\\\d_1=\dfrac{D}{\sqrt2}[/tex]
So, the correct option is (a).
An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N, the spring is stretched by 15.5 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 10.4 cm from that position. (in J)
A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant k is
43.8 N = k (0.155 m) ==> k = (43.8 N) / (0.155 m) ≈ 283 N/m
The total work done on the spring to stretch it to 15.5 cm from equilibrium is
1/2 (283 N/m) (0.155 m)² ≈ 3.39 J
The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be
1/2 (283 N/m) (0.259 m)² ≈ 9.48 J
Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.
I need help solving my homework problem , can somebody help me please . This class is physics two . My question is “ where the potential difference across ab is 49.5 V “ ?
Explanation:
a) [tex]Q_{Total} = C_{eq}V = (3.5\:\mu \text{F})(49.5V) = 1.73×10^{-4}\:\text{C}[/tex]
b) The individual capacitors in series circuit carry the same amount of charge as the [tex]Q_{Total}[/tex] so
[tex]Q_{10} = Q_{Total} = 1.73×10^{-4}\:\text{C}[/tex]
c) Likewise, [tex]C_9[/tex] will carry the same amount of charge as [tex]C_{10}[/tex]
g to generate electricity, solar panels typically absorb visible light. How many photons of light with a frequency of 5045x10 14 hz does a solar panel absorb to create 360 kj
Answer:
the number of photons absorbed by the solar panel is 1.08 x 10²¹
Explanation:
Given;
frequency of each photon absorbed, f = 5045 x 10¹⁴ Hz
energy to be created by the solar panel, E = 360 kJ = 360,000 J
The energy of each photon absorbed is calculated as;
[tex]E_{photon} = hf\\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\E_{photon} = (6.626 \times 10^{-34} )(5045 \times 10^{14})\\\\E_{photon} = 3.343 \times 10^{-16} \ J[/tex]
let the number of photons absorbed = n
[tex]n(E_{photon}) = 360,000 \ J\\\\n = \frac{360,000 \ J}{3.343 \times 10^{-16} \ J} \\\\n = 1.08 \times 10^{21} \ photons[/tex]
Therefore, the number of photons absorbed by the solar panel is 1.08 x 10²¹
The total power input to a pumped storage power station is 600 MW
The useful power output is 540 MW calculate the efficiency of this pumped storage power station.
Calculate how much power is wasted by the pumped storage power station.
Answer:
60MW wasted
Explanation:
600-540
=60MW
A steel playground slide is 5.25 m long and is raised 2.75 m on one end. A 45.0 kg child slides down from the top starting at rest. The final speed of the child at the bottom is 6.81 m/s. Find the average force of friction between the child and the slide.
Answer:
[tex]F=32.24N[/tex]
Explanation:
From the question we are told that:
Height [tex]h= 2.75 m[/tex]
Length[tex]l = 5.25 m[/tex]
Mass [tex]m=45kg[/tex]
Final speed [tex]v_f=6.81[/tex]
Generally the equation for Potential Energy P.E is mathematically given by
[tex]P.E=mgh[/tex]
Therefore
Initial potential energy
[tex]P.E_1=45*9.8*2.75 \\\\P.E_1= 1212.75 J[/tex]
Generally the equation for Kinetic Energy K.E is mathematically given by
[tex]K.E=0.5mv^2[/tex]
Therefore
Final kinetic energy
[tex]K.E_2= 1/2*45*6.81*6.81 \\\\K.E_2= 1043.46J[/tex]
Generally the equation for Work_done is mathematically given by
[tex]W=P.E_1-K.E_2\\\\W=169.3[/tex]
Therefore
[tex]F=\frac{W}{d}\\\\F=\frac{169.3}{5.25}[/tex]
[tex]F=32.24N[/tex]
What does this circle graph tell you about water on Earth? (2 points)
a pie graph with a big blue section covering seventy one percent and small gray section covering twenty nine percent with a key indicating that blue is water and gray is land
Fresh water covers 71 percent of Earth's surface.
Oceans covers 71 percent of Earth's surface.
Salt water covers 71 percent of Earth's surface.
Water covers 71 percent of Earth's surface.
Answer:
ocean covers 71 percent of the earth
Answer:
the ocean covers 71 percent of Earth's surface.
Explanation:
SOMEONE HELP ME PLSS
Answer:
Explanation:
1 minute =60 second
power=work done/time taken
=2500/60
=41.66 watt
work=force *displacement
=500 N * 100 m
=50000 joule
A satellite has a mass of 6463 kg and is in a circular orbit 4.82 × 105 m above the surface of a planet. The period of the orbit is 2.0 hours. The radius of the planet is 4.29 × 106 m. What would be the true weight of the satellite if it were at rest on the planet’s surface?
Answer:
The weight of the planet is 29083.5 N .
Explanation:
mass of satellite, m = 6463 kg
height of orbit, h = 4.82 x 10^5 m
period, T = 2 h
radius of planet, R = 4.29 x 10^6 m
Let the acceleration due to gravity at the planet is g.
[tex]T = 2\pi\sqrt\frac{(R+h)^3}{gR^2}\\\\2\times 3600 = 2\times3.14\sqrt\frac{(4.29+0.482)^3\times10^{18}}{g\times 4.29\times 4.29\times 10^{12} }\\\\24.2 g =108.67\\\\g = 4.5 m/s^2[/tex]
The weight of the satellite at the surface of the planet is
W = m g = 6463 x 4.5 = 29083.5 N