The answer is A- Venus has phases.
The Copernicus's heliocentric (Sun-centered) theory stated that the Sun is at rest near the center of the Universe, and that the Earth, spinning on its axis once daily, revolves annually around the Sun. He proved this theory by discovering that Venus went through phases just like the moon by recording his observations through the telescope.He also disproved the theory that the sun and other planets revolve around the earth through his recorded observations seen via the telescope.
The discovery Galileo made to support the theory that the planets, including Earth, orbit the Sun is that Venus has phases.
What are Phases?This is defined as a planet's area that reflects sunlight when viewed from a given point with the duration being inclusive.
Galileo discovered that Venus has phases which depicts that planets, including Earth, orbit the Sun.
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Part of a neutralization reaction between an acid and a base is shown below. What is the second product of the reaction?
Answer:
Salt Compound
Explanation:
acid + base →→→ water +salt compound
a foot ball with a mass of 0.42kg. the aveage acceleration of the football was 14.8 m/s^2. how much force did the kicker supply to the football
a. 6.22m/s^2
b. 622 N
c. 35.24 m/s^2
d. 34.24 N
Answer: 6.22 N(the units have been mentioned wrongly in the question)
Given:
mass of the ball(m)=0.42 Kg
acceleration of the ball(a)=14.8m/s^2
F=mxa
Where m is the mass of the ball.
a is the acceleration of the ball.
F is the force applied on the ball.
F=0.42 X 14.8
F= 6.26 N
what role does density play in the movement of convection curruents?
Convection occurs through heat transfer due to a difference in density in the fluids.
Consider a pot of water being heated. The water in the pot gets heated rises ,it provides energy for the particles in the water to move and thus the water expands which results in the density of the water becoming less. These particles rise up till the top of the pot owing to the property of very less density of the particles.
Then after some time, it cools down which results in increase of the density of the particles. The heavier denser particles sink to the bottom and is once again heated and rises and this process continues unless all the water particles in the pot are heated and warmed.This transfer of energy by the movement of particles cause convection.
During the day the heat surface above the earth exposed to the sun's rays is heated. This rise in temperature,decreases the density of the air and the warm air rises.
This air cools down and becomes denser. This dense cool air sink back and forces the warm air to rise again.
This transfer of movement of particles causes convection. This cycle is the cause of winds and thus energy is transferred through the atmosphere.
Final answer:
Density plays a crucial role in convection currents as it dictates that less dense, warmer fluid will rise, while denser, cooler fluid will sink, facilitating heat transfer through fluids.
Explanation:
The role density plays in the movement of convection currents is foundational to how heat is transferred through fluids (gases and liquids). Density is inversely related to temperature: as the temperature of a fluid increases, it becomes less dense due to thermal expansion. This property causes warmer, less dense fluid to rise, while cooler, denser fluid sinks under the influence of gravity, creating a convection current. This process results in the transfer of heat from warmer regions to cooler ones. An example of this can be seen when water is heated in a pot on the stove—hotter water expands and rises to the top, while cooler water sinks to the bottom, creating a cycle that evenly distributes the heat.
Two masses are separated by a distance r. If this distance is doubled, is the force of interaction between the two masses doubled, halved, or changed by some other amount? explain
If the interaction is gravitational or electrical, it gets multiplied by (1/2-squared) or 1/4 .
Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
This is mathematically represented as
F= (G X m1 x m2) /r∧2
where F is the force acting between the charged particles
r is the distance between the two charges measured in m
G is the gravitational constant which has a value of 6.674×10^-11 Nm^2 kg^-2
m1 and m2 are the masses of the objects measured in Kg
Now if the distance between the is doubled then r becomes 2r. Substituting this in the above formula we get the new Force as
Force (new) = (G X m1 x m2) /(2r)∧2
Thus dividing Force(new)/Force we get
Force(new)/Force = 1/4.
Thus the gravitational force becomes 1/4th of the original value if the distance between the two masses are doubled.
list all the storage forms of energy that forms of energy that you are familiar with. for each storage form, give an example of an object or situation that is storing this form of energy.
Storage form of energy:
Potential energyNuclear energyelectrical energythermal energymagnetic energyPotential energy:
All stationary objects are having potential energy stored in it. This energy can be transferred in form of kinetic energy when it comes in the motion from rest. Example, An object placed at height h having potential energy in it. When it comes in motion from the rest the potential energy is converted into kinetic energy.
Nuclear energy:
Nuclear energy is energy that is stored in nucleus of any element. Example, fusion reaction on sun gives earth solar energy.
Electric energy:
Electrical energy is due to movement of the electrical charges. Example, In elctrical batteries electrical energy is stored.
Thermal energy:
Thermal energy is the internal energy of a substance that is transferred to other substance in the form of heat. Example, on heating water is a beaker stem energy is developed.
Magnetic energy:
Magnetic energy is the potential energy stored in the magnetic field. Example, using magnetic energy electric field is produced according to Faraday's law.
Energy has various stored forms, including chemical, mechanical, radiant, electrical, and gravitational potential energy.
Explanation:Energy can be categorized into multiple storage forms. We often see energy transforming from one form to another in everyday situations.
Chemical Energy: This is stored in the bonds between atoms and molecules. For instance, the energy stored in the bonds of a glucose molecule that our bodies break down for fuel.
Mechanical Energy: It's the sum of kinetic and potential energy associated with the motion and position of an object. A working windmill is an example of mechanical energy.
Radiant Energy: It's the energy of electromagnetic waves. For example, the light emitted by the sun.
Electrical Energy: This involves the movement of charged particles. An example would be the energy that drives your computer or mobile device.
Gravitational potential energy: It's depends on an object's height above the ground. The potential energy of a book on a shelf, for example, becomes kinetic energy when the book falls, which proves the energy's existence.
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A pendulum has 294 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom of its swing
Newton's law of conservation states that energy of an isolated system remains a constant. It can neither be created nor destroyed but can be transformed from one form to the other.
Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.
Mathematically also potential energy is represented as
Potential energy= mgh
Where m is the mass of the pendulum.
g is the acceleration due to gravity
h is the height from the bottom z the ground.
At the bottom of the swing,the height is zero, hence the potential energy is also zero.
The kinetic energy is represented mathematically as
Kinetic energy= 1/2 mv^2
Where m is the mass of the pendulum
v is the velocity of the pendulum
At the bottom the pendulum has the maximum velocity. Hence the kinetic energy is maximum at the bottom.
Energy can neither be created e destroyed. It can only be transferred from one form to another. Implying this law and the above explainations we conclude that at the bottom of the pendulum,the potential energy=0 and the kinetic energy=294J as the entire potential energy is converted to kinetic energy at the bottom.
Please i need answer ASAP... Please guys
3 a )Given:
u( initial velocity):60Km/hr=16.67m/sec
v(final velocity):120Km/hr=33.33m/sec
a(acceleration):20 m/s^2
Consider s as the distance traveled by the car. We can calculate s from the below formula.
v^2 - u^2= 2as
Where v is the final velocity measured in m/s
u is the initial velocity measured in m/s
a is the acceleration measured in m/s^2
s is the distance traveled by the car.
Substituting the given values in the above formula we get
33.33^2- 16.67^2= 2 x 20 x s
832.99= 40 s
s = 250.12 m
3b) Consider t as the time taken for the car to travel the above distance. We can calculate t from the below formula.
s = ut +1/2(at^2)
250.12= 16.67 X t + 1/2(20 x t^2)
500.233 = 33.33t + 20t^2
Solving the above quadratic equation we get t= 1026 secs.
4) Given
v( final velocity) = 0.
time taken to cover the distance= 25 secs
Distance traveled(s)=40Km= 40000m
Now consider the below equation
v = u + at
Where v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
Substituting the given values in the above equation we get
0= u+ax25
u= -25a
Now we already know that
s = ut + 1/2(at^2)
Where s is the distance traveled
u is the initial velocity
a is the acceleration
t is the time
Substituting the given values in the above formula we get
40000 = u25 +1/2(ax25x25)
Now as solved above -25a =u. Substituting this in the above formula we get
40000= 25u +1/2(-25u)
40000= 12.5u
Thus u = 40000/12.5
u = 3200m/s
As per the above derived equation
We know in this case
-25 a = u
a= -128 m/s^2
Which of the following do the objects in the four categories of small objects in the solar system all have in common?
A.
They all orbit the sun.
B.
They can all be found in the Oort cloud.
C.
They all contain a nucleus.
D. They all are made of gas and ice.
A) They all orbit the sun
because they all are attracted through the forces of gravity.
Limestone originates because of: chemical precipitation chemical weathering physical weathering biological processes
Answer:chemical weathering
Explanation:I just got a 100 percent on my quiz
A ball is thrown straight up into the air, with an initial speed of 18.2 m/s. How far has it gone after 1.00 s?
It's vertical position in the air at time [tex]t[/tex] is
[tex]y=\left(18.2\,\dfrac{\mathrm m}{\mathrm s}\right)t-\dfrac g2t^2[/tex]
where [tex]g=9.80\,\dfrac{\mathrm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. After 1.00 second has passed, its position is
[tex]\left(18.2\,\dfrac{\mathrm m}{\mathrm s}\right)(1.00\,\mathrm s)-\dfrac g2(1.00\,\mathrm s)^2=13.3\,\mathrm m[/tex]
Answer:
The ball is going to reach 13.3 meters after 1 second.
Explanation:
The vertical movement of the ball can be described by the kinematic equations of y component for the position:
[tex]y=yo+vy_{o}t+1/2*ayt^2[/tex]
Where yo is the initial position and depends on where the system of reference is located. For this purpose is useful to consider the origin of the system of reference at the ground level, thus yo will be 0. The vyo coefficient is the initial vertical velocity and is given as 18.2 m/s. The ay coefficient is the vertical acceleration and due there is only the ball weight acting, the acceleration is taken of -9.8 m/s^2 (acceleration of gravity ). The time t must be expressed in seconds to obtaining the position in meters.
[tex]y=18.2\frac{m}{s}*t+1/2*(-9.8\frac{m}{s^2})*t^2[/tex]
Evaluating the equation for a time of 1 second:
[tex]y=18.2\frac{m}{s}*(1 s)+1/2*(-9.8\frac{m}{s^2})* (1 s)^2[/tex]
[tex]y=18.2 m+ (-4.9 m)[/tex]
[tex]y=13.3 m[/tex]
For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velocity vector at time t1, and let v⃗ 2 be the velocity vector at time t2. Consider Δv⃗ =v⃗ 2−v⃗ 1 and Δt=t2−t1. Recall that a⃗ av=Δv⃗ /Δt.
A) For Δt = 0.4 s calculate the magnitude (to four significant figures) of the average acceleration a⃗ av.
B) For Δt = 0.4 s calculate the direction (relative to v⃗ 1) of the average acceleration a⃗ av.
C) For Δt = 0.2 s calculate the magnitude (to four significant figures) of the average acceleration a⃗ av.
D) For Δt = 0.2 s calculate the direction (relative to v⃗ 1) of the average acceleration a⃗ av.
E) For Δt = 7x10^-2 s calculate the magnitude (to four significant figures) of the average acceleration a⃗ av.
F) For Δt = 7x10^-2 s calculate the direction (relative to v⃗ 1) of the average acceleration a⃗ av.
At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is
[tex]\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}[/tex]
so that its period is [tex]T=3.644\,\frac{\mathrm s}{\mathrm{rev}}[/tex] (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration [tex]\Delta t[/tex]. Denote by [tex]\theta[/tex] the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then
[tex]\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ[/tex]
[tex]\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ[/tex]
[tex]\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ[/tex]
We can then compute the magnitude of the velocity vector differences [tex]\Delta\vec v[/tex] for each time interval by using the law of cosines:
[tex]|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta[/tex]
[tex]\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}[/tex]
and in turn we find the magnitude of the average acceleration vectors to be
[tex]\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}[/tex]
So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.
The average acceleration, a⃗ av, of the poodle running in a circle can be calculated using the formula a⃗ av = v⃗ ^2 / r. The direction of the acceleration is always towards the center of the circle in uniform circular motion. The calculations remain the same regardless of the time interval Δt.
Explanation:This question is about the physics of circular motion, specifically calculating the acceleration of an object moving in a circular path. The poodle running in a circle at a constant speed indicates that this is a case of uniform circular motion. In such a situation the acceleration, a⃗ av, is always directed towards the center of the circle. This type of acceleration is also known as centripetal acceleration.
The formula to calculate the magnitude of the average acceleration in circular motion is a⃗ av = v⃗ ^2 / r, where v⃗ is the velocity (given as 5.00 m/s) and r is the radius (given as 2.9 m).
A) For Δt = 0.4 s, a⃗ av = (5.00 m/s) ^2 / 2.9 m = 8.62069 m/s^2, to four significant figures.
B) In uniform circular motion, the direction of the acceleration is always towards the center of the circle which is perpendicular to v⃗ 1.
C) & D) The calculations are identical for any Δt in uniform circular motion. so a⃗ av remains same and direction is also same.
E) & F) Again a⃗ av = 8.62069 m/s^2, to four significant figures and the direction is towards the center of the circle.
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Small-plane pilots regularly compete in "message drop" competitions, dropping heavy weights (for which air resistance can be ignored) from their low-flying planes and scoring points for having the weights land close to a target. a plane 70 m above the ground is flying directly toward a target at 44 m/s . at what distance from the target should the pilot drop the weight?
plane is flying at an altitude of 70 m
now if an object is dropped from it then time taken by object to drop on ground will be given as
[tex]y = v_i* t + \frac{1}{2}at^2[/tex]
here initial speed in vertical direction must be zero as plane is moving horizontal
given that
y = 70 m
a = 9.8 m/s^2
[tex]70 = 0 + \frac{1}{2}*9.8*t^2[/tex]
[tex]t = 3.77 s[/tex]
now since the plane is moving horizontally with speed v = 44 m/s
so the horizontal distance moved by the object will be
[tex]d = v_x * t[/tex]
[tex]d = 44 * 3.77 [/tex]
[tex]d = 166.3 m[/tex]
so the distance moved by the box is 166.3 m
Answer:
1.7 × 10² m
Explanation:
The movement of the weight can be decomposed in a vertical component and a horizontal component.
The vertical movement is uniformly accelerated motion (constant acceleration) and is the one that we will use to find the time of flight (t). The initial vertical speed is zero, and the vertical distance (y) traveled is 70 m. The acceleration is that of gravity.
y = 1/2 . a. t²
t = √(2y/a) = √(2 . 70 m/ 9.8 m/s²) = 3.8 s
The horizontal movement is a uniform motion (constant speed). The horizontal speed is that of the plane. The horizontal distance at what the pilot should drop the weight is:
d = v . t = 44 m/s . 3.8s = 1.7 × 10² m
An 80-kg fireman slides 5.0m down a fire pole. He holds the pole, which exerts a 500-N steady resistive force on thr fireman. At the bottom he slows to a stop in 0.40m by bending his knees. What can you determine using thos information?
Quite a few things can be determined using this information.
(i) First, we can calculate the Weight of the fireman using W = mg.
We get its magnitude as W = (80)(9.8) = 784 N
(ii) While his Weight is responsible for pulling him down, the pole exerts a constant Resistive Force that is given to be 500 N.
We can calculate the Net Force acting on the fireman as
[tex]F_{net} = W - Resistive Force[/tex]
We get its numerical value to be [tex]F_{net} = 284N[/tex]
(iii) Using this, we can calculate the acceleration with which he slides down the pole from Newton's 2nd law equation as
[tex]F_{net} =ma[/tex]
Therefore, [tex]a = \frac{284}{80} =3.55 m/s^{2}[/tex]
(iv) With this acceleration, he slides down a distance of 5.0 m - 0.4 m = 4.6 m before he starts applying an additional force with his knees.
We can calculate the velocity he attains just before bending his knees using the following data:
Initial Velocity at the top of the pole [tex]V_{i} =0[/tex]
Vertical displacement down the pole D = 4.6 m
Acceleration [tex]a = 3.55 m/s^{2}[/tex]
Final Velocity [tex]V_{f} = ?[/tex]
Using the equation [tex]V^{2} _{f} =V^{2} _{i} +2aD[/tex]
Plugging in the numbers, we have [tex]V^{2} _{f} =0+2(3.55)(4.6)[/tex]
Thus, we get the value of [tex]V_{f}[/tex] as 5.72 m/s
(v) This velocity serves as the initial velocity for the part of the journey with his knees bent.
We can calculate the acceleration he has using the following data:
Initial Velocity [tex]V_{i}=5.72[/tex] m/s
Final Velocity [tex]V_{f} =0[/tex]
Displacement during the last part of the journey D = 0.4 m
Acceleration a = ?
Again using the equation [tex]V^{2} _{f} =V^{2} _{i} +2aD[/tex], and plugging in the known numbers, we get
[tex](0)^{2} =(5.72)^{2} +2(a)(0.4)[/tex]
Hence, [tex]a=-40.898 m/s^{2}[/tex]
(vi) We can calculate the Net Force acting on him during this part of the journey as
[tex]F_{net} =(80)(-40.898)=-3271.84N[/tex]
(vii) Since the Net Force is the vector sum of Weight, Resistive Force, and the additional Knee Force, we can write them as
[tex]W-Resistive Force-Knee Force=-3271.84N[/tex]
Solving for Knee Force gives its magnitude to be 3555.84 N.
Thus, the fireman begins applying an additional 3,556 N force to stop himself in 0.4 m.
(viii) We can even calculate the time taken for the entire journey. We will deal with it in two parts.
For part one, the following information can be used:
Initial Velocity [tex]V_{i} =0[/tex]
Final Velocity [tex]V_{f} =5.72m/s[/tex]
Acceleration [tex]a = 3.55m/s^{2}[/tex]
Time taken t = ?
Using the equation [tex]V_{f} =V_{i} +at[/tex], we get the time taken as 1.6 seconds.
For the second part of the journey, the following information can be used:
Initial Velocity [tex]V_{i} =5.72m/s[/tex]
Final Velocity [tex]V_{f} =0[/tex]
Acceleration [tex]a=-40.898m/s^{2}[/tex]
Time taken t = ?
Using the equation [tex]V_{f} =V_{i} +at[/tex] again, and plugging in the appropriate values, we get the time taken as 0.14 seconds.
Hence, the total time the fireman took to slide down the 5.0 m pole is 1.74 seconds.
Hope this helps!
The calculation using the physics concept of work and energy determines the work done by a fireman when he slides down and stops, and quantifies the force he exerts when slowing to a stop.
Explanation:This is a classic problem in physics, specifically in the area of kinetic energy and work-energy theorem. First, as the fireman slides down, he is doing work and transforming potential energy into kinetic energy. We can calculate this work using the equation work = force x distance. In this case, the force is the resistive force of 500N, and the distance is 5.0m. So the work done is 500N x 5.0m = 2500J.
Next, when he bends his knees to slow to a stop, he is doing another amount of work to remove this kinetic energy. We can use the same formula as before, but this time the distance is 0.40m. We won't know the force directly, but we know the work done needs to be equal to the kinetic energy obtained earlier. So, we can solve for the 'force': 2500J = force x 0.40m, thus the force exerted would be around 6250N.
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When an electric soldering iron is used in a 110 V circuit, the current flowing through the iron is 2 A. What is the resistance of the iron?
Now we know by Ohm's law that
Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
Introducing the constant of proportionality, the resistance,the Ohm's law can be mathematically represented as
V=I x R
Where V is the voltage measured in volts
I is the current measured in amperes
R is the resistance measured in ohms
Given:
I = 2 A
V= 110 V
Applying Ohm's law and substituting the given values in the above formula we get
V=I x R
110 = 2 X R
R = 55 ohms
Final answer:
The resistance of the electric iron that is connected to a 110 V circuit and has a current of 2 A flowing through it is 55 ohms.
Explanation:
To calculate the resistance of an electric iron that is used in a 110 V circuit with a current of 2 A, we use Ohm's law, which is stated as V = IR, where V is voltage, I is current, and R is resistance. Rearranging the formula to solve for resistance gives us R = V/I. Plugging in the values provided, we get R = 110 V / 2 A, which gives us a resistance of 55 ohms.
An elevator is moving upward at 0.91 m/s when it experiences an acceleration 0.31 m/s2 downward, over a distance of 0.61 m. What will its final velocity be?
Given:
u= 0.91 m/s
a=0.31m/s∧2
s= 0.61 m
s = ut +1/2(at∧2)
where s is the displacement of the object
u is the initial velocity
t is the time
a is the acceleration
Substituting the values
0.61=0.91×t +(0.31 ×t∧2)/2
0.61=0.91 t + 0.155 t∧2
t=0.61 secs
Consider the equation
v=u +at
where v is the initial velocity
u is the initial velocity
a is the acceleration
t is the time
Substituting the values we get
v= 0.91 +(0.31×0.61)
v= 1.099 m/s
A body is accelerated continuously. What is the form of the graph?
A. Cubic
B. Inverse
C. Linear
D. Quadratic
Note :the graph shows distance and time
If a boy is continuously accelerated then we can assume it an example of uniform acceleration
Here we can say
[tex]d = v_i*t + \frac{1}{2}at^2[/tex]
now here we have
[tex]d = v*t + 0.5a t^2[/tex]
if we draw the graph between d and t then we can say that this graph will be a quadratic graph as it is the equation with power 2
So here in this case the graph will be QUADRATIC
Option "D" is correct option
This organelle functions much like a recycling center.
lysosomes
nucleus
mitochondria
cell wall
The lysosomes functions like a recycling center. The pH inside the lysosome is acidic around 5 due to the hydrogen ions and protons produced inside. These lysosomes contain hydrolase which is an enzyme which breaks down the molecular bonds in a chemical compound.This enzyme is active only in acidic conditions.
the answer is mitochondria
a water wave squeezing through a small hole is an example of
Answer;
Diffraction
Explanation;
Diffraction involves a change in direction of waves as they pass through an opening or around a barrier in their path. Water waves have the ability to travel around corners, around obstacles and through openings. This ability is most obvious for water waves with longer wavelengths.
Diffraction of water waves is observed in a harbor as waves bend around small boats and are found to disturb the water behind them. The same waves however are unable to diffract around larger boats since their wavelength is smaller than the boat.
A shiny sports car at the top of a vertical cliff has a poten tial energy of 100MJ relative to the ground below. Unfortunately, a mishap occurs and it falls over the edge. When it is halfway to the ground, its kinetic energy is A. the same as the potential energy at that point. B. negligible. C. about 60 MJ. D. more than 60 MJ.
When car is at the top of the hill its whole energy is stored in the form of gravitational potential energy
[tex]U = mgh[/tex]
so when height of the car becomes half then its potential energy is given as
[tex]U_f = \frac{mgh}{2}[/tex]
so final potential energy when car falls down by half of the height will become half of the initial potential energy
So it is U = 50 MJ after falling down
Now by energy conservation we can say that final potential energy + final Kinetic energy must be equal to the initial potential energy of the car
So here at half of the height kinetic energy of car = 100 - 50 = 50 MJ
so we can say at this point magnitude of potential energy and kinetic energy will be same
A. the same as the potential energy at that point.
A train travels 75 kilometers in 3 hours, and then 66 kilometers in 1 hours. What is its average speed?
Initial distance (D1):75 Km
Time taken to cover this distance= 3 hrs.
Speed= Distance /time
Speed1= 75/3= 25Km/hr
Later the train travels
Distance (D2):66 Km
Time taken to cover this: 1hr
Speed= Distance/time
Speed2= 66/1= 66Km/hr
Average speed= (speed1+ speed 2)/2
Average speed= (25+66)/2
Average speed= 45.5 Km/hr
if energy is conserved, why do people get tired , and why are we running short on “energy supplies?”
A mass and spring system has a resonance at f0 = 1.40 hz with a full-width of ffw = 0.021 hz. what is the quality factor q?
Quality Factor is defined as the ratio of Resonance frequency per unit band width
it is given as
[tex]Q = \frac{\omega_0}{\Delta \omega}[/tex]
[tex]Q = \frac{2\pi f_0}{2\pi f_{fw}}[/tex]
now here it is given that
[tex]f_0 = 1.40 hz[/tex]
[tex]f_{fw} = 0.021 hz[/tex]
now we have
[tex]Q = \frac{2\pi*1.40}{2\pi*0.021}[/tex]
so we have
[tex]Q = 66.67[/tex]
So quality factor of given AC is 66.67
A 200N force pushes forward on the 20kg truck. The truck pulls two crates connected together by ropes, as shown. The gravitational constant is 10N/kg. Determine the acceleration of the crates and truck and the tension in each rope for one of the following sets of conditions.
Please help me! I am really confused on how to draw the diagrams and how to find the x component forms and the unknowns! I've tried working on it but ended up erasing it cause it seemed wrong.
Force on left crate, right crate and truck we need to draw here
First we will write the equation for left crate
[tex]T_{left} = m*a[/tex]
[tex]T_{left} = 20*a[/tex]
Now similarly for right crate
[tex]T_{right} - T_{left} = 20*a[/tex]
now for truck
[tex]F - T_{right} = 20*a[/tex]
Now we know that F = 200 N
[tex]200 - T_{right} = 20*a[/tex]
now add all three equation
[tex]T_{left} + T_{right} - T_{left} + 200 - T{right} = 20a + 20 a + 20a[/tex]
[tex]200 = 60 a[/tex]
[tex]a = 10/3 m/s^2[/tex]
now we will find all tension using this value of acceleration
[tex]T_{left} = 20 * \frac{10}{3} = \frac{200}{3} N[/tex]
[tex]T_{right} - \frac{200}{3} = 20*\frac{10}{3}[/tex]
[tex]T_{right} = \frac{400}{3} N[/tex]
3 Why does the frequency of a wave increase as the wavelength decreases?
A) because the wave speed remains constant
B) because the amplitude of the wave must remain constant
C) because the frequency must get larger to increase the speed
D) because the wavelength must get smaller to decrease the speed Elimin
A skier traveling 11.0 m/s reaches the foot of a steady upward 17 â incline and glides 15 m up along this slope before coming to rest. part a what was the average coefficient of friction?
The attached free-body diagram shows the forces that are responsible for the skier coming to rest eventually on the incline.
We see that the component of his Weight along the incline [tex]mgSin \alpha[/tex] and the Friction both act in tandem to stop him.
In order to calculate the Friction, we can make use of Newton's 2nd law, which states that [tex]F_{net} = ma[/tex]
The [tex]F_{net}[/tex] here is given by [tex]mgSin \alpha + F_{k}[/tex], where [tex]F_{k}[/tex] is the Kinetic Friction.
We also know that the magnitude of Friction Force can be calculated using the equation [tex]F_{k} =[/tex] μ.[tex]F_{N}[/tex], where [tex]F_{N}[/tex] is the Normal Force acting perpendicular to the incline as shown in the figure.
We see that [tex]F_{N} = mgCos \alpha[/tex] since they both balance each other out.
Hence, putting all these together, we have [tex]mgSin \alpha +[/tex]μ.[tex]mgCos \alpha = ma[/tex]
Simplifying this, we get [tex]gSin \alpha +[/tex] μ[tex]gCos \alpha = a[/tex]
We clearly see that we need to calculate the acceleration before we can obtain the value of the Coefficient of Friction μ
And for that, we make use of the following data obtained from the question:
Initial Velocity [tex]V_{i} = 11.0 m/s[/tex]
Final Velocity [tex]V_{f} = 0[/tex]
Displacement along the incline [tex]D = 15m[/tex]
Acceleration a = ?
Using the equation [tex]V_{f} ^{2} = V_{i} ^{2} + 2aD[/tex], and
Plugging in known numerical values, we get [tex]0 = (11)^{2} + 2a(15)[/tex]
Solving for a gives us, [tex]a = -4.03 m/s^{2}[/tex]
Since the negative sign indicates that this is deceleration, we can ignore the sign and consider the magnitude alone.
Thus, plugging in [tex]a = 4.03 m/s^{2}[/tex] in the force equation we wrote above, we have
[tex](9.8)Sin (17) +[/tex] μ.[tex](9.8)Cos (17) = 4.03[/tex]
Solving this for μ, we get its value as μ = 0.124
Thus, the average coefficient of friction on the incline is 0.12
To find the average coefficient of friction for the skier, you need to set the work done by friction equal to the loss in kinetic energy, and solve for the coefficient of friction. This situation is independent of the skier's mass. Using given information about speed, distance travelled up the incline, angle of incline, and acceleration due to gravity, you can calculate the coefficient of friction.
Explanation:To calculate the average coefficient of friction in this scenario involving a skier on an incline, you need to use the principle that the work done by friction is equal to the kinetic energy lost by the object. The skier's initial kinetic energy is given by (1/2)mv², where m is the mass of the skier and v is the speed (11 m/s). The work done by friction is given by Fd, where F is the force of friction and d is the distance travelled up the incline (15 m).
Since the force of friction is equal to μN (where μ is the coefficient of friction and N is the normal force), we substitute N = mg cos θ (where θ is the angle of incline, g is the acceleration due to gravity, and m is again the mass of the skier).
After setting the work done by friction equal to the loss in kinetic energy and solving for μ, one can obtain the average coefficient of friction.
Note that in this set up, the mass of the skier cancels out, meaning the result is independent of the skier's mass. Assuming the acceleration due to gravity to be approximately 9.8 m/s², and the angle of inclination to be 17 degrees, the resulting average coefficient of friction would be determined through this method.
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Which planet is small , with a rocky surface and an atmosphere?
Venus is your best bet looking at your answers.
Answer:
Venus
Explanation:
Solar system has 8 planets namely Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. Out of these the first four planets are rocky while the last four are gaseous. When you compare the size of the rocky planet with gaseous planets, gaseous planets are way too bigger than rocky planets.
Using the same information, out of the given options Venus is the correct answer as it is smaller than other three planets and has a rocky surface and atmosphere.
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are mA = 340 kg, mB = 567 kg, and mC = 139 kg. Take the positive direction to be to the right. Find the net gravitational force, including sign, acting on (a) particle A, (b) particle B, and (c) particle C.
Formula of the gravitational force between two particles:
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
[tex]G=6.67 \cdot 10^{-11} Nm^2 kg^{-2}[/tex] is the gravitational constant
m1 and m2 are the masses of the two particles
r is their distance
(a) particle A
The gravitational force exerted by particle B on particle A is
[tex]F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N[/tex] to the right
The gravitational force exerted by particle C on particle A is
[tex]F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N[/tex] to the right
So the net gravitational force on particle A is
[tex]F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N[/tex] to the right
(b) Particle B
The gravitational force exerted by particle A on particle B is
[tex]F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N[/tex] to the left
The gravitational force exerted by particle C on particle B is
[tex]F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N[/tex] to the right
So the net gravitational force on particle B is
[tex]F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N[/tex] to the right
(c) Particle C
The gravitational force exerted by particle A on particle C is
[tex]F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N[/tex] to the left
The gravitational force exerted by particle B on particle C is
[tex]F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N[/tex] to the left
So the net gravitational force on particle C is
[tex]F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N[/tex] to the left
Particle A,B, and C is [tex]\rm \bold{ 5.6 \times 10^-^5N}[/tex],[tex]\rm \bold{ 3.27 \times 10^-^5N}[/tex], [tex]\rm \bold{ -8.97 \times 10^-^5N}[/tex] respectively. Negative sign represents left direction.
The gravitational force between two bodies
[tex]\rm \bold { F= G\frac{m^1 m^2}{r^2} }[/tex]
Where,
G- gravitational constant = [tex]\rm \bold{6.67\times 10^1^1 Nm^2kg^-^2 }[/tex]
[tex]\rm \bold { {m^1 m^2} }[/tex] = mass of bodies
r - is distance between them
Net gravitational force on Particle A
[tex]\rm \bold{F_a = F_b+ F_c}[/tex]
The gravitational force exerted by particle B on particle A is [tex]\rm \bold{ 5.14 \times 10^-^5N}[/tex] to the right .
The gravitational force exerted by particle C on particle A is [tex]\rm \bold{ 5.6 \times 10^-^6N}[/tex] to the right.
Hence, net Gravitational force on A is [tex]\rm \bold{ 5.6 \times 10^-^5N}[/tex]
Net gravitational force on Particle B
[tex]\rm \bold{F_b = F_a+ F_c}[/tex]
The gravitational force exerted by particle A on particle B is [tex]\rm \bold{ -5.14 \times 10^-^5N}[/tex] on to the left.
The gravitational force exerted by particle C on particle B is [tex]\rm \bold{ 8.41 \times 10^-^5}[/tex] to the right.
Hence net gravitational force on particle B will be [tex]\rm \bold{ 3.27 \times 10^-^5N}[/tex]
Net gravitational force on Particle C is
[tex]\rm \bold{F_c = F_a+ F_b}[/tex]
The gravitational force exerted by particle A on particle C is [tex]\rm \bold{ -5.6 \times 10^-^6N}[/tex] to the left.
The gravitational force exerted by particle B on particle C is [tex]\rm \bold{ -8.41 \times 10^-^5N}[/tex] to the left.
Hence, net gravitational force on particle C is [tex]\rm \bold{ -8.97 \times 10^-^5N}[/tex] to the left.
Therefore we can conclude that particle A,B, and C is [tex]\rm \bold{ 5.6 \times 10^-^5N}[/tex],[tex]\rm \bold{ 3.27 \times 10^-^5N}[/tex], [tex]\rm \bold{ -8.97 \times 10^-^5N}[/tex] respectively.
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What average braking force is required to stop a 1134-kg car traveling at a speed of 83 km/hr before it reaches a stop sign that is 98 m away?
Given:
m(mass of the car)=1134 Kg
u(Initial velocity)=83Km/HR=23m/s
s(distance traveled by the car)=98m
v(final velocity)=0(as it is given the car stops).
Now we know,
v=u+at
Where v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
0=23+at
at=-23
Also
s=ut+1/2(at^2)
s is the distance covered by the car
u is the initial velocity
t is the time necessary for the car to cover a particular distance.
a is the acceleration
Now substituting these values we get
98=23t-1/2(23t)
98=23t-11.5t
11.5t=98
t=8.52secs
Now we have already derived
at=-23
ax8.52=-23
a=-23/8.52
a=-2.75 m/s^2
F=mxa
Where F is the force acting on the car.
m is the mass of the car.
a is the acceleration.
F=1134 x-2.75
F=-3119N
A heat transfer of 9.8 ✕ 105 J is required to convert a block of ice at -12°C to water at 12°C. What was the mass of the block of ice?
Final answer:
The mass of the ice block that requires a heat transfer of 9.8 x 10^5 J to convert it from -12°C ice to 12°C water is approximately 2.94 kg.
Explanation:
The question relates to the heat transfer required to convert a block of ice at -12°C to water at 12°C. To calculate the mass of the ice, we can use the formula for heat transfer — Q = mLf, where Q is the heat in joules, m is the mass in kilograms, and Lf is the latent heat of fusion for ice, which is 334 kJ/kg (or 334 x 10³ J/kg).
To find the mass of the ice, we set the total heat transfer equal to Q = mLf. The total heat transfer given is 9.8 × 10µ J. Using the latent heat of fusion value, we have:
9.8 × 10µ J = m × 334 × 10³ J/kg
Solving for m, we get:
m = µ ≟ 334 × 10³ kg
m ≈ 2.94 kg.
John and Caroline go out for a walk one day. This graph represents their distance from home.
Which statement accurately describes their walk?
A) They walked quickly, then they were still, then they walked slowly.
B) They walked forward fast, then they were still, then they walked backward slow.
C) They walked in one direction, then they were still, then they walked in the opposite direction.
D) They walked faster and faster until they walked at a constant rate, then they walked slower and slower.
Answer:
Option C is the correct answer.
Explanation:
We have velocity of a body = Change in position/ Time.
Considering first portion of graph,
Change in position = 30 - 0 = 30 m
Time = 0.75 hours = 45 minutes = 2700 seconds
Velocity = 30/2700 = 0.011 m/s
Considering second portion of graph,
Change in position = 30 - 30 = 0 m
Time = 0.5 hours = 30 minutes = 1800 seconds
Velocity = 0/1800 = 0 m/s
Considering third portion of graph,
Change in position = 0 - 30 = -30 m
Time = 0.75 hours = 45 minutes = 2700 seconds
Velocity = -30/2700 = -0.011 m/s
So firstly they walked in one direction(positive direction), then they were still(velocity is zero), then they walked in the opposite direction( velocity is negative).
Option C is the correct answer.
Answer:
I had this same question the answer is C 100% sure
Explanation: