Answer:b
Explanation:I’m just trynna get more money dude
state Ohm`s law as applied in electricity
Answer:
Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists. When spelled out, it means voltage = current x resistance, or volts = amps x ohms, or V = A x Ω.
A truck has a bed that is 4.50 metres long,and 2.50 metres wide, and 1.50 metres high. What is maximum volume of sand can the truck carry within this dimensions?
Answer:
since the bed is a cuboid, we find the volume by L×W×H
4.50 × 2.50 × 1.50 = 16.875m³
HOPE THIS HELPS
Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (mb). Note how the air pressure changes as you move StationB towards the center of the high-pressure system.
A. What do you notice?
B. Why do you think this is called a high-pressure system?
Answer:
a) When moving towards a high pressure center the pressure values increase in the equipment
b) This area is called high prison since the weight of the atmosphere on top is maximum
Explanation:
A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.
When moving towards a high pressure center the pressure values increase in the equipment
B) This area is called high prison since the weight of the atmosphere on top is maximum
in general they are areas of good weather
A rod of mass M = 154 g and length L = 35 cm can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 11 g, moving with speed V = 9 m/s, strikes the rod at angle θ = 29° a distance D = L/3 from the end and sticks to the rod after the collision.Calculate the rotational kinetic energy, in joules, of the system after the collision.
Answer:
Explanation:
moment of inertia of the rod = 1/3 mL² , m is mass and L is length of rod.
1/3 x .154 x .35²
= .00629
moment of inertia of putty about the axis of rotation
= m d² , m is mass of putty and d is distance fro axis
= .011 x( .35 / 3 )²
= .00015
Total moment of inertia I = .00644 kgm²
angular momentum of putty about the axis of rotation
= mvRsinθ
m is mass , v is velocity , R is distance where it strikes the rod and θ is angle with the rod at which putty strikes
= .011 x 9 x .35 / 3 x sin 29
= .0056
Applying conservation of angular momentum
angular momentum of putty = angular momentum of system after of collision
.0056 = .00644 ω where ω is angular velocity of the rod after collision
ω = .87 rad /s .
Rotational energy
= 1/2 I ω²
I is total moment of inertia
= .5 x .00644 x .87²
= 2.44 x 10⁻³ J .
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal range of the ball from the base of the platform is 20.0m. What is the velocity of the ball just before it touches the ground
Answer:
v = 46.99 m/s
Explanation:
The velocity of the ball just before it touches the ground, is given by the following formula:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex] (1)
vx: horizontal component of the velocity
vy: vertical component of the velocity
The vertical component vy is calculated by using the following formula:
[tex]v_y^2=v_{oy}^2+2gh[/tex] (2)
vy: final velocity
voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)
g: gravitational acceleration = 9.8 m/s^2
h: height = 1.60m
You replace the values of the parameters in the equation (2):
[tex]v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}[/tex]
vx is calculated by using the information about the horizontal range of the ball:
[tex]R=v_o\sqrt{\frac{2h}{g}}[/tex] (3)
R: horizontal range of the ball = 20.0 m
You solve the previous equation for vo, the initial horizontal velocity:
[tex]v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}[/tex]
The horizontal component of the velocity is constant in the complete trajectory, hence, you have that
vx = vo = 35 m/s
Finally, you replace the values of vx and vy in the equation (1):
[tex]v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}[/tex]
The velocity of the ball just before it touches the ground is 46.99 m/s
A 2500 kg truck moving at 10.00 m/s strikes a car waiting at the light. Assume there is no friction on the road. The hook bumpers continue to move at 7.00 m/s. What is the mass of the struck car
A certain type of laser emits light that has a frequency of 4.6 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 3.1 x 10-11s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse
Answer:
14,260
Explanation:
Relevant data provided for computing the wavelengths are in one pulse is here below:-
The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]
Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]
The number of wavelengths are in one pulse is shown below:-
[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]
[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]
= 14,260
Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.
A force of 640 newtons stretches a spring 4 meters. A mass of 40 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion.
a research submarine what is the maximum depth it can go
Answer: 36, 200 feet deep according to information on google
Explanation:
A small submarine, the bathyscape Trieste, made it to 10,916 meters (35,813 feet) below sea level in the deepest point in the ocean, the Challenger Deep in the Marianas Trench, a few hundred miles east of the Philippines. This part of the ocean is 11,034 m (36,200 ft) deep, so it seems that a submarine can make it as deep as it's theoretically possible to go
Astronauts are testing the gravity on a new planet. A rock is dropped between two photogates that are 0.5 meters apart. The first photogate reads a velocity of 1.2 m/s and the the second photogate reads a velocity of 4.3 m/s . What is the acceleration of gravity on this new planet?
Answer:
a = 17 m / s²
Explanation:
For this experiment that the astronauts are carrying out, the value of the relation of gravity is cosecant, therefore we can use the kinematic relations
v² = v₀² + 2a y
They indicate the initial speed 1.2 m / s the final speed 4.3 m / s and the distance remembered 0.5 m
we clear
a = (v² - v₀²) / 2y
we calculate
a = (4.3² -1.2²) / 2 0.5
a = 17 m / s²
this is the gravity of the new planet
A car of mass 410 kg travels around a flat, circular race track of radius 83.4 m. The coefficient of static friction between the wheels and the track is 0.286. The acceleration of gravity is 9.8 m/s 2 . What is the maximum speed v that the car can go without flying off the track
Answer:
The maximum speed v that the car can go without flying off the track = 15.29 m/s
Explanation:
let us first lay out the information clearly:
mass of car (m) = 410 kg
radius of race track (r) = 83.4 m
coefficient of friction (μ) = 0.286
acceleration due to gravity (g) = 9.8 m/s²
maximum speed = v m/s
For a body in a constant circular motion, the centripetal for (F) acting on the body is given by:
F = mass × ω
where:
F = maximum centripetal force = mass × μ × g
ω = angular acceleration = [tex]\frac{(velocity)^2}{radius}[/tex]
∴ F = mass × ω
m × μ × g = m × [tex]\frac{v^{2} }{r}[/tex]
410 × 0.286 × 9.8 = 410 × [tex]\frac{v^{2} }{83.4}[/tex]
since 410 is on both sides, they will cancel out:
0.286 × 9.8 = [tex]\frac{v^{2} }{83.4}[/tex]
2.8028 = [tex]\frac{v^{2} }{83.4}[/tex]
now, we cross-multiply the equation
2.8028 × 83.4 = [tex]v^{2}[/tex]
[tex]v^{2}[/tex] = 233.754
∴ v = √(233.754)
v = 15.29 m/s
Therefore, the maximum speed v that the car can go without flying off the track = 15.29 m/s
Countries create quotas and tariffs to increase the volume of trade with their neighbors.
Oooooo, that statement is not true. Countries create quotas and tariffs to LIMIT the volume of trade with other countries, including their neighbors.
Answer:
False
Explanation:
I took the text :)
What is the power of a child that has
done work of 50J in 10 seconds.
(a)50W (b)20W (c)30W (d)5W
_____________________________
Solution,
Work=50 Joule
Time=10 seconds
Power=?
Now,
Power=Work/time
= 50/10
= 5 Watt.
So the right answer is 5 W
Hope it helps..
Good luck on your assignment
__________________________
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 28.0 m/s. Then the vehicle moves for 41.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.
(a) How long is the ride-sharing car in motion (in s)?
(b) What is the average velocity of the ride-sharing car for the motion described? (Enter the magnitude in m/s.)
Answer:
Explanation:
Time taken to accelerate to 28 m /s
= 28 / 2 = 14 s
a ) Total length of time in motion
= 14 + 41 + 5
= 60 s .
b )
Distance covered while accelerating
s = ut + 1/2 at²
= 0 + .5 x 2 x 14²
= 196 m .
Distance covered while moving in uniform motion
= 28 x 41
= 1148 m
distance covered while decelerating
v = u - at
0 = 28 - a x 5
a = 5.6 m / s²
v² = u² - 2 a s
0 = 28² - 2 x 5.6 x s
s = 28² / 2 x 5.6
= 70 m .
Total distance covered
= 196 + 1148 + 70
= 1414 m
total time taken = 60 s
average velocity
= 1414 / 60
= 23.56 m /s .
A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?
Answer:
(a) q = 2.357 x 10⁻⁵ C
(b) Φ = 2.66 x 10⁶ N.m²/C
Explanation:
Given;
diameter of the sphere, d = 1.1 m
radius of the sphere, r = 1.1 / 2 = 0.55 m
surface charge density, σ = 6.2 µC/m²
(a) Net charge on the sphere
q = 4πr²σ
where;
4πr² is surface area of the sphere
q is the net charge on the sphere
σ is the surface charge density
q = 4π(0.55)²(6.2 x 10⁻⁶)
q = 2.357 x 10⁻⁵ C
(b) the total electric flux leaving the surface of the sphere
Φ = q / ε
where;
Φ is the total electric flux leaving the surface of the sphere
ε is the permittivity of free space
Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)
Φ = 2.66 x 10⁶ N.m²/C
Light bulb 1 operates with a filament temperature of 3000 K, whereas light bulb 2 has a filament temperature of 2000 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.
Answer:
A₁/A₂ = 0.44
Explanation:
The emissive power of the bulb is given by the formula:
P = σεAT⁴
where,
P = Emissive Power
σ = Stefan-Boltzman constant
ε = Emissivity
A = Surface Area
T = Absolute Temperature of Surface
FOR BULB 1:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₁T₁⁴ ----------- equation 1
where,
A₁ = Surface Area of Bulb 1
T₁ = Temperature of Bulb 1 = 3000 k
FOR BULB 2:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₂T₂⁴ ----------- equation 2
where,
A₂ = Surface Area of Bulb 2
T₂ = Temperature of Bulb 1 = 2000 k
Dividing equation 1 by equation 2, we get:
P/P = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁(3000)²/A₂(2000)²
A₁/A₂ = (2000)²/(3000)²
A₁/A₂ = 0.44
the distance between 2 station is 5400 m find the time taken by a train to cover this distance, if the train travels with speed 60m/s
Answer:
I dont know bro
Explanation:
Ask an expert
Answer:
Time=90s
Explanation:
Speed=distance /time
[tex]60 = \frac{5400}{t} where \: t \: is \: time \\60t = 5400 \\ t = \frac{5400}{60} \\ t =90 \\ hope \: this \: helps..good \: luck [/tex]
a 15-nC point charge is at the center of a thin spherical shell of radius 10cm, carrying -22nC of charge distributed uniformly over its surface. find the magnitude and direction of the electric field (a) 2.2cm,(b)5.6cm,and (c)14 cm from the point charge.
Answer:
A) E = 278925.62 N/C with direction; radially out.
B) E = 43048.47 N/C with direction radially out.
C) E = -3214.29 N/C with direction radially in.
Explanation:
From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;
E = kQ/r²
where;
Q is the net charge within the distance r.
We are given the charge Q = 15-nC and
spherical shell of radius 10cm
A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C
While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²
E = 278925.62 N/C
This will be radially out ,since the net charge is positive.
B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C
While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²
E = 43048.47 N/C
This will be radially out ,since the net charge is positive.
C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;
Q = 15 nC - 22 nC
Q = -7 nC = -7 x 10^(-9) C
and;
E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²
E = -3214.29 N/C
This will be radially in, since the net charge is negative. You can indicate this with a negative answer.
A) When The distance r is = 2.2 cm = 0.022 m is between the surface and also the point charge, also that so only the point charge lies within this distance and also Q = 15 NC = 15 x 10^(-9) C
Then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²Then E = 278925.62 N/CThen This will be radially out since the net charge is positive.
B) When The distance r = 5.6 cm = 0.056 m is between the surface and also the point charge, so only the point charge lies within this distance and also Q = 15 nC = 15 x 10^(-9) C
then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²Then E = 43048.47 N/CAfter that This will be radially out since the net charge is positive.
C) Then when The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;
Then Q = 15 nC - 22 nCAfter that Q = -7 nC = -7 x 10^(-9) CWhen E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²Then E = -3214.29 N/C Thus, This will be radially in, since the net charge is negative.Find out more information about magnitude here:
https://brainly.com/question/13502329
A certain metal with work function of Φ = 1.7 eV is illuminated in vacuum by 1.4 x 10-6 W of light with a wavelength of λ = 600 nm. 1)How many photons per second, N, are incident on the metal? N = photons per second Submit 2)What is KEmax, the maximum kinetic energy of the electron that is emitted from the metal? KEmax = eV
Answer:
1) n = 4.47*10^12 photons
2) K = 0.25 eV
Explanation:
This is a problem about the photoelectric effect.
1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:
[tex]P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}[/tex] (1)
Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:
[tex]E_p=h\frac{c}{\lambda}[/tex] (2)
c: speed of light = 3*10^8 m/s
h: Planck's constant = 6.626*10^-34 Js
λ: wavelength = 600*10^-9 m
You replace the values of the parameters in the equation (2):
[tex]E_p=(6.626*10^{-34}Js)\frac{3*10^8m/s}{600*10^{-9}m}=3.131*10^{-19}J[/tex]
Next, you calculate the quotient between the power of the light (equation (1)) and the energy of the photon:
[tex]n=\frac{P}{E_p}=\frac{1.4*10^{-6}J/s}{3.131*10^{-19}J}=4.47*10^{12}photons[/tex]
The number of photons is 4.47*10^12
2) The kinetic energy of the electrons emitted by the metal is given by the following formula:
[tex]K=E_p-\Phi[/tex] (3)
Ep: energy of the photons
Φ: work function of the metal = 1.7 eV
You first convert the energy of the photons to eV:
[tex]E_p=3.131*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.954eV[/tex]
You replace in the equation (3):
[tex]K=1.95eV-1.7eV=0.25eV[/tex]
The kinetic energy of the electrons emitted by the metal is 0.25 eV
(1). The Number of photons per second is,[tex]4.23*10^{12}[/tex]
(2). The maximum kinetic energy of the electron is 0.37eV.
(1). The power of light is given as,
[tex]P=1.4*10^{-6}W[/tex]
Energy is given as,
[tex]E=\frac{hc}{\lambda} =\frac{6.626*10^{-34}*3*10^{8} }{600*10^{-9} } \\\\E=3.313*10^{-19} Joule\\\\E=\frac{3.313*10^{-19}}{1.6*10^{-19} }=2.07eV[/tex]
Number of photons per second is,
[tex]N=\frac{P}{E}=\frac{1.4*10^{-6} }{3.313*10^{-19} } =4.23*10^{12}[/tex]
(2). the maximum kinetic energy of the electron is,
[tex]K.E=E-\phi[/tex]
Where [tex]\phi[/tex] is work function.
[tex]K.E=2.07-1.7=0.37eV[/tex]
Learn more:
https://brainly.com/question/12337396
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes the point x=385 with a speed of 45.0 m/s at t=20.0 s. Find the average velocity and the average acceleration between t=3.0 s and 20.0 s.
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
Determined to test the law of gravity for himself, a student walks off a skyscraper 180 m high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student.
a) Superman leaves the roof with an initial velocity that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. What must the value of the initial velocity be so that Superman catches the student just before they reach the ground?
b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a).
c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?
Answer:
a) v₀ = - 164.62 m / s , c) y = 122.5 m
Explanation:
We can solve this exercise using the free fall kinematic relations.
We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m
For the boy
y = y₀ + v₀ t - ½ g t²
with free fall its initial speed is zero
y = ½ g t2
For superman
y = y₀ + v₀ (t-5) - ½ g (t-5)²
how superman grabs the lot just before hitting the ground
we look for the time it takes the boy down
t = √ (2 y₀ / g)
t = √ (2 180 / 9,8)
t = 6.06 s
in the equation for superman, we clear the volume and calculate
v₀ (t-5) = -y₀ + ½ g (t-5)²
v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²
v₀ 1.06 = -174.49
v₀ = - 174.49 / 1.06
v₀ = - 164.62 m / s
the negative sign indicates that the initial speed is down
b) to graph the position of the two we use the table
t (s) Y_boy (m) Y_superman (m)
0 180 180
1 175.1 180
5 57.5 180
6 3.6 10.18
see attachment for the two curves
c) calculate the height that falls a lot in the 5 seconds (t = 5)
y = -1/2 g t²
y = ½ 9.8 5²
y = 122.5 m
for this height superman has not yet left the skyscraper, so the boy hits the ground
a) Write the names of the materials used in the ohm law according to the Figure 1?
b) If the voltage of a circuit is 12 V and the resistance is 40 , What is the generated power?
Answer:
a. i. conducting wire
ii high-pass and low-pass filters
iii. Cobra-4 Xpert-link
iii. voltage source
b. Power generated is 3.6 W.
Explanation:
Ohm's law state that the current passing through a metallic conductor, e.g wire is directly proportional to the potential difference across its ends, provided temperature is constant.
i.e V = IR
i. conducting wire
ii high-pass and low-pass filters
iii. Cobra-4 Xpert-link
iii. voltage source
b. Given that; V = 12 V and R = 40 Ohm's.
P = IV
From Ohm's law, I = [tex]\frac{V}{R}[/tex]
So that;
P = [tex]\frac{V^{2} }{R}[/tex]
= [tex]\frac{12^{2} }{40}[/tex]
= [tex]\frac{144}{40}[/tex]
= 3.6 W
The power is 3.6 W.
Work out the velocity v at the end of a rollercoaster ride (0). (rearrange the equation for KE to make velocity v the subject)
KE=1/2mv^2
Explanation:
If the kinetic energy of an object is given and we need to find its velocity of motion, then we can find it by using the formula of kinetic energy as :
[tex]K=\dfrac{1}{2}mv^2[/tex]
m is mass of the object
We can rearrange the above equation such that,
[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]
Hence, this is the velocity at the end of a rollercoaster ride.
A sphere of diameter 6.0cm is moulded into a thin uniform wire of diameter 0.2mm. Calculate the length of the wire in metres (Take π = 22/7) *
Answer:
2025m
Explanation:
Since all materials of the sphere is made to a cylindrical wire, it implies the volume of the sphere material is same as that of the cylinder. This is expressed mathematically thus.
Volume of Sphere= volume of cylinder
4/3 ×π×R^3= π× r2× L
4/3 ×R^3= r^2×L
Hence
L = 3/4 × R^3/ r^2
But R = 6.0/2 = 3.0cm{ Diameter is twice raduis}
r= 0.2/2 = 0.1mm=>0.01cm{ Diameter is twice raduis and unit converted by dividing by 10 since 10mm = 1cm}
Substituting R and r into the expression for L, we have :
L = 3/4 × 3^3/ 0.01^2= 0.75 ×27/0.0001 = 202500cm
202500/100= 2025m{ we divide by 100 because 100cm=1m}
Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?
Answer:
1.25kg
Explanation:
Simply multiply volume and density together
1. (a) The battery on your car has a rating stated in ampere-minutes which permits you to
estimate the length of time a fully charged battery could deliver any particular current
before discharge. Approximately how much energy is stored by a 50 ampere-minute 12
volt battery?
Answer:
Energy Stored = 36000 J = 36 KJ
Explanation:
The power of a battery is given by the formula:
P = IV
where,
P = Power delivered by the battery
I = Current Supplied to the battery
V = Potential Difference between terminals of battery = 12 volt
Now, we multiply both sides by the time period (t):
Pt = VIt
where,
Pt = (Power)(Time) = Energy Stored = E = ?
It = Battery Current Rating = 50 A.min
Converting this to A.sec;
It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec
Therefore,
E = (12 volt)(3000 A.sec)
E = 36000 J = 36 KJ
A hydraulic lift is made by sealing an ideal fluid inside a container with an input piston of cross-sectional area 0.004 m2 , and an output piston of cross-sectional area 1.2 m2 . The pistons can slide up or down without friction while keeping the fluid sealed inside. What is the maximum weight that can be lifted when a force of 60 N is applied to the input piston
Answer:
Maximum weight that can be lifted = 18,000 N
Explanation:
Given:
Cross-sectional area of input (A1) = 0.004 m²
Cross-sectional area of the output (A2) = 1.2 m ²
Force (F) = 60 N
Computation:
Pressure on input piston (P1) = F / A1
Assume,
Maximum weight lifted by piston = W
Pressure on output piston (P2) = W / A2
We, know that
P1 = P2
[F / A1] = [W / A2]
[60 / 0.004] = [W / 1.2]
150,00 = W / 1.2
Weight = 18,000 N
Maximum weight that can be lifted = 18,000 N
A bus travelling at a speed of 40 kmph reaches its destination in 8 minutes and 15 seconds. How far is the destination? a. 5.43 km b. 5.44 km c. 5.50 km d. 9.06 km
Answer:
c. 5.50 km
Explanation:
8 min * 1h/(60 min) = 8/60 = 2/15 h
15 sec* 1 min/60 sec = 1/4 min * 1h/(60 min) = 1/240 h
8 min 15 sec = (2/15+1/240)h
40 km/h *(2/15 +1/240)h =5.50 km
Answer: 5.50 km
Explanation:
what is the speed of light in quartz
Answer:
1.95 x 10^8 m/s.
Explanation:
Answer:
the answer is 1.95 x 10^8 m/s
Explanation:
Question 10
Air with a density of 1.20 kg/m3 flows through a 75.0 cm diameter pipe with a velocity of 2.00 m/s. What is the mass flow rate?
Answer:
75.0 cm
Explanation:
becouse i don,t no the right answer
