Answer:
[tex]u=34cm[/tex]
Explanation:
From the question we are told that:
Far point is [tex]V=34 cm[/tex]
Near point is [tex]u=17 cm[/tex]
Therefore
Focal Length
[tex]f=-34cm[/tex]
Generally the equation for the Lens is mathematically given by
[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]
[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]
[tex]u=34cm[/tex]
Convert the unit of 0.00023 kilograms into grams. (Answer in scientific notation)
Answer:
2.3 × [tex]10^{-1}[/tex]
Explanation:
1 kg = 1000 g.
0.00023 kg x 1000 g = 0.23 grams
Answer:
0.23×10⁴
Explanation:
kilogram to gram ÷ 1000
0.00023kg ÷ 1000
=0.23g
scientific notation=0.23×10⁴
What is the Ah rating of a battery that can provide 0.8 A for 76 h?
Answer:
6.08
Explanation:
Given that,
Current, I = 0.8 A
Time, t = 76 h
We need to find the Ah rating of a battery. It can be calculated by taking the product of current and time. So,
Ah = (0.8)(76)
= 6.08 Ah
So, the Ah rating of the battery is 6.08.
A 69.0-kg astronaut is floating in space, luckily he has his trusty 28.0-kg physics book. He throws his physics book and accelerates at 0.0130 m/s2 in the opposite direction. What is the magnitude of the acceleration of the physics book?
Answer:
0.032 [tex]m/s^2[/tex]
Explanation:
Given :
Weight of the astronaut = 69 kg
Weight of the physics book = 28 kg
Acceleration of the astronaut = 0.0130 [tex]m/s^2[/tex]
The force that is applied on the astronaut :
[tex]F=ma[/tex]
[tex]$=69 \times 0.013$[/tex]
= 0.897 N
Therefore, by Newton's 3rd law, we know that the force applied on the physics book is also F = 0.897 N
Therefore, the acceleration of the physics book is given by :
[tex]$a = \frac{\text{Force on physics book}}{\text{mass of physics book}}$[/tex]
[tex]$a = \frac{0.897}{28}$[/tex]
a = 0.032 [tex]m/s^2[/tex]
Hence, the acceleration of the physics book is 0.032 [tex]m/s^2[/tex].
Answer:
The acceleration of astronaut is 5.27 x 10^-3 m/s^2.
Explanation:
mass of astronaut, M = 69 kg
Mass of book, m = 28 kg
acceleration of book, a = 0.013 m/s^2
Let the acceleration of astronaut is A.
According to the Newton's third law, for every action there is an equal and opposite reaction.
So, the force acting on the book is same as the force acting on the astronaut but the direction is opposite to each other.
M A = m a
69 x A = 28 x 0.013
A = 5.27 x 10^-3 m/s^2
A cylinder that is 18 cm tall is filled with water. If a hole is made in the side of the cylinder, 5.0 cm below the top level. Assume that the cylinder is large enough so that the level of the water in the cylinder does not drop significantly. How far will the stream land from the base of the cylinder?
Answer:
The distance is 22.45 cm.
Explanation:
Height of cylinder, H = 14 cm
depth of hole, h = 5 cm
The distance of landing of stream from the base of cylinder is
[tex]R = 2\sqrt{H(H-h)}\\\\R = 2\sqrt{14(14-5)}\\\\R = 22.45 cm[/tex]
A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum has a mass of 2.00 kg, and the pivot is located 0.340 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.
Answer:
Therefore, the moment of inertia is:
[tex]I=0.37 \: kgm^{2} [/tex]
Explanation:
The period of an oscillation equation of a solid pendulum is given by:
[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex] (1)
Where:
I is the moment of inertiaM is the mass of the pendulumd is the distance from the center of mass to the pivotg is the gravityLet's solve the equation (1) for I
[tex]T=2\pi \sqrt{\frac{I}{Mgd}}[/tex]
[tex]I=Mgd(\frac{T}{2\pi})^{2}[/tex]
Before find I, we need to remember that
[tex]T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s[/tex]
Now, the moment of inertia will be:
[tex]I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}[/tex]
Therefore, the moment of inertia is:
[tex]I=0.37 \: kgm^{2} [/tex]
I hope it helps you!
Two friends, Al and Jo, have a combined mass of 194 kg. At the ice skating rink, they stand close together on skates, at rest and facing each other. Using their arms, they push on each other for 1 second and move off in opposite directions. Al moves off with a speed of 7.9 m/sec in one direction and Jo moves off with a speed of 6.7 m/sec in the other. You can assume friction is negligible.
What is Al's mass? 110.58 What is Jo's mass? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Al on Jo? If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Jo on Al?
Answer:
The mass of Al is 89.027 kilograms.
The mass of Jo is 104.973 kilograms.
The magnitude of the force of Jo on Al is 596.481 newtons.
Explanation:
Given the absence of external forces, this situation can be described will by Principle of Linear Momentum Conservation and Impact Theorem on each skater:
Al:
[tex]m_{1}\cdot (v_{1, f}-v_{1, o}) = -F \cdot \Delta t[/tex] (1)
Jo:
[tex]m_{2}\cdot (v_{2,f}-v_{2,o}) = F\cdot \Delta t[/tex] (2)
Total mass:
[tex]m_{1} + m_{2} = 194\,kg[/tex]
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the skaters, in kilograms.
[tex]v_{1,o}[/tex], [tex]v_{1,f}[/tex] - Initial and final velocities of Al, in meters per second.
[tex]v_{2,o}[/tex], [tex]v_{2,f}[/tex] - Initial and final velocities of Jo, in meters per second.
[tex]F[/tex] - Impact force between skaters, in newtons.
[tex]\Delta t[/tex] - Impact time, in seconds.
If we know that [tex]v_{1,o} = 0\,\frac{m}{s}[/tex], [tex]v_{1,f} = -7.9\,\frac{m}{s}[/tex], [tex]\Delta t = 1\,s[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex] and [tex]v_{2,f} = 6.7\,\frac{m}{s}[/tex], then the masses of the skaters are, respectively:
[tex](194-m_{2})\cdot (-7.9) = -F[/tex] (1b)
[tex]m_{2} \cdot 6.7 = F[/tex] (2b)
(2b) in (1b):
[tex](194-m_{2})\cdot (-7.9) = -m_{2}\cdot 6.7[/tex]
[tex]-1532.6 +7.9\cdot m_{2} = -6.7\cdot m_{2}[/tex]
[tex]14.6\cdot m_{2} = 1532.6[/tex]
[tex]m_{2} = 104.973\,kg[/tex]
[tex]m_{1} = 194\,kg - 104.973\,kg[/tex]
[tex]m_{1} = 89.027\,kg[/tex]
And the magnitude of the force is:
[tex]F = 6.7\cdot m_{2}[/tex]
[tex]F = 596.481\,N[/tex]
The mass of Al is 89.027 kilograms.
The mass of Jo is 104.973 kilograms.
The magnitude of the force of Jo on Al is 596.481 newtons.
A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s. (a) Determine the magnitude of the emf induced in the loop. (b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B
This question is incomplete, the complete question is;
A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s.
(a) Determine the magnitude of the emf induced in the loop.
(b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 1.5 T, if the induced emf is to be zero? (Give the magnitude of the rate of change of the area.) (m2/s)
Answer:
a) the magnitude of the emf induced in the loop is 0.003 V
b) dA/dt = 0.002 m²/s
Explanation:
Area of the loop wire A = 0.015 m²
magnitude of the field is increasing dB/dt = 0.20 T/s
a)
Determine the magnitude of the emf induced in the loop.
V = A( dB/dt )
we substitute
V = 0.015 m² × 0.20 T/s
V = 0.003 V
Therefore, the magnitude of the emf induced in the loop is 0.003 V
b) the induced emf is;
V = B( dA/dt ) + A( dB/dt )
given that; induced emf is 0, B = 1.5
so we substitute
0 = [ 1.5T × ( dA/dt ) ] + [ 0.015 m² × 0.20 T/s ]
-[ 1.5T × ( dA/dt )] = 0.003 m²T/s
dA/dt = -[ 0.003 m²T/s / 1.5T ]
dA/dt = -0.002 m²/s
the negative shows that the area is decreasing
hence, dA/dt = 0.002 m²/s
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 24 ft/s2. What is the distance (in ft) traveled before the car comes to a stop? (Round your answer to one decimal place.)
The car has initial speed 50 mi/h ≈ 73 ft/s, so it covers a distance x such that
0² - (73 ft/s)² = 2 (-24 ft/s²) x
==> x ≈ 111.0 ft
I NEEED HELP IN PHYSICS PLEASE!
Answer:
in which topic you need help
Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the sound intensity increase and then decrease periodically due to wave interference. How long does it take between successive instances of the sound intensity increasing
Answer:
[tex]T=1.54s[/tex]
Explanation:
From the question we are told that:
Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]
Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]
Therefore
Frequency of Motor 1 [tex]f_1=13.75[/tex]
Frequency of Motor 2 [tex]f_2= 13.1[/tex]
Generally the equation for Time Elapsed is mathematically given by
[tex]T=\frac{1}{df}[/tex]
Where
[tex]df=f_1-f_2[/tex]
[tex]df=13.75-13.1[/tex]
[tex]df=0.65Hz[/tex]
Therefore
[tex]T=\frac{1}{65}[/tex]
[tex]T=1.54s[/tex]
a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy when it's 2.00 m above the ground
Answer:
KE_2 = 3.48J
Explanation:
Conservation of Energy
E_1 = E_2
PE_1+KE_1 = PE_2+KE_2
m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²
(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2
4.10J+0.914J = 1.53J + KE_2
5.01J = 1.53J + KE_2
KE_2 = 3.48J
In a rolling race, two objects are released from the top of two identical ramps. They then roll without slipping to the bottom of the ramp. If the two objects are 2 hoops of the same radius but different masses, which reaches the bottom first?
a. The lighter one reaches the bottom first
b. The heavier one reaches the bottom first
c. We don’t have enough information
d. They reach the bottom at the same time
Answer:
b. The heavier one reaches the bottom first.
Answer:
B
Explanation:
The answer is B the heavier item has more g force pushing it making it roll faster reaching the bottom of the ramp first.
If you dive underwater, you notice an uncomfortable pressure on your eardrums due to the increased pressure. The human eardrum has an area of about 70 mm217 * 10-5 m22, and it can sustain a force of about 7 N without rupturing. If your body had no means of balancing the extra pressure (which, in reality, it does), what would be the maximum depth you could dive without rupturing your eardrum
Answer:
[tex]h=10m[/tex]
Explanation:
From the question we are told that:
Area [tex]a=70 x 10^{-6}[/tex]
Force [tex]F=7N[/tex]
Generally the equation for Pressure is mathematically given by
Pressure = Force/Area
[tex]P=\frac{F}{A}[/tex]
[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]
[tex]P= 1*10^{5} Pa[/tex]
Generally the equation for Pressure is also mathematically given by
[tex]P=hpg[/tex]
Therefore
[tex]h=\frac{P}{hg}[/tex]
[tex]h=\frac{10000}{1000*9.8}[/tex]
[tex]h=10m[/tex]
Each rarefraction on a longitudinal wave correspond to what point on a transverse wave?
Joule is a SI unit of power
Measuring cylinder is used to measure the volume of a liquid
Answer:
The SI unit of power is watt
A man throw a ball vertically up word with an intial speed 20m/s. What is the maximum height rich by the ball and how long does it take to return to the point it was trow
Answer:
u=20 m/s, T=4s
Explanation:
Given final velocity v= 0 m/s and displacement h= 20 m; acceleration due to gravity = 10 m/ s 2
From equation of motion
v2=u2+2gs−u2=−2(10).20u=20m/s
and time t can be determined by the formula
t=gv−u=−10−20=2s
total time = 2× time of ascend=2×2=4s
it is helpful for you
A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
Answer:
0
Explanation:
The speed of the ball when it reaches the floor is 0 because any object at rest or in uniform motion has no speed or velocity
In a double-slit experiment, the slit separation is 1.75 mm, and two coherent wavelengths of light, 425 nm and 510 nm, illuminate the slits. At what angle from the centerline on either side of the central maximum will a bright fringe from one pattern first coincide with a bright fringe from the other pattern
Answer:
the required angle is 0.0834879⁰
Explanation:
Given the data in the question;
slit separation; d = 1.75 mm = 1.75 × 10⁻³ m
wavelength λ₁ = 425 nm = 425 × 10⁻⁹ m
wavelength λ₂ 510 nm = 510 × 10⁻⁹ m
Now, we know that, the angle at which a particular bright fringe occurs on either side of the central bright fringe will be;
tanθ = [tex]y_m[/tex] / D = mλ/d
since they both coincides;
tanθ₁ = tanθ₂
m₁λ₁/d = m₂λ₂/d
multiply both sides by d
so,
m₁/m₂ = λ₂/λ₁
we substitute
m₁/m₂ = 510 nm / 425 nm
m₁/m₂ = 510 nm / 425 nm
divide through by 85
m₁/m₂ = 6 / 5
hence m₁ and m₂ are 6 and 5
so, from the previous formula
tanθ₂ = m₂λ₂/d
we substitute
tanθ₂ = [ 5 × ( 510 × 10⁻⁹ m ) ] / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 0.00145714
θ₂ = tan⁻¹( 0.00145714 )
θ₂ = 0.0834879⁰
Therefore, the required angle is 0.0834879⁰
The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?
Answer:
[tex]v_1=87.40m/s[/tex]
Explanation:
From the question we are told that:
Mass of arrow [tex]m=52g[/tex]
Mass of rock [tex]m_r=1.50kg[/tex]
Height [tex]h=0.47m[/tex]
Generally the equation for Velocity is mathematically given by
[tex]v = \sqrt{(2gh)}[/tex]
[tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]
[tex]v= 3.035m/s[/tex]
Generally the equation for conservation of momentum is mathematically given by
[tex]m_1v_1=m_2v_2[/tex]
[tex]0.052kg * v = 1.5 * 3.03m/s[/tex]
[tex]v_1=87.40m/s[/tex]
A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates are separated by a distance d = 2 mm and a voltage applied between the plates. The electric field strength within the plates is E = 4000 V/m. The energy stored in the capacitor is
Answer:
The energy stored is 1.4 x 10^-9 J.
Explanation:
Side of square, L = 10 cm = 0.1 m
Distance, d = 2 mm = 0.002 m
Electric field, E = 4000 V/m
The energy stored in the capacitor is
[tex]U = 0.5 C V^2[/tex]
The capacitance is given by
[tex]C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J[/tex]
The area around a charged object that can exert a force on other charged objects is an electric ___
These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor
Answer:
Following are the solution to the given question:
Explanation:
For charging plates that are connected in a similar manner:
Calculating the total charge:
[tex]\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C[/tex]
Calculating the common potential:
[tex]\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\[/tex]
Calculating the charge after redistribution:
[tex]When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2[/tex]
[tex]\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C[/tex]
A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.
Answer:
The resolving power remains same.
Explanation:
The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.
As the diameter is same but the focal length is doubled so the resolving power remains same.
After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving
This question is incomplete, the complete question is;
A bat flies towards a moth at 7.1 m/s while the moth is flying towards the bat at 4.4 m/s. The bat emits a sound wave of 51.7 kHz.
After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving
Answer:
the second moth is moving at 5.062 m/s
Explanation:
Given the data in the question;
Using doppler's effect
[tex]f_{moth[/tex] = f₀( [tex]v_{s[/tex] ± [tex]v_{observer[/tex] / [tex]v_{s[/tex] ± [tex]v_{source[/tex] )
f₁ = f₀( ([tex]v_{s[/tex] + v₂) / ( [tex]v_{s[/tex] - v₁ ) )
frequency reflected from the moth,
Now, moth is the source and the bat is the receiver
f₂ = f₁( ([tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) )
hence, f = f₀[ ( ( [tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) ) ( ( [tex]v_{s[/tex] + u₂ ) / ( [tex]v_{s[/tex] - u₁ ) )
we know that, the velocity of sound [tex]v_{s[/tex] = 343 m/s.
given that v₁ and v₂ { velocity of bat } = 7.1 m/s, f₀ = 51.7 kHz and f = 55.5 kHz.
we substitute
55.5 = 51.7[ ( ( 343 + 7.1 ) / ( 343 - 7.1 ) ) ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 51.7[ ( 350.1 / 335.9 ) ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 51.7[ 1.04227 ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 53.885359 ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 / 53.885359 = ( 343 + u ) / ( 343 - u )
1.02996 = ( 343 + u₂ ) / ( 343 - u )
( 343 + u₂ ) = 1.02996( 343 - u )
343 + u = 353.27628 - 1.02996u
u + 1.02996u = 353.27628 - 343
2.02996u = 10.27628
u = 10.27628 / 2.02996
u = 5.062 m/s
Therefore, the second moth is moving at 5.062 m/s
You have 150 W/m^2 hitting your roof each day. You can convert 13% of it into
usable energy, and you need 3.5 kW to run your house for a day. Show the MATH,
answer and units, to determine the size solar panel you will need to succeed.
Answer:
Energy = .13 W / m^2 energy of incident energy
N = 3500 Watts / day power needed
N = 3500 Watts (3600 * 24 sec) = .0405 Watts/sec
The problem must mean that one needs 3.5 Kw-days
3.5 Kw-days = 3500 watts * 86400 sec = 3.02E8 joules
150 J/sec-m^2 * .13 = 19.5 J / sec-m^2 usable energy
In one day 19.5 J/sec-m^2 = 1.68E6 J/m^2 usable energy received
Area = 3.028E8 J / 1.68E6 J/m2 = 180 m^2
One would need 180 m^2 of solar panels
That's quite a lot of energy
A 1100 watt microwave oven uses 1.1 kW while running so 3.5 kW for 24 hours seems to be quite a lot.
A redox reaction is always a single-displacement reaction, but a single-
displacement reaction isn't always a redox reaction.
A. True
B. False
SUBMIT
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
Which nucleus completes the following equation?
39 17 CI-> 0 -1 e+?
Answer:
[tex]_{18}^{39} } Ar[/tex]
Explanation:
The given equation shows the disintegration of an unstable isotope of chlorine to beta particle and Argon nucleus. The nucleus undergoes the emission of a beta particle to form a more stable nucleus of Argon.
[tex]_{17} ^{39} Cl[/tex] ⇒ [tex]_{-1}^{0} e[/tex] + [tex]_{18}^{39} } Ar[/tex]
Argon is a stable gas and is found in the group 8 on the periodic table of elements.
Answer:
Answer is below
Explanation:
39 18 Ar
Derive the explicit rule for the pattern
3, 0, -3, -6, -9,
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
