Weight of the object on surface of any planet is given as the gravitational pull on the object due to planet
This gravitational pull is defined as formula below
[tex]F = mg[/tex]
m = mass of object
g = acceleration due to gravity of the planet
now here it is given that weight on the planet is 370 N which is defined as the force due to planet
Also the mass is given as 100 kg
now using the formula above
[tex]370 = 100 * g[/tex]
[tex]g = \frac{370}{100}[/tex]
[tex]g = 3.7 m/s^2[/tex]
so the gravitational strength of the mars will be 3.7 m/s^2
Answer:
The gravitational strength of the mars will be 3.7 m/s^2
Explanation:
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.
The cause and effect of breaking a egg
Hold up raw egg and say “CAUSE” as you crack the egg on the side of the bowl.
2. Say “EFFECT” as you break open the egg into the bowl.
3. Hold up the hardboiled egg and say “CAUSE” as you go through exaggerated motions of tossing the egg to a student.
4. As they either catch it or drop it, say “EFFECT”.
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
What will the stopping distance be for a 2,000-kg car if -2,000 N of force are applied when the car is traveling 20 m/s?
A. 400,000 m
B. 200 m
C. 100 m
D. 20 m
Answer is B- 200 m
Given:
m (mass of the car) = 2000 Kg
F = -2000 N
u(initial velocity)= 20 m/s.
v(final velocity)= 0.
Now we know that
F= ma
Where F is the force exerted on the object
m is the mass of the object
a is the acceleration of the object
Substituting the given values
-2000 = 2000 × a
a = -1 m/s∧2
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
0= 20 -t
t=20 secs
s = ut +1/2(at∧2)
where s is the displacement of the object
u is the initial velocity
t is the time
v is the final velocity
a is the acceleration
s= 20 ×20 +(-1×20×20)/2
s= 200 m
Answer:
Stopping distance, s = 200 meters
Explanation:
Mass of the car, m = 2000 kg
Force acting in the car, F = -2000 N
Initial speed of car, u = 20 m/s
Finally, it stops, v = 0
Using second equation of motion as :
[tex]F=ma[/tex]
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{-2000}{2000}[/tex]
[tex]a=-1\ m/s^2[/tex]
Let s is the stopping distance. Now using third equation of motion as :
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{0-(20)^2}{2\times -1}[/tex]
s = 200 meters
So, the stopping distance of the car is 200 meters. Hence, this is the required solution.
Use conservation of energy arguments to find the speed of the ball when it reaches the top of the circular path about the peg. your answer should be expressed in terms of l, g and
d.
Final answer:
Conservation of energy dictates that the speed of the ball at the top of the circular path decreases as its kinetic energy is converted to potential energy. The final speed can be calculated using the formula v = √(2gh).
Explanation:
Conservation of energy states that the total energy of a system remains constant. In the case of the ball on a string winding around a post, as the ball reaches the top, its kinetic energy will be converted to potential energy, resulting in a decreased speed.
To find the speed of the ball at the top of the circular path, we can set the initial kinetic energy equal to the final potential energy. This can be expressed as:
[tex]1/2 mv^2 = mgh[/tex], where 'v' is the final speed, 'm' is the mass of the ball, 'h' is the height above the peg, and 'g' is the acceleration due to gravity.
Therefore, the speed of the ball at the top of the circular path is v = √(2gh), where 'g' is the acceleration due to gravity and 'h' is the height of the circular path above the peg.
The given vector represents a hot air balloon that experiences a 0.25m/s^2 up and 0.25 m/s^2 to the right at 45 degrees. Find the magnitude of the resultant
when two vectors are inclined at some angle to each other then the magnitude of resultant of two vectors is given as
[tex]R = \sqrt{A^2 + B^2 + 2AB cos\theta}[/tex]
here A and B are the magnitude of two vectors and theta is the angle between them
now here we know that two vectors are of magnitude 0.25 m/s^2 each and they are inclined at angle of 45 degree
now we will plug in all data above
A = B = 0.25
[tex]\theta = 45[/tex]
[tex]a_{net} = \sqrt{0.25^2 + 0.25^2 + 2*(0.25)(0.25)cos45}[/tex]
[tex]a_{net} = 0.46 m/s^2[/tex]
So the magnitude of resultant acceleration will be 0.46 m/s^2
The magnitude of the resultant is 0.354 m/s^2.
Explanation:The magnitude of the resultant can be found using the Pythagorean theorem. Since the balloon experiences an upward acceleration of 0.25 m/s^2 and a rightward acceleration of 0.25 m/s^2 at 45 degrees, we can find the magnitude of the resultant using:
R = sqrt((0.25)^2 + (0.25)^2) = 0.354 m/s^2.
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List four conditions that may result in the emission of electrons from a conductor
There are 4 ways in which electrons are emitted from the conductor.
i. Thermionic emission
ii. Electric field electron emission
iii. Photoelectric emission
iv. Secondary emission
In thermionic emission large amount of external energy in the form of heat is supplied to release free electrons from the metal.
In electric field electron emission, electrons are emitted from the metal surface when the metals are placed in a very strong electric field.
During photoelectric emission, light is absorbed by the metals and this provides energy to the valence electrons which break their bond with the parent atom and which are then released from the atom.
Valence electrons do have some kinetic energy, but they don't have enough energy to escape from the atom. During secondary emission, a high-speed electron is bombarded with an atom, which provides the energy for the valence electrons to break their bonds with their parent atom which are then released from the atom.
Answer:
Conditions that result in the emission of electrons from a conductor:
Heating the conductor to a suitable temperature
Exposing the conductor to a strong light
Subjecting the conductor to a very high applied voltage
Subjecting the conductor to high-speed electrons from another source
Explanation:
if energy is conserved, why do people get tired , and why are we running short on “energy supplies?”
Which of these chemical compounds is generated during the process of cellular respiration
Cellular respiration is of two types namely aerobic and anaerobic respiration.
During aerobic respiration,the food(glucose) is broken down in the mitochondria to produce ATP,water and CO2.This is carried out in 3 steps glycolysis followed by Tricarboxylic acid cycle and electron transport chain.
During anaerobic respiration, glucose is broken down in the absence of oxygen to produce ethanol,CO2 and water. Anaerobic respiration occurs in the cytoplasm.
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
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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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]
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:
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.
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
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 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
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.
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
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
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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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.
To know more about Gravitational force, refer to the link:
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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.
As a pendulum swings, the amounts of potential and kinetic energy change. Does the amount of mechanical energy change?
No, the mechanical energy does not change if it is ideal.
We define mechanical energy as the sum of kinetic energy and potential energy
ME = PE + KE
When the pendulum swings, potential energy is converted to kinetic energy and vise versa. But the sum of both remains constant, which is the mechanical energy.
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.
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
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
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