When the rock is suspended in the air, the net force on it is
∑ F₁ = T₁ - m₁g = 0
where T₁ is the magnitude of tension in the string and m₁g is the rock's weight. So
T₁ = m₁g = 37.8 N
When immersed in water, the tension reduces to T₂ = 32.0 N. The net force on the rock is then
∑ F₂ = T₂ + B₂ - m₁g = 0
where B₂ is the magnitude of the buoyant force. Then
B₂ = m₁g - T₂ = 37.8 N - 32.0 N = 5.8 N
B₂ is also the weight of the water that was displaced by submerging the rock. Let m₂ be the mass of the displaced water; then
5.8 N = m₂g ==> m₂ ≈ 0.592 kg
If one takes the density of water to be 1.00 g/cm³ = 1.00 × 10³ kg/m³, then the volume of water V that was displaced was
1.00 × 10³ kg/m³ = m₂/V ==> V ≈ 0.000592 m³ = 592 cm³
and this is also the volume of the rock.
When immersed in the unknown liquid, the tension reduces further to T₃ = 20.2 N, and so the net force on the rock is
∑ F₃ = T₃ + B₃ - m₁g = 0
which means the buoyant force is
B₃ = m₁g - T₃ = 37.8 N - 20.2 N = 17.6 N
The mass m₃ of the liquid displaced is then
17.6 N = m₃g ==> m₃ ≈ 1.80 kg
Then the density ρ of the unknown liquid is
ρ = m₃/V ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³
A string has its 4th harmonic at 31.5 Hz. What is the frequency of its third harmonic?
Answer:
The answer would be 7.5 Hz.
What is hydroelectric power ?
Answer quickly..!
Answer:
It's electricity produced from hydropower. It's also a form of energy that controls the power of water motion.
Explanation:
One pro about hydroelectric power is that it's renewable energy. But one con about hydroelectric power is that it can impact the environment in a negative way.
When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pulls on the cell phone from you, the Earth, and the Sun. Rank the gravitational forces on the phone from largest to smallest. Assume the Sun is roughly 109 times further away from the phone than you are, and 1028 times more massive than you. Rank the following choices in order from largest gravitational pull on the phone to smallest. To rank items as equivalent, overlap them.
a. Pull phone from you
b. Pull on phone from earth
c. Pull on phone from sun
Answer:
The answer is "Option b, c, and a".
Explanation:
Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.
We now understand the effects of gravity:
[tex]F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}[/tex]
The force of the sun is, therefore, [tex]10^{10}[/tex] times greater and the proper sequence, therefore, option steps are:
b. Pull-on phone from earth
c. Pull-on phone from sun
a. Pull phone from you
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How far horizontally will it travel in 2 seconds?
A. 30 m
B. 90 m
C. 45 m
D. 60 m
Answer:
It will travel Vx * t = 30 m/s * 2 s = 60 m
what do we mean by thrust?
Answer:
the answer is push example: she thrust her hand into her pocket
An ink-jet printer steers charged ink drops vertically. Each drop of ink has a mass of 10-11 kg, and a charge due to 500,000 extra electrons. It goes through two electrodes that gives a vertical acceleration of 104 m/s2. The deflecting electric field is _____ MV/m.
Answer:
E = 1.25 MV / m
Explanation:
For this exercise let's use Newton's second law
F = m a
where the force is electric
F = q E
we substitute
q E = m a
E = m a / q
indicate there are 500,000 excess electrons
q = 500000 e
q = 500000 1.6 10⁻¹⁹
q = 8 10⁻¹⁴ C
the mass is m = 10⁻¹¹ kg and the acceleration a = 10⁴ m / s²
let's calculate
E = 10⁻¹¹ 10⁴ / 8 10⁻¹⁴
E = 0.125 10⁷ V / m = 1.25 10⁶ V / m
E = 1.25 MV / m
You are asked to build an LC circuit that oscillates at 13 kHz. In addition, you are told that the maximum current in the circuit can be 0.14 A and the maximum energy stored in the capacitor must be 1.4x10-5 J. What value of inductor and capacitor should you choose
Answer:
a)[tex]L=0.00142H[/tex]
b) [tex]C=2.65*10^{-12}[/tex]
Explanation:
From the question we are told that:
Frequency[tex]F=13kHz[/tex]
Current [tex]I=0.14A[/tex]
Capacitor[tex]C_e=1.4*10^{-5}J[/tex]
Generally the equation for Energy in the inductor is mathematically given by
Where L is now subject
[tex]L=\frac{2C_e}{I^2}[/tex]
[tex]L=\frac{2*1.4*10^{-5}}{(0.14)^2}[/tex]
[tex]L=0.00142H[/tex]
Generally the equation for Value of Capacitor is mathematically given by
[tex]C=\frac{1}{(2 \pi f)^2} L[/tex]
[tex]C=\frac{1}{(2 3.142 13*10^3Hz)^2} *0.00142[/tex]
[tex]C=2.65*10^{-12}[/tex]
Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. Part A When she wears 2.0 D reading glasses, what is her near point
Answer:
The right answer is "0.273 m".
Explanation:
Given:
Power (P),
[tex]\frac{1}{f} = 2D[/tex]
Near point,
u = 0.6 m
As we know,
⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]
By substituting the values, we get
⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]
[tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]
[tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]
[tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]
By applying cross-multiplication, we get
[tex]0.6=2.2 \ v[/tex]
[tex]v = \frac{0.6}{2.2}[/tex]
[tex]S_{near} = 0.273 \ m[/tex]
A wire long and with mass is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude . What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards
The question is incomplete. The complete question is :
A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.
Solution :
Given :
Length of the wire, L = 0.6 m
Mass of the wire length, m = 11 g
= [tex]11 \times 10^{-3}[/tex] kg
Magnetic field , B = 0.4 T
Know we know that :
ILB = mg
or [tex]$I=\frac{mg}{BL}$[/tex]
[tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]
[tex]I=0.44963\ A[/tex]
[tex]I = 449.63 \ mA[/tex]
A large metal sphere has three times the diameter of a smaller sphere and carries three times the charge. Both spheres are isolated, so their surface charge densities are uniform. Compare (a) the potentials (relative to infinity) and (b) the electric field strengths at their surfaces.
Answer:
A. Equals to that of the smaller sphere
B. 3 times less than that of the smaller sphere
Explanation:
(a) Equals to that of the smaller sphere
The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R has the same potential
b) 3 times less than that of the smaller sphere
However, the electric field at the surface of the smaller sphere is ?=? 0 = kQ=R2 , so tripling Q and R reduces the surface field by a factor of 1/3
Express 6revolutions to radians
Answer:
About 37.70 radians.
Explanation:
1 revolution = 2[tex]\pi[/tex] radians
∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)
6 revolutions = 37.6991 or ≈ 37.70 radians
If 56.5 m3 of a gas are collected at a pressure of 455 mm Hg, what volume will the gas occupy if the pressure is changed to 632 mm Hg? *
Assuming ideal conditions, Boyle's law says that
P₁ V₁ = P₂ V₂
where P₁ and V₁ are the initial pressure and temperature, respectively, and P₂ and V₂ are the final pressure and temperature.
So you have
(455 mm Hg) (56.5 m³) = (632 mm Hg) V₂
==> V₂ = (455 mm Hg) (56.5 m³) / (632 mm Hg) ≈ 40.7 m³
Give reason why think before you use a simple cell ?
The disadvantages of simple cell are: It is not rechargeable. The battery needs to be disposed of after all the power has been used up. It can't produce electricity anymore. That is why, why think before you use a simple cell.
What are the benefits and drawbacks of simple cell?A battery designed to be used only once is called a simple cell. Small gadgets used in the house are frequently powered by simple cells.
The benefits of a simple cell include:
A simple cell can be used to power small electronic devices because of its modest size. (Games, lightsabers, radios on the go, cameras, hearing aids)Simple cell electrolyte is not very detrimental to the environment.Simple cells are reasonably priced.Among the drawbacks of a simple cell are:
The biggest drawback of a simple cell is that once it runs out of electricity, it cannot be replenished.Learn more about cell here:
https://brainly.com/question/30046049
#SPJ2
A point charge of +35 nC is above a point charge of –35 nC on a vertical line. The distance between the charges is 4.0 mm. What are the magnitude and direction of the dipole moment ?
Answer:
Magnitude = 140 x 10⁻¹² Cm
Direction = upwards
Explanation:
A pair of two equal and opposite point charges forms an electric dipole.
The magnitude of the moment of such dipole is the product of the magnitude of any of the charges (since the charges are the same in magnitude) and the distance of separation between them. i.e
p = q x d ----------(i)
Where;
p = dipole moment
q = magnitude of any of the charges
d = distance between the charges.
The direction of the dipole moment is from the negative charge to the positive charge.
(a) From the question, the charges are +35 nC and -35 nC, and the distance between them is 4.00mm.
This implies that;
q = 35 nC = 35 x 10⁻⁹C
d = 4.00mm = 4.0 x 10⁻³ m
Substitute the values of q and d into equation (i) to give;
p = 35 x 10⁻⁹C x 4.00 x 10⁻³ m
p = (35 x 4.0) x (10⁻⁹ x 10⁻³) C m
p = 140 x 10⁻¹² Cm
The magnitude of the dipole moment is 140 x 10⁻¹² Cm
(b) From the question, the +35nC charge is above the -35nC charge on a vertical line as shown below;
o +35nC
|
|
|
|
|
|
o -35nC
Since the direction should point from the negative charge to the positive charge, this means that the direction of the dipole moment of the two charges is upwards (due North).
o +35nC
↑
|
|
|
|
|
|
o -35nC
3. Some guitarists like the feel of a set of strings that all have the same tension. For such a guitar, the G string (196 Hz) has a mass density of 0.31 g/m. What is the mass density of the A string (110 Hz)
Answer:
0.98 g/m
Explanation:
Note: Since Tension and frequency are constant,
Applying,
F₁²M₁ = F₂²M₂............... Equation 1
Where F₁ = Frequency of the G string, F₂ = Frequency of the A string, M₁ = mass density of the G string, M₂ = mass density of the A string.
make M₂ the subject of the equation
M₂ = F₁²M₁/F₂²............... Equation 2
From the question,
Given: F₁ = 196 Hz, M₁ = 0.31 g/m, F₂ = 110 Hz
Substitute these values into equation 2
M₂ = 196²(0.31)/110²
M₂ = 0.98 g/m
A lead ball is dropped into a lake from a diving board 5.20 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 s after it is released. How deep is the lake?
Answer:
35.047 m
Explanation:
The time it takes the lead ball to reach the surface of the water is
s = ut+gt²/2............. Equation 1
Where t = time it takes the lead ball to reach the surface of water, u = initial velocity of the lead ball, g = acceleration due to gravity, s = heigth.
From the question,
Given: s = 5.20 m, u = 0 m/s (dropped from a height)
Constant: g = 9.8 m/s²
5.2 = 0+9.8t²/2
t² = (5.2×2)/9.8
t² = 10.4/9.8
t² = 1.06
t = √(1.06)
t = 1.03 s
Hence, time taken for the lead ball to reach the bottom of the lake is
t' = 4.5-1.03
t' = 3.47 seconds
v² = u²+2gs............... Equation 2
Where v = final velocity of the lead ball
Substitute into equation 2
v² = 0+2(9.8)(5.2)
v² = 101.92
v = √(101.92)
v = 10.1 m/s
Therefore, depth of the lake is
D = vt'
D = 10.1(3.47)
D = 35.047 m
An airplane propeller is 2.16 m in length (from tip to tip) with mass 100 kg and is rotating at 2900 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. What is its rotational kinetic energy? Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?
a) The rotational kinetic energy of the airplane propeller is 1792152.287 joules.
b) The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute.
How to determine the angular speed of a airplane propeller
Let consider the airplane propeller a rigid body, the rotational kinetic energy of the propeller (K), in joules, is described by the following formula:
K = 0.5 · I · ω² (1)
Where:
I - Moment of inertia of the airplane propeller, in kilogram-square meters.ω - Angular speed, in radians per secondIn addition, the moment of inertia of a slender rod rotating around its center is:
I = 0.0833 · M · L² (2)
Where:
M - Mass of the propeller, in kilogramsL - Length of the propeller, in metersa) If we know that M = 100 kg, L = 2.16 m and ω = 303.687 rad/s, then the rotational kinetic energy of the propeller is:
K = 0.5 · [0.0833 · (100 kg) · (2.16 m)²] · (303.687 rad/s)²
K = 1792152.287 J
The rotational kinetic energy of the airplane propeller is 1792152.287 joules. [tex]\blacksquare[/tex]
b) By (1) and (2) we know that the mass of the propeller is inversely proportional to the square of the angular speed. Therefore, we have the following relationship:
[tex]M_{o}\cdot \omega_{o}^{2} = M_{f}\cdot \omega_{f}^{2}[/tex]
[tex]\omega_{f} = \sqrt{\frac{M_{o}}{M_{f}} }\cdot \omega_{o}[/tex] (3)
If we know that [tex]\omega_{o} = 2900\,\frac{rev}{min}[/tex], [tex]M_{o} = 100\,kg[/tex] and [tex]M_{f} = 75\,kg[/tex], then the angular speed of the airplane propeller is:
[tex]\omega_{f} = \left(2900\,\frac{rev}{min} \right)\cdot \sqrt{\frac{100\,kg}{75\,kg} }[/tex]
[tex]\omega_{f} \approx 3348.631\, \frac{rev}{min}[/tex]
The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute. [tex]\blacksquare[/tex]
To learn more on rotational kinetic energy, we kindly invite to check this verified question: https://brainly.com/question/20261989
What is an internal resistance?
Explanation:
some thing inside a resistor
01.04 Law of Conservation of Energy
science question
Answer:
law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.
how to make an uncharged particle positively charged
Answer:
If a neutral atom gains electrons, then it will become negatively charged. If a neutral atom loses electrons, then it become positively charged.
PLEASE ANSWER IF YOU CAN AND NOT FOR THE SAKE OF GAINING POINTS!
Để sử dụng nguồn điện xoay chiều 220V/50Hz thắp sáng bóng đèn 12V/3W, ta chọn điện trở giảm áp có giá trị:
Explanation:
Hi Linda,
How's it going?
Sorry I haven't been in touch for such a long time but I've had exams so I've been studying every free minute. Anyway, I'd love to hear all your news and I'm hoping we can get together soon to catch up. We just moved to a bigger flat so maybe you can come and visit one weekend?
How's the new job?
Looking forward to hearing from you!
Helga
A wire, 0.60 m in length, is carrying a current of 2.0 A and is placed at a certain angle with respect to the magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field
Answer:
[tex]\theta=30 \textdegree[/tex]
Explanation:
From the question we are told that:
Current [tex]I=2.0A[/tex]
Length [tex]L=0.60m[/tex]
Magnetic field [tex]B=0.30T[/tex]
Force [tex]F=0.18N[/tex]
Generally the equation for Force is mathematically given by
[tex]F = BIL sin\theta[/tex]
[tex]sin\theta=\frac{F}{BIL}[/tex]
[tex]\theta=sin^{-1}\frac{0.18}{0.3*2*0.6}[/tex]
[tex]\theta=30 \textdegree[/tex]
just me or does brainly just want people to watch ads or pay for answers that are sometimes wrong dont get it
Answer:
i think bro is becuse they want to get money also paying to this app if i put paying right im latin but that is my opinion and also becuse they want we learn somethings
Explanation:
An observer on Earth sees spaceship 1 fly by at 0.80c. 6.0 years later, the observer on Earth sees spaceship 2 fly by at 0.80c, traveling in the same direction as the first. Both spaceships continue to travel with constant velocities. An observer in spaceship 1 observes Earth to pass spaceship 2 ____ years after Earth passed spaceship 1.
Answer:
[tex]t_2=10[/tex]
Explanation:
From the question we are told that:
Velocity of both spaceships [tex]V=0.8c[/tex]
Time [tex]t_1=6[/tex]
Generally the equation for time of spaceship 2 through earth is mathematically given by
[tex]t_2=\frac{t_1}{\sqrt{1-v^2}}[/tex]
[tex]t_2=\frac{6}{\sqrt{1-0.8^2}}[/tex]
[tex]t_2=10[/tex]
Two train 75 km apart approach each other on parallel tracks, each moving at
15km/h. A bird flies back and forth between the trains at 20km/h until the trains pass
each other. How far does the bird fly?
Answer:
The correct solution is "37.5 km".
Explanation:
Given:
Distance between the trains,
d = 75 km
Speed of each train,
= 15 km/h
The relative speed will be:
= [tex]15 + (-15)[/tex]
= [tex]30 \ km/h[/tex]
The speed of the bird,
V = 15 km/h
Now,
The time taken to meet will be:
[tex]t=\frac{Distance}{Relative \ speed}[/tex]
[tex]=\frac{75}{30}[/tex]
[tex]=2.5 \ h[/tex]
hence,
The distance travelled by the bird in 2.5 h will be:
⇒ [tex]D = V t[/tex]
[tex]=15\times 2.5[/tex]
[tex]=37.5 \ km[/tex]
A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?
Answer:
The mass of the second block=0.457 kg
Explanation:
We are given that
m1=1.5 kg
v1=1.3m/s
v2=4.3 m/s
V=2.0 m/s
We have to find the mass of the second block.
[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]
Let m2=m
Substitute the values
[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]
[tex]1.95+4.3m=3+2m[/tex]
[tex]4.3m-2m=3-1.95[/tex]
[tex]2.3m=1.05[/tex]
[tex]m=\frac{1.05}{2.3}[/tex]
[tex]m=0.457 kg[/tex]
Hence, the mass of the second block=0.457 kg
A strontium vapor laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an angle equal to the angle of incidence. However, fringes are also observed. If the wall is 1.2 m from the CD, and the second bright fringe is 0.803 m from the central maximum, what is the spacing (in m) of grooves on the CD
Answer:
[tex]d=1.29*10^{-6}m[/tex]
Explanation:
From the question we are told that:
Distance of wall from CD [tex]D=1.4[/tex]
Second bright fringe [tex]y_2= 0.803 m[/tex]
Let
Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m
Generally the equation for Interference is mathematically given by
[tex]y=frac{n*\lambda*D}{d}[/tex]
Where
[tex]d=\frac{n*\lambda*D}{y}[/tex]
[tex]d=\frac{2*431 *10^{-9}m*1.4}{0.803}[/tex]
[tex]d=1.29*10^{-6}m[/tex]
S.I unit for distance =______
(A) m (B)cm
(c) km (d) mm
Answer:
opinion a
Explanation:
the si units of distance is metre (m)
Answer:
A
Explanation:
There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12 V until it is loaded completely. Then the capacitor is disconnected from the voltage source. After this the two plates of the capacitor are driven apart until their distance is 5 mm. Now a positive test charge of 1 nC is brought from the positively charged plate to the negatively charges plate. How large is the kinetic energy of the test charge? The test charge of 1 nC can be regarded to be so small that it does not influence the electric field between the two plates of the capacitor.
Answer:
K = 2 10⁻⁸ J
Explanation:
Let's solve this exercise in parts, we start by finding the charge on each plate of the capacitor
C = Q / ΔV
C = ε₀ A / d
ε₀ A / d = Q / ΔV
Q = ε₀ A ΔV / d (1)
indicate the potential difference ΔV₁ = 12 V, the distance between the plates d₁ = 3 mm = 0.003 m,
as the power supply is disconnected and the capacitor is ideal the charge remains constant
in the second part we separate the plates at d₂ = 5 mm = 0.005 m, using equation 1
ΔV₂ = [tex]\frac{Q d_2}{ \epsilon_o A}[/tex]
we substitute the equation for Q
ΔV₂ = [tex]\frac{d_2}{\epsilon_o A} \ \frac{\epsilon_o A \Delta V }{d_1}[/tex]
ΔV₂ = [tex]\frac{d_2}{d_1} \ \Delta V_1[/tex]
in the third part we use the concepts of energy
starting point. Test charge near positive plate
Em₀ = U = q ΔV₂
final point. Test charge near negative plate
Em_f = K
energy is conserved
Em₀ = Em_f
q ΔV₂ = K
K = q ΔV₁ [tex]\frac{d_2}{d_1}[/tex]
we calculate
K = 1 10⁻⁹ 12 0.005/0.003
K = 2 10⁻⁸ J
