Answer:
because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.
g Consider a mass-spring system where the spring constant is 5 N/m and the mass on the spring is 0.5 kg. The mass is moved a distance of -0.9 m from its equilibrium position. How much work is done by the spring
Answer:
The work done by the spring is 2.025 J
Explanation:
Given;
mass on the spring, m = 0.5 kg
spring constant, k = 5 N/m
extension of the spring, x = 0.9 m
The work done by the spring is calculated as;
[tex]W = \frac{1}{2} kx^2\\\\W = \frac{1}{2} \times 5 \times (0.9)^2\\\\W = 2.025 \ J[/tex]
Therefore, the work done by the spring is 2.025 J
trình bày nguyên lý Đa lăm be
how can scientific method solve real world problems examples
If you change the motor in your vehicle you need to notify the DMV within ____,
days of this change.
-20
-25
-10
-15
when you change your motor on your vehicle you need to notify the DMV within 10 days
If you change the motor in your vehicle you need to notify the DMV within 10 days of this change.
An engine or motorAn engine or motor is a machine designed to convert one or more forms of energy into mechanical energy.
Available energy sources include potential energy (e.g. energy of the Earth's gravitational field as exploited in hydroelectric power generation), heat energy (e.g. geothermal), chemical energy, electric potential, and nuclear energy (from nuclear fission or nuclear fusion). Many of these processes generate heat as an intermediate energy form, so heat engines have special importance. Some natural processes, such as atmospheric convection cells convert environmental heat into motion (e.g. in the form of rising air currents). Mechanical energy is of particular importance in transportation but also plays a role in many industrial processes such as cutting, grinding, crushing, and mixing.
Mechanical heat engines convert heat into work via various thermodynamic processes. The internal combustion engine is perhaps the most common example of a mechanical heat engine, in which heat from the combustion of fuel causes rapid pressurization of the gaseous combustion products in the combustion chamber, causing them to expand and drive a piston, which turns a crankshaft. Unlike internal combustion engines, a reaction engine (such as a jet engine) produces thrust by expelling reaction mass, by Newton's third law of motion.
Learn more about motor
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which has higher eneergy electron r proton
Answer:
proton have higher energy than electron
Explanation:
tag me brainliest
Answer:
proton
Explanation:
proton is higher energy than the electron
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of with the horizontal
Complete Question
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of
40° 16' with the horizontal.
Answer:
[tex]d=8.01m[/tex]
Explanation:
From the question we are told that:
Length of ladder [tex]l=12.5m[/tex]
Angle [tex]\theta=40° 16'=20.26 \textdegree[/tex]
Generally the Trigonometric equation for distance d it goes up the wall is mathematically given by
[tex]d=l sin \theta[/tex]
[tex]d=12.5 sin 40.26[/tex]
[tex]d=8.01m[/tex]
Hi can someon help me how to answer this?
Btw I'm from Philippines
Answer:
Test 1
1.True
2.True
3.True
4.False
5.True
6.True
7.False
8.True
9.True
10.True
yung iba nasa pic
A wheel has a diameter of 10m and weight 360N what minimum horizontal force is necessary to pull the wheel over a brick 0.1m when a force is applied at the wheel
The angular velocity of an object is given by the following equation: ω(t)=(5rads3)t2\omega\left(t\right)=\left(5\frac{rad}{s^3}\right)t^2ω(t)=(5s3rad)t2 What is the angular displacement of the object (in rad) between t = 2 s and t = 4 s?
Answer:
The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.
Explanation:
The angular velocity of the object ([tex]\omega[/tex]), in radians per second, is given by the following expression:
[tex]\omega(t) = 5\cdot t^{2}[/tex] (1)
Where [tex]t[/tex] is the time, measured in seconds.
The change in the angular displacement ([tex]\Delta \theta[/tex]), in radians, is found by means of the following definite integral:
[tex]\Delta \theta = \int\limits^{4}_{2} {5\cdot t^{2}} \, dt[/tex] (2)
Then we proceed to integrate on the function in time:
[tex]\Delta \theta = \frac{5}{3}\cdot (4^{2}-2^{2})[/tex]
[tex]\Delta \theta = 20\,rad[/tex]
The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.
You are stranded in a stationary boat. Your friend is on a dock, but the boat is just beyond his reach. There is a 5 kg anchor in the boat. You'd like to get the boat to move closer to the dock so your friend can rescue you. Select from the following list what effect each change will have on the position of the boat relative to the dock. A. The boat will move closer to the dock. B. The boat will move away from the dock. C. The position of the boat relative to the dock will not change.
Answer:
running away and launching the anchor that will give a greater speed towards the dock v₄.
Explanation:
To try to bring the boat closer to the dock, several cases can be carried out.
* move inside the ship so that the center of mass changes and since moving away you have a speed v, the ship will approach the dock at a speed v₂,
* Throw the anchor in the opposite direction to the dock so that using the conservation of the moment the boat moves towards it, it moves at a speed v₃
* A combination of the two processes running away and launching the anchor that will give a greater speed towards the dock v₄.
In all cases, the friction must be zero.
All other movements move the ship away from the dock
The diagram here shows an image being formed by a convex lens. Compared to the object at right, the image at left is-
larger and upright.
smaller and upright.
smaller and upside down.
larger and upside down.
Answer:
Smaller and upside down
Explanation:
To answer the question, we must recognise that the characteristics of the image formed by a convex lens depends on the position of the object from the lens.
From the diagram given in the question above, the following data were obtained:
1. The image is smaller than the object.
2. The image is inverted i.e upside down.
3. The image is closer to the lens
4. The image between 2f and f
Now, considering the options given in question above, the correct answer to the question is:
The image is smaller and upside down.
A TV satellite dish is designed to receive radio waves of wavelength
0.0644 meters. What is the frequency of the waves it receives? _______GHz
Give your answer in gigahertz (GHz). 1 GHz = 10^9 Hz.
Give your answer to the nearest tenth of a GHz (one place after the decimal). Just enter the number; do NOT use scientific notation.
Answer:
4.7 GHz
Explanation:
Applying,
v = λf................. Equation 1
Where v = velocity of the radio wave, λ = wavelength, f = frequency
make f the subject of the equation
f = v/λ.............. Equation 2
Note: A radio wave is an electromagnetic wave, as such it moves with a velocity of 3.00 x 10⁸ m/s
From the question,
Given: λ = 0.0644 meters
Constant: v = 3.00 x 10⁸ m/s
Substitute these values into equation 2
f = (3.00 x 10⁸)/0.0644
f = 4.66×10⁹ Hz
f = 4.7 GHz
b) Assume the rod is 0.60 m long and has a mass of 0.50 kg, and the clay blob has a mass of 0.20 kg and moves at an initial velocity of 8.0 m/s. Calculate the final angular velocity of the rod. Be sure to put units in your calculation and show the resulting units in your answer.
Answer:
The correct answer is "6.96 rad/s".
Explanation:
The given values are:
Length,
L = 0.6 m
Mass,
m₁ = 0.5 kg
m₂ = 0.2 kg
Initial velocity,
V = 8 m/s
Now,
The final angular velocity will be:
⇒ [tex]\omega =\frac{6m_1V}{(4m_1+3m_2)L}[/tex]
By substituting the values, we get
⇒ [tex]=\frac{6\times 0.2\times 8}{(4\times 0.2+3\times 0.5)0.6}[/tex]
⇒ [tex]=\frac{9.6}{1.38}[/tex]
⇒ [tex]=6.96 \ rad/s[/tex]
Why is the temperature constant during the melting of water?
[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}[/tex]
Why is the temperature constant during the melting of water?
[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]
[tex] \orange{\underline{\huge{\bold{\textit{\green{\bf{ REASON}}}}}}}[/tex]
THE HEAT WE R SUPPLYING TO THE WATER TO RAISE THE TEMP OF THE WATER IS USED BY THE MOLECULES TO BREAK INTERMOLECULAR BONDS WHICH HELP IN THE CHANGING OF THE LATTICE (STRUCTURE) OF THE WATER .
ICE HAS A HEXAGONAL RING LIKE STRUCTURE WHICH IS CONVERTED INTO REGULAR CRYSTALLINE STRUCTURE WHICH CAN ONLY BE FORMED WITH THE HELP OF FORMATION OF NEW BONDS AND BREAKDOWN OF OLDER ONES
THE AMOUNT OF ENERGY WHICH IS USED IN CONVERSATION OF THE STATE OF FROM SOLID TO LIQUID IS KNOWN AS LATENT HEAT OF FUSION.
SO TEMP REMAIN CONSTANT DURING CHANGE IN STATE .
[tex] \red \star{Thanks \: And \: Brainlist} \blue\star \\ \green\star If \: U \: Liked \: My \: Answer \purple \star[/tex]
2. Below what depth would a submarine have to submerge so that it would not be swayed by surface waves with a wavelength of 24 meters?
Answer: Below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves
Explanation:
To avoid the surface waves, a submarine has to submerge below the wave base. It is the position below which the motion of the waves is negligible.
This wave base is equal to half of the wavelength. The equation becomes:
Wave base = [tex]\frac{\text{Wavelength}}{2}[/tex]
We are given:
Wavelength = 24 m
Putting values in above equation, we get:
Wave base = [tex]\frac{24m}{2}=12m[/tex]
Hence, below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves
when blueshift occurs,the preceived frequency of the wave would be?
Answer:
When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.
Explanation:
As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.
Blueshifts happens when the source of the wave and the observer are moving closer towards one another.
Assume that the wave is of frequency [tex]f\; {\rm Hz}[/tex] at the source. In other words, the source of the wave sends out a peak after every [tex](1/f)\; {\text{seconds}}[/tex].
Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.
The source of this wave sends out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. It would appear to the observer that consecutive peaks arrive every [tex](1/f)\; {\text{seconds}}\![/tex]. That would correspond to a frequency of [tex]f\; {\rm Hz}[/tex].
On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of [tex]v \; {\rm m \cdot s^{-1}}[/tex].
Again, the source of this wave would send out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. However, by the time the source sends out the second peak, the source would have been [tex]v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}[/tex] closer to the observer then when the source sent out the first peak.
When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than [tex](1/f)\; {\text{seconds}}[/tex]. The observed frequency of this wave would be larger than the original [tex]f\; {\rm Hz}[/tex].
Which image illustrates reflection?
A
B
с
D
Answer: I beleive A
Explanation:
Answer:
A
Explanation:
We can see the light being reflected off the mirror.
What quantity of heat is transferred when a 150.0g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of the heat flow?
Answer:
Heat is flowing into the metal.
Explanation:
From the question given above, the following data were obtained:
Mass (M) of iron = 150 g
Initial temperature (T₁) = 25.0°C
Final temperature (T₂) = 73.3°C
Direction of heat flow =?
Next, we shall determine the change in the temperature of iron. This can be obtained as follow:
Initial temperature (T₁) = 25.0 °C
Final temperature (T₂) = 73.3 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 73.3 – 25
ΔT = 48.3 °C
Next, we shall determine the heat transfered. This can be obtained as follow:
Mass (M) of iron = 150 g
Change in temperature (ΔT) = 48.3 °C
Specific heat capacity (C) of iron = 0.450 J/gºC
Heat (Q) transfered =?
Q = MCΔT
Q = 150 × 0.450 × 48.3
Q = 3260.25 J
Since the heat transferred is positive, it means the iron metal is absorbing the heat. Thus, heat is flowing into the metal.
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll.
What will be the car’s speed as it coasts into the gas station on the other side of the valley?
Answer:
v_f = 17.4 m / s
Explanation:
For this exercise we can use conservation of energy
starting point. On the hill when running out of gas
Em₀ = K + U = ½ m v₀² + m g y₁
final point. Arriving at the gas station
Em_f = K + U = ½ m v_f ² + m g y₂
energy is conserved
Em₀ = Em_f
½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂
v_f ² = v₀² + 2g (y₁ -y₂)
we calculate
v_f ² = 20² + 2 9.8 (10 -15)
v_f = √302
v_f = 17.4 m / s
Steel wire rope is used to lift a heavy object. We use a 3.1m steel wire that
is 6.0mm in diameter and lift a 1700kg object. Then, the wire elongates
0.17m. Calculate the Young’s modulus for the rope material.
Answer:
Young's modulus for the rope material is 20.8 MPa.
Explanation:
The Young's modulus is given by:
[tex] E = \frac{FL_{0}}{A\Delta L} [/tex]
Where:
F: is the force applied on the wire
L₀: is the initial length of the wire = 3.1 m
A: is the cross-section area of the wire
ΔL: is the change in the length = 0.17 m
The cross-section area of the wire is given by the area of a circle:
[tex] A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2} [/tex]
Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:
[tex] F = T_{w} = W_{o} [/tex]
Where:
[tex] T_{w} [/tex]: is the tension of the wire
[tex]W_{o} [/tex]: is the weigh of the object = mg
m: is the mass of the object = 1700 kg
g: is the acceleration due to gravity = 9.81 m/s²
[tex] F = mg = 1700 kg*9.81 m/s^{2} = 16677 N [/tex]
Hence, the Young's modulus is:
[tex] E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa [/tex]
Therefore, Young's modulus for the rope material is 20.8 MPa.
I hope it helps you!
Momentum is defined as mass ___ velocity
Differentiate between Scalar quantity and Vector quantity and give two examples each
At which point is the kinetic energy of the pendulum the greatest?
C
A
D
B
Answer:
Point C
Explanation:
Greatest Kinetic Energy means lowest potential energy since energy is conserved. Lowest potential energy means lowest height which is at Point C.
Electrical charges are of two types. True False
Answer:
Electrical charges r of 2 types its true.they are positive and negative.hope it helps.stay safe healthy and happy..Answer: Think its true
find the weight of a body of mass 200kg on the earth at a latitude 30°.(R=6400 km ,g=9.8m/s²,ω=7.27×10⁻⁵ rad/sec)
Answer:
................ftf6x
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas. During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles, while during the adiabatic expansion the volume increases by a factor of 5.7. The work output of the engine is 940 J in each cycle. Compute the temperatures of the two reservoirs between which this engine
operates.
Answer:
The hot temperature is 157.5 K
The cold temperature is 48.8 K
Explanation:
Step 1: Data given
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas.
The volume increases by a factor of 5.7
The work output of the engine is 940 J in each cycle.
During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles. This means V2 = 2*V1 (and V4 = 2*V3)
Step 2:For a carnot engine:
V2/V1 = V4/V3
Work = nR((T1)ln(V2/V1) - (T2)ln(V4/V3))
⇒with Work = the work done in the cycle = 940J
⇒with n = the number of moles = 1.50 moles
⇒with R = the gas constant = 8.314 J/mol*K
⇒with T1 = the hot temperature
⇒With T2⇒ the cold temperature
where R = 8.31 J/mol K Gas Constant
940J = 1.5moles * 8.314 J/mol*K * (T1*ln(2) - T2*ln(2)))
940 = 1.5 * 8.314 ln(2) * (T1-T2)
(T1-T2) = 940 / (1.5*8.314*ln(2))
(T1-T2) = 108.7K
For the reversible adiabatic expansion: T2 = T1*(V1/V2)^(R/Cv). Where V2/V1 = 5.7 (Because during the adiabatic expansion the volume increases by a factor of 5.7)
For a monatomic ideal gas, Cv = 3/2R
When we combine both, we'll have:
T2 = T1*(1/5.7)^(R/3/2R)
T2 = T1*(1/5.7)^(2/3)
T2= T1 * 0.31
Since we know that (T1-T2) = 108.7K
we have:
T1 - 0.31T1= 108.7K
0.69T1 = 108.7K
T1 = 157.5K
T2 = 157.5*0.31 = 48.8K
Container A and container B hold samples of the same ideal gas. The volume and the pressure of container A is equal to the volume and pressure of container B, respectively. If Container A has half as many molecules of the ideal gas in it as Container B does, then which of the following mathematical statements is correct regarding the absolute temperatures TA and TB in Container A and Container B. respectively?
A. TA = TB/2.
B. TA = 4TB.
C. TA = TB/4.
D. TA = 2TB.
E. TA = TB
Answer:
A. TA = TB/2.
Explanation:
Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:
[tex]V_A = \frac{1}{2}V_B[/tex]
Now, from Charle's Law:
[tex]\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}[/tex]
Hence, the correct option is:
A. TA = TB/2.
You work in the special effects department of a movie studio. You are currently working on a superhero movie where the hero is very
strong and cannot be hurt by normal weapons such as a sword. In the next scene, the villain
is going to hit the hero with a lead pipe.
The hero's costume is mostly made of foam and carbon so that is it very light and easy to move around in but will crumble if hit with
anything hard. For this scene, a section of the
costume needs to be replaced with a different material that will only dent, not crumble,
when hit with the pipe. You are on a deadline and look around the props
department for ideas, what could you use?
A. You find sheets of copper (Cu) used in an induced rescue factory scene
B. You find a barrel of phosphorus (P) used as a component in explosive powder
C. You find a tank of helium (He) used for balloons
D. You find large desk of Chlorine (Cl) used to kill bacteria and pools
Answer:
A. You find sheets of copper (Cu) used in an induced rescue factory scene
Explanation:
I would use a sheets of copper (Cu) used in an induced rescue factory scene because, copper is a metal and the only material out of the other options that would only dent and not crumble when hit by the lead pipe.
The other options in B - D are non-metals.
Phosphorus is a non-metal and is used as an explosive and thus cannot be used.
Helium is a gas used in filling balloons and thus cannot be used.
Chlorine is used in killing bacteria and thus cannot be used.
So, copper is the only option available since it is a metal and can thus accommodate a large force.
So, option A is the answer.
Which is the most difficult subject?
Answer:
Quantum Mechanics
Explanation:
Well, that's what I think personally.
A uniform steel rod of length 0.9 m and mass 3.8 kg has two point masses of 2.3 kg each at the two ends. Calculate the moment of inertia of the system about an axis perpendicular to the rod, and passing through its center.
Answer: [tex]2.4705\ kg.m^2[/tex]
Explanation:
Given
length of the rod is L=0.9 m
Mass of the rod m=3.8 kg
Point masses has mass of m=2.3 kg
Moment of Inertia of the rod about the center is
[tex]\Rightarrow I_o=\dfrac{1}{12}ML^2[/tex]
Moment of inertia of combined system is the sum of rod and two point masses.
[tex]\Rightarrow I=I_o+2mr^2[/tex]
[tex]\Rightarrow I=\dfrac{1}{12}3.8\times 0.9^2+2\times 2.3\times \left(\dfrac{0.9}{2}\right)^2\\\\\Rightarrow I=1.539+0.9315\\\Rightarrow I=2.4705\ kg-m^2[/tex]