It depends on the liquid the fluid is being put in. If the liquid is more dense than the fluid then it will sink but if the fluid is denser than the liquid then it will float.
if the fluid becomes more dense as it is put in the liquid it can sink as long as it reaches a high spenough density to be denser than this liquid it is in.
You and your two younger siblings are playing on a seesaw. Your little brother, whose mass is 25 kg, sits 2.5 meters to the left of the pivot point, and your little sister, whose mass is 35 kg, sits 1 meter to the left of the pivot point. Because you weigh 70 kg, where should you sit to make the seesaw balance
Answer: 1.4 m
Explanation:
Given
Mass of siblings are 25kg and 35 kg who sits at 2.5 m and 1 m left to pivot.
Suppose the 70 kg mass sits at a distance x from the pivot towards right.
Torque applied by two siblings must cancel the torque of 70 kg mass i.e.
[tex]\Rightarrow 25\times 2.5+35\times1=70x\\\Rightarrow 62.5+35=70x\\\\\Rightarrow x=\dfrac{97.5}{70}\\\\\Rightarrow x=1.392\approx 1.4\ m[/tex]
An astronomer who loved reading the Guinness Book of World Records when she was a child becomes obsessed with quasars and wants desperately to find the most distant quasar ever (the one with the largest redshift.) Where should she be looking to have the best chance of finding such a quasar
Answer: there is simply no way that she can improve her chances of finding such a quasar; such discoveries are completely random
Explanation:
The speed of sound is approximately 340 m/s.what is the wavelength of a sound wave with a frequency of 1000Hz
Answer:
34cm
Explanation:
When the balloon sticks to the wall (assuming it sticks to the wall). It is
because the balloon is negatively charged and the wall carries an extra
positive charge.
A.false
B.true
What happens to matter when it reaches absolute zero?
A. Its particles speed up.
O B. Its particles have no kinetic energy.
O C. Its particles vibrate slightly.
D. Its particles gain kinetic energy.
Answer:
O C. Its particles vibrate slightly.
Explanation:
At absolute zero, atoms would occupy the lowest energy state.They move much less than at higher temperatures, but they still have small vibrations at absolute zero.
Answer:
The answer is B its particles have no kinetic energy
Explanation:
trust the big brain
5. Which of the following is velocity? *
2 points
A. 20 m/s
B. 40 m/s east
C. 40 m
D. 20 m
what happens to an object as its thermal energy increases
Answer:
As the average kinetic energy of its particles increases, so does the thermal energy of the object. Therefore, the thermal energy of the object increases with increasing temperature
Answer: It gets hotter
Explanation:
When thermal energy increases, the object's particles gain kinetic energy, meaning they move faster. Therefore the particles move farther apart, causing the object to heat up.
When the balloon sticks to the wall (assuming it sticks to the wall). It is
because the balloon is negatively charged and the wall carries an extra
positive charge.
1.false
2.true
Sarah and her bicycle have a total mass of 40 kg. Her speed at the top of a 10 m high and 100m long hill is 5 m/s. If the force of friction on her way down is 20 N, at what speed will she be going when she reaches the bottom
Answer:
She will be going at 11.01 m/s when she reaches the bottom.
Explanation:
We can find the speed at the bottom by equating the total work with the change in energy:
[tex] W = E_{f} - E_{i} [/tex] (1)
There is no energy conservation because there is a force of friction on her way down.
By entering [tex]W = -F_{\mu}*d[/tex], where [tex]F_{\mu}[/tex] is the force of friction (is negative because it is in the opposite direction of motion) and d is the displacement, into equation (1) we have:
[tex]-F_{\mu}*d = E_{f} - E_{i}[/tex]
In the initial state, we have kinetic and potential energy and in the final state, we have only kinetic energy.
[tex]-F_{\mu}*d = \frac{1}{2}mv_{f}^{2} - (\frac{1}{2}mv_{i}^{2} + mgh)[/tex]
Where:
m: is the total mass = 40 kg
[tex]v_{f}[/tex]: is the final speed =?
[tex]v_{i}[/tex]: is the intial speed = 5 m/s
g: is the gravity = 9.81 m/s²
h: is the height = 10 m
[tex] -20 N*100 m = \frac{1}{2}40 kg*v_{f}^{2} - \frac{1}{2}*40 kg*(5 m/s)^{2} - 40 kg*9.81 m/s^{2}*10 m [/tex]
By solving the above equation for [tex]v_{f}[/tex] we have:
[tex] v_{f} = 11.01 m/s [/tex]
Therefore, she will be going at 11.01 m/s when she reaches the bottom.
I hope it helps you!
When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 53.5 N and it is acting with an effective lever arm of 2.25 cm , what is the torque that the muscle produces on the wrist
Answer:
1.20Nm
Explanation:
Given data
Force= 53.5N
Perpendicular distance= 2.25cm= 2.25/100= 0.0225m
The expression for the torque is given as
Torque= Force* Perpendicular distance
Torque= 53.5*0.0225
Torque= 1.20Nm
Hence the toque produced is 1.20Nm
What does a Lewis structure diagram represent?
Answer:
Its answer is
The atomic symbol of an element surrounded by valence electrons.
Explanation:
HELPPP 40 POINTS!!!! Mr. Tedesco has two metal cubes, one made of
tin and the other made of silver. He heats the tin
cube to 80°C and places the silver one in the
freezer until it reaches 5°C. He places the cubes
in a beaker containing water at 20°C. The cubes
do not touch. Which best describes how heat
will flow in the system?
Answer: Heat energy is transferred from warmer objects to cooler objects.
Explanation:
A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon weigh 150n with what force must the child pull on the handle if the handle is parallel to the incline?
Answer:
F = 51.3°
Explanation:
The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:
[tex]F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\[/tex]
where,
W = Weight of Wagon = 150 N
θ = Angle of Inclinition = 20°
Therefore,
[tex]F = (150\ N)Sin\ 20^o[/tex]
F = 51.3°
A guy wire 1000 feet long is attached to the top of a tower. When pulled taut it makes a 42˚ angle with the ground. How tall is the tower? How far away from the base of the tower does the wire hit the ground?
Answer:
669.1 ft and 743.1 ft
Explanation:
Given data
Lenght of wire= = 1000ft
Angle = 42˚
Applying SOH CAH TOA
1. How tall is the tower?
SOH
sine ∅= opp/hyp
sin 42= opp/1000
0.6691= Opp/1000
0.6691*1000= Opp
Opp=669.1 ft
Applying SOH CAH TOA
2.How far away from the base of the tower does the wire hit the ground?
CAH
cos ∅= adj/hyp
cos 42= adj/1000
0.7431*1000=adj
adj=743.1 ft
Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates <0, 4.00 cm> and <0, -4.00 cm>. A particle with charge 10.0 nC is fixed at the origin. (a) Find the electric potential energy of the configuration of the three fixed charges. (b) A fourth particle, with a mass of 2.00 x 10-13 kg and charge of 40.0 nC, is released from rest at the point <3.00 cm, 0>. Find its speed after it has moved freely to a very large distance away.
Answer:
Explanation:
Potential energy of the system of charges
= 9 x 10⁹ x [ q₁q₂ / r₁₂ + q₂q₃ / r₂₃ + q₁q₃ / r₁₃ ]
here r₁₂ , r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.
r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.
q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C
Potential energy = 9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08 + -200x10⁻¹⁸ / .04 + 200 x 10¹⁸ / .04 ]
= 9 x 10⁹ x - 400 x 10⁻¹⁸ / .08
= 45 x 10⁻⁶ J .
b)
Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre
9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]
= 9 x 10⁹ x 10 x 10⁻⁹ / .03
= 3000 V .
potential energy of fourth particle = charge x potential
= 3000 x 40 x 10⁻⁹ = 12 x 10⁻⁵ J .
kinetic energy at infinity = 12 x 10⁻⁵ J
1/2 m v² = 12 x 10⁻⁵ J
.5 x 2 x 10⁻¹³ x v² = 12 x 10⁻⁵
v² = 12 x 10⁸
v = 3.46 x 10⁴ m/s
= 9 x 10⁹
An object falls from rest at a rate of 10 m/s / s . How fast was it traveling after 5 seconds of fall?
Answer:
[tex]v = u + at \\ v = 10 + 9.81 \times 5 \\ v = 59.05m {s}^{ - 1} [/tex]
What is the acceleration of a race car if it has a mass of 1200kg and is moving with an engine force of 400N
Answer:
0.34 m/s^2
Explanation:
force=mass × acceleration
400 =1200 × acceleration
acceleration=400/1200
=0.34 m/s^2
A sample of gas, initially with a volume of 1.0 L, undergoes a thermodynamic cycle. Find the work done by the gas on its environment during each stage of the cycle described below. (Enter your answers in J.)
(a) First, the gas expands from a volume of 1.0 L to 6.0 L at a constant pressure of 6.5 atm.
(b) Second, the gas is cooled at constant volume until the pressure falls to 1.0 atm.
(c) Third, the gas is compressed at a constant pressure of 1.0 atm from a volume of 6.0 L to 1.0 L. (Note: Be careful of signs.)
(d) Finally, the gas is heated until its pressure increases from 1.0 atm to 6.5 atm at a constant volume. (e) What is the net work done by the gas on its environment during the complete cycle described above?
Answer:
(a) W = 3293 J = 3.293 KJ
(b) W = 0 KJ
(c) W = -506.6 J = -0.507 KJ
(d) W = 0 KJ
Net Work = 2.786 KJ = 2786 J
Explanation:
(a)
The work done is given as:
[tex]W = P\Delta V\\W = P(V_2-V_1)[/tex]
where,
P = Constant Pressure = (6.5 atm)(101325 Pa/1 atm) = 6.59 x 10⁶ Pa
V₁ = initial volume = (1 L)(0.001 m³/1 L) = 0.001 m³
V₂ = final volume = (6 L)(0.001 m³/1 L) = 0.006 m³
Therefore,
[tex]W = (6.59\ x\ 10^6\ Pa)(0.006\ m^3-0.001\ m^3)[/tex]
W = 3293 J = 3.293 KJ
(b)
Since the volume is constant in this stage. Therefore,
ΔV = 0
As a result:
W = 0 KJ
(c)
The work done is given as:
[tex]W = P\Delta V\\W = P(V_2-V_1)[/tex]
where,
P = Constant Pressure = (1 atm)(101325 Pa/1 atm) = 1.01 x 10⁵ Pa
V₁ = initial volume = (6 L)(0.001 m³/1 L) = 0.006 m³
V₂ = final volume = (1 L)(0.001 m³/1 L) = 0.001 m³
Therefore,
[tex]W = (1.01\ x\ 10^5\ Pa)(0.001\ m^3-0.006\ m^3)[/tex]
W = -506.6 J = -0.507 KJ
negative sign show that the work is done on the gas by the environment.
(d)
Since the volume is constant in this stage. Therefore,
ΔV = 0
As a result:
W = 0 KJ
Net work will be the sum of all the works:
Net Work = 3.293 KJ + 0 KJ - 0.507 KJ + 0 KJ
Net Work = 2.786 KJ = 2786 J
Extra glucose is _____.
stored in roots, stems, and leaves
stored inside chloroplasts
changed back into carbon dioxide
released into the environment
Answer: stored in roots, stems, and leaves
Explanation:
As a result of the process of photosynthesis where carbon dioxide and water and used to synthesize glucose, the plant will find itself with extra glucose which is still needed but not at that point.
It will therefore store the glucose in roots, stems and leaves. It will however, convert them to starch first so that they are not affected by osmosis. Starch is not soluble in water which is why osmosis will not affect them and cause they to swell too much which would reduce the space the plant has.
2 weeks late :')
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the person above me is right tho
A 620 nm light falls on a photoelectric surface and electrons with the maximum kinetic energy of 0.14 eV are emitted. (a) Determine the work function (in eV). eV (b) Determine the cutoff frequency of the surface (in THz). THz (c) What is the stopping potential (in V) when the surface is illuminated with light of wavelength 420 nm
Answer:
(a) The work function is 1.86 eV.
(b) The cut off frequency is 450 THz.
(c) The stopping potential is 1.16 V.
Explanation:
incident wavelength = 620 nm
Kinetic energy, K = 0.14 eV
According to the photoelectric equation
E = W + KE
where, W is the work function, KE is the kinetic energy.
(a) Let the work function is W.
[tex]W = E - KE\\W = \frac{h c}{\lambda }- KE\\W =\frac{6.63\times 10^{-34}\times3\times 10^{8}}{620\times 10^{-9}\times 1.6\times 10^{-19}}-0.14\\\\W =1.86 eV[/tex]
(b) Let the cut off frequency is f.
W = h f
[tex]1.86\times 1.6\times 10^{-19} = 6.63\times 10^{-34}\times f\\f = 4.5\times 10^{14} Hz =450 THz[/tex]
(c) Let the stopping potential is V.
[tex]E = W + eV\\\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{420\times 10^{-9}\times 1.6\times 10^{-19}}=1.8 + eV\\\\V = 1.16 V[/tex]
A spring with spring constant 35 N/m is attached to the ceiling, and a 5.5-cm-diameter, 1.3 kg metal cylinder is attached to its lower end. The cylinder is held so that the spring is neither stretched nor compressed, then a tank of water is placed underneath with the surface of the water just touching the bottom of the cylinder. When released, the cylinder will oscillate a few times but, damped by the water, quickly reach an equilibrium position.
Requried:
When in equilibrium, what length of the cylinder is submerged?
Answer:
Explanation:
Let length of cylinder submerged be L .
Upward force due to restoration force of spring = k L
= 35 L .
Buoyant force by water in upward direction = Vρg
V is volume of water displaced , ρ is density of water , g is acceleration due to gravity .
V = π R²L , R is radius of cylinder
V = 3.14 x (2.75 x 10⁻²)² x L
= 23.75 x 10⁻⁴ L
Buoyant force by water in upward direction =23.75 x 10⁻⁴ L X 1000 X 9.8
= 23.275L N
Total upward force = 35 L + 23.275L
= 58.275 L
For equilibrium ,
Total upward force = weight of cylinder
58.275 L = 1.3 x 9.8
L = .2186 m
= 21.86 cm
Let's apply the torque equation to a circular coil of wire in a magnetic field. The coil has an average radius of 0.0435 m , has 40 turns, and lies in a horizontal plane. It carries a current of 7.25 A in a counterclockwise sense when viewed from above. The coil is in a uniform magnetic field directed toward the right, with magnitude 1.05 T . Find the magnetic moment and the torque on the coil. Which way does the coil tend to rotate
Answer:
[tex]M=1.72Am^2[/tex]
[tex]T=1.810N-m[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=0.0435[/tex]
Number of turns [tex]N=40turns[/tex]
Current [tex]I= 7.25 A[/tex]
Magnetic Field [tex]B=1.05T[/tex]
Generally the equation for magnetic moment M is mathematically given by
[tex]M=NIA[/tex]
Where
[tex]A= Surface Area[/tex]
[tex]A=\pi(r)^2[/tex]
[tex]A=3.142*(0.0435)^2[/tex]
[tex]A=5.95*10^{-3}[/tex]
Therefore the magnetic moment is given as
[tex]M=NIA[/tex]
[tex]M=40*7.25*5.95*10^{-3}[/tex]
[tex]M=1.72Am^2[/tex]
Generally the equation for Torque on the coil T is mathematically given by
[tex]T=M*B[/tex]
[tex]T=1.72*1.05[/tex]
[tex]T=1.810N-m[/tex]
Therefore torque on coil T is given a
[tex]T=1.810N-m[/tex]
what is motion? give 3 examples
Answer:
Motion definition: the action or process of moving or being moved.
Examples:
1. moving a hand
2. riding a bicycle
3. running
A car battery dies not so much because its voltage drops, but because chemical reactions increase its internal resistance. A good car battery typically has a terminal voltage of about 12.5 V and an internal resistance of about 0.020 Ω. When the battery dies its voltage can drop slightly, let's say it drops to about 10.1 V and the internal resistance increases to around 0.100 Ω.
Required:
a. How much current could the good battery alone drive through the starter motor?
b. How much current is the dead battery alone able to drive through the starter motor?
Answer:
(a) I = 625 A
(b) I = 101 A
Explanation:
The relationship between current, voltage and resistance can be given by Ohm's Law:
V = IR
where,
V = Voltage
I = Current
R = Resistance
(a)
Here,
V = 12.5 V
R = 0.02 Ω
Therefore,
[tex]12.5\ V = I(0.02\ \Omega)\\\\I = \frac{12.5\ V}{0.02\ \Omega}[/tex]
I = 625 A
(b)
Here,
V = 10.1 V
R = 0.1 Ω
Therefore,
[tex]10.1\ V = I(0.1\ \Omega)\\\\I = \frac{10.1\ V}{0.1\ \Omega}[/tex]
I = 101 A
A train moving at 30km/h accelerates to a speed of 50km/h in 30s. find the average acceleration in m/s²
Answer:
Explanation:
a= (v-u)/t where v = the final velocity at 50m/s, u= the inital velocity at 30m/s and t= the time taken at 30 seconds
a=(50-30)/30
a= 2/3m/s^-2 or 0.67m/s^-2 (m/s^-2 is the units for accelaration)
As part of a physics experiment, you carry a bathroom scale calibrated in newtons onto an elevator and stand on it. At rest, you check the scale and it reads 588 N. Then the elevator starts accelerating upward at 2.0 m/s2 and you check the reading again. Now what does the scale show
Answer: 708 N
Explanation:
Given
At rest, Elevator reads 588 N
When it starts moving upward at [tex]2\ m/s^2[/tex], apparent weight changes
i.e. weight can be given by
[tex]\Rightarrow W'=m(g+a)\\\Rightarrow W'=mg+mg\cdot \dfrac{a}{g}\\\\\Rightarrow W'=W\left(1+\dfrac{a}{g}\right)\\\\\Rightarrow W'=588\left(1+\dfrac{2}{9.8}\right)\\\\\Rightarrow W'=707.99\approx 708\ N[/tex]
The apparent weight is 708 N
A windmill captures 700W of wind power for 17h. The kinetic energy is converted to electric energy and stored in a battery, before being used by a 20W light bulb. During the process 68% of the energy is lost.
wind energy captured: ___W h
loss: ___W h
energy delivered to light bulb: ___W h
How long could the light bulb be used? ____hours
Answer:
this might help
Explanation:
Wind power or wind energy is the use of wind to provide mechanical power through wind turbines to turn electric generators for electrical power. Wind power is a popular sustainable, renewable source of power that has a much smaller impact on the environment compared to burning fossil fuels. Wind farms consist of many individual wind turbines, which are connected to the electric power transmission network. Onshore wind is an inexpensive source of electric power, competitive with or in many places cheaper than coal or gas plants. Onshore wind farms have a greater visual impact on the landscape than other power stations, as they need to be spread over more land and need to be built away from dense population. Offshore wind is steadier and stronger than on land and offshore farms have less visual impact, but construction and maintenance costs are significantly higher. Small onshore wind farms can feed some energy into the grid or provide power to isolated off-grid locations. The wind is an intermittent energy source, which cannot be dispatched on demand. Locally, it gives variable power, which is consistent from year to year but varies greatly over shorter time scales. Therefore, it must be used together with other power sources to give a reliable supply. Power-management techniques such as having dispatchable power sources (often gas-fired power plant or hydroelectric power), excess capacity, geographically distributed turbines, exporting and importing power to neighboring areas, grid storage, reducing demand when wind production is low, and curtailing occasional excess wind power, are used to overcome these problems. As the proportion of wind power in a region increases the grid may need to be upgraded. Weather forecasting permits the electric-power network to be readied for the predictable variations in production that occur. In 2019, wind supplied 1430 TWh of electricity, which was 5.3% of worldwide electrical generation, with the global installed wind power capacity reaching more than 651 G…
How many complete wavelengths are present in the sound wave shown? O 1 o 2 4 O 6 09.
Answer:
2
Explanation:
You have just landed your first job as a structural engineer and you have been asked to design a regional airport for small planes. Your manager informs you that the planes that will use this runaway must reach a speed before takeoff of at least 28.1 m/s, and can accelerate at 2 m/s2. What is the minimum length your runaway must have to facilitate a safe take off
Answer: [tex]197.40\ m[/tex]
Explanation:
Given
final velocity at takeoff [tex]v=28.1\ m/s[/tex]
Acceleration of the plane can be [tex]a=2\ m/s^2[/tex]
Initial velocity is zero for the plane i.e. [tex]u=0[/tex]
Using the equation of motion
[tex]\Rightarrow v^2-u^2=2as\quad [\text{s=displacement}]\\\text{Insert the values}\\\Rightarrow (28.1)^2-0=2\times 2\times s\\\\\Rightarrow s=\dfrac{789.61}{4}\\\\\Rightarrow s=197.40\ m[/tex]
Thu,s the minimum length must be [tex]197.40\ m[/tex]
One long wire lies along an x axis and carries a current of 39 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 4.4 m, 0), and carries a current of 47 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 0.80 m, 0)
Answer:
Explanation:
Magnetic field due to a long current carrying wire can be calculated as follows .
B = 10⁻⁷ x 2I / d where B is magnetic field , I is current .
The wire is along x -axis and the point is on y-axis at a distance of 0.8 m
Magnetic field at point of .8 m on y -axis
B₁ = 10⁻⁷ x 2 x 39 / 0.8
= 97.5 x 10⁻⁷ T .
Second wire is parallel to z-axis and passes through point on y-axis at a distance of 4.4 m . So the given point is at a distance of 4.4 - .8 = 3.6 m
Magnetic field
B₂ = 10⁻⁷ x 2 x 47 / 3.6
= 26.11 x 10⁻⁷ T .
Both these magnetic fields are perpendicular to each other so
Resultant magnetic field
B = √ ( 26.11² + 97.5² ) x 10⁻⁷ T
= √( 681.73 + 9506.25 ) x 10⁻⁷ T
= √( 10187.98) x 10⁻⁷ T
= 100.93 x 10⁻⁷ T .
