Answer:hh
The acceleration may point in the opposite direct as the velocity
Explanation:
By definition free-fall is an uniform accelerate movement so speed always is increasing in fact acelertion and velocity always point in the same direction
Answer:
T h e a c c e l . e r a t i o n m a y p o i n t i n t h e o p p o s i t e d i r e c t a s t h e v e l o c i t y .
Explanation:
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 8.5 kg gibbon has an arm length (hand to shoulder) of 0.60 m. We can model its motion as that of a point mass swinging at the end of a 0.60-m-long, massless rod. At the lowest point of its swing, the gibbon is moving at 3.2 m/s .
What upward force must a branch provide to support the swinging gibbon?
Express your answer to two significant figures and include the appropriate units.
Answer:
230 N
Explanation:
At the lowest position , the velocity is maximum hence at this point, maximum support force T is given by the branch.
The swinging motion of the ape on a vertical circular path , will require
a centripetal force in upward direction . This is related to weight as follows
T - mg = m v² / R
R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s
T = mg - mv² / R
T = 8.5 X 9.8 + 8.5 X 3.2² / .60 { R is length of hand of ape. }
T = 83.3 + 145.06
= 228.36
= 230 N ( approximately )
The branch must provide an upward force of 227 Newtons to support a gibbon weighing 8.5 kg, swinging with a speed of 3.2 m/s, and a radius of 0.6 m.
Explanation:To calculate the upward force that the branch must provide, we need to take into account both the gravitational force acting on the gibbon and the centrifugal force due to its circular motion. The weight of the gibbon (gravitational force) is its mass times the acceleration due to gravity (8.5 kg * 9.8 m/s^2), which equals 83.3 N.
Centrifugal force is given by mass times velocity squared divided by radius of the motion (m*v^2/r), which results in (8.5 kg * (3.2 m/s)^2) / 0.6 m = 144 N.
By adding these two forces together, we find the total force the branch must supply at the lowest point: 83.3 N + 144 N = 227 N.The branch must provide an upward force of 227 Newtons to support the swinging gibbon.
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A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 12.0 m/s and accelerates at the rate of 0.650 m/s2 for 7.00 s. (a) What is his final velocity (in m/s)?
Answer:
Final speed of the racer, v = 16.55m/s
Explanation:
It is given that,
Initial velocity of the racer, u = 12 m/s
Acceleration, [tex]a=0.65\ m/s^2[/tex]
time taken, t = 7 s
We need to find the final velocity of the racer. Let it is given by v. It can be calculated using first equation of motion as :
[tex]v=u+at[/tex]
[tex]v=12+0.65\times 7[/tex]
v = 16.55 m/s
So, the final speed of the racer is 16.55 m/s. Hence, this is the required solution.
Compared to pure water, a salt water solution will have a
a. higher freezing point and a lower vapor pressure and boiling point.
b. lower vapor pressure and freezing point and a higher boiling point.
c. higher vapor pressure, freezing point and boiling point.
d. lower vapor pressure, freezing point and boiling point.
Answer:
lower vapor pressure and freezing point and a higher boiling point
Explanation:
All the properties listed above are colligative properties, they depend on the amount of solute present in a solution.
It is known that a solution has a lower vapour pressure than the corresponding pure solvent due to the presence of a solute. Also the freezing point of a solution is decreased while its boiling point is increased when compared with those of the pure solvent.
A salt water solution, compared to pure water, has a lower vapor pressure and freezing point, but a higher boiling point due to the impact of the salt, a non-volatile solute.
Explanation:Compared to pure water, a salt water solution has lower vapor pressure, lower freezing point, and a higher boiling point. The presence of salt, a non-volatile solute, reduces the vapor pressure of the solution, leading to a decreased freezing point and an increased boiling point. This is due to a phenomenon known as colligative properties, which states that adding a solute to a solvent affects the physical properties of the solvent, regardless of the nature of the solute.
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In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 80 kg, are pulled at a constant speed by a tow rope that maintains a constant tension of 360 N. How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?
Answer:
[tex]E_{th} = 19680 J[/tex]
Explanation:
GIVEN DATA:
Total mass ( rider + his tube) = 80 kg
tension Force = 360 N
height of slope = 30 m
length of slope = 120 m
we know that thermal energy is given as
E_{th} = W- Ug
W= F*d = (360N*120m)= 43200 J
Ug= m*g*h = (80kg*9.8m/s2*30m) = 23520 J
[tex]E_{th} = 43200J - 23520 J[/tex]
[tex]E_{th} = 19680 J[/tex]
The amount of thermal energy that is created in the slope and the tube during the ascent is 19680 Joules.
Given the following data:
Total mass = 80 kgTension = 360 NewtonHeight = 30 metersDisplacement = 120 metersTo find the amount of thermal energy that is created in the slope and the tube during the ascent:
By applying the Law of Conservation of Energy:
[tex]Work\;done = K.E + P.E + T.E[/tex]
Where:
K.E is the kinetic energy.P.E is the potential energy.T.E is the thermal energy.Since the rider and his tube are pulled at a constant speed, K.E is equal to zero (0).
Therefore, the formula now becomes:
[tex]Work\;done = 0 + P.E + T.E\\\\T.E = Work\;done-P.E[/tex]
For the work done:
[tex]Work\;done = Tensional\;force \times displacement\\\\Work\;done = 360 \times 120[/tex]
Work done = 43,200 Joules.
For P.E:
[tex]P.E = mgh\\\\P.E = 80\times9.8\times30[/tex]
P.E = 23,520 Joules.
Now, we can find the amount of thermal energy:
[tex]T.E = Work\;done-P.E\\\\T.E =43200-23520[/tex]
T.E = 19680 Joules.
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A uniform thin rod of length 0.700 m and mass 4.10 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 15 rad/s immediately after the collision, what is the bullet's speed just before impact? m/s
Answer:
[tex]v_o = 2761 m/s[/tex]
Explanation:
Since there is no external Force or Torque on the system of bullet and rod
So we can say that total angular momentum of the system will remain conserved about its center
So we will have
[tex]mv_o\frac{L}{2}sin\theta = (I_{rod} + I_{bullet})\omega[/tex]
here we know that
[tex]I_{rod} = \frac{mL^2}{12}[/tex]
[tex]I_{rod} = \frac{4.10\times 0.70^2}{12} [/tex]
[tex]I_{rod} = 0.167 kg m^2[/tex]
[tex]I_{bullet} = mr^2[/tex]
[tex]I_{bullet} = (0.003)(0.35^2)[/tex]
[tex]I_{bullet} = 3.675 \times 10^{-4} kg m^2[/tex]
[tex]\omega = 15 rad/s[/tex]
[tex]\theta = 60 [/tex]
now we have
[tex]0.003(v_o)(0.35)sin60 = (0.167 + 3.675 \times 10^{-4})15[/tex]
[tex]v_o = 2761 m/s[/tex]
This physics question is about calculating the speed of a bullet before it strikes a rotating rod, based on the conservation of angular momentum. After setting up the equations for initial and final angular momentum, by equating these two, we can find the speed of the bullet just before it hits the rod.
Explanation:This is a physics problem regarding the conservation of angular momentum during a collision. We are given a scenario where a bullet impacting and lodging into a rotating rod which is initially at rest. According to the conservation of angular momentum, the initial angular momentum prior to the collision should equal the angular momentum after the collision, if no external torque is acting on the system.
The initial angular momentum (just before the collision) is the product of the bullet's mass, speed, and distance from the axis of rotation (which is half of the rod's length), and the cos(θ), where θ is the angle the bullet's path makes with the rod. In this case, angular momentum is mbrvb cos(θ)r
The final angular momentum (just after the collision) is the moment of inertia of the system (bullet plus rod) times the ensuing angular velocity, which is (mb(r^2) + (1/12)M(r^2))ω
By setting the initial and the final angular momentum equations equal to each other and arranging for vb, we will find the speed of the bullet just before impact.
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A ball starts from rest and accelerates at 0.465 m/s2 while moving down an inclined plane 8.20 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 16.7 m, it comes to rest. What is the speed of the ball at the bottom of the first plane?
Answer:
2.76m/s
Explanation:
The position x for a given time t and constant acceleration a is:
(1) [tex]x=\frac{1}{2}at^2[/tex]
The velocity v:
(2) [tex]v=at[/tex]
Solving equation 2 for time t:
(3) [tex]t=\frac{v}{a}[/tex]
Combining equations 1 and 3:
(4) [tex]x=\frac{v^2}{2a}[/tex]
Solving equation 4 for velocity v:
(5) [tex]v=\sqrt{2ax}[/tex]
For a = 0.465m/s² and x = 8.2m:
v = 2.76m/s
At the beginning of a basketball game, the referee tosses the ball straight up with a speed of 4.6m/s. A player cannot touch the basketball until it has reached maximum height and begins to fall back down. What is the minimum time a player must wait before touching the ball?
Answer:
t=0.47s
Explanation:
the ball has uniformly accelerated movement due to gravity
Vo=initial speed=4.6m/s
g=gravity=-9.8m/s^2
Vf=final speed=0, the player must wait for the ball to stop. so the final speed will be 0
we can use the following ecuation
T=(Vf-Vo)/g
T=(0-4.6)/-9.8m/s^2
T=0.47s
A 35.1 kg wagon is towed up a hill inclined at 18.3 ◦ with respect to the horizontal. The tow rope is parallel to the incline and has a tension of 125 N in it. Assume that the wagon starts from rest at the bottom of the hill, and neglect friction. The acceleration of gravity is 9.8 m/s 2 . How fast is the wagon going after moving 75.4 m up the hill? Answer in units of m/s.
Answer:
4.933m/s
Explanation:
the wagon has a weight of 35.1kg*9.81m/s2 = 343.98N
of that weight 343.98N*sin(18.3)=108N are parallel to the hill and oposit tothe tension of the rope.
then, the force that is moving the wagon is 125N-108N=17N
F=m*a then 17N=35.1kg*a
a=0.4843 m/s2
we have two equations
[tex]v=a.t\\x=v.t+\frac{1}{2} . a . t^{2}[/tex]
then
[tex]x=a.t^{2} +\frac{1}{2}. a t^{2} \\x=\frac{3}{2}. a t^{2}[/tex]
[tex]t=\sqrt{\frac{2x}{3a} }[/tex]
t=10.188s
[tex]v=a.t[/tex]
v=4.933 m/s
From January 26, 1977, to September 18, 1983, George Meegan of Great Britain walked from Ushuaia, at the southern tip of South America, to Prudhoe Bay in Alaska, covering 30 600 km. In meters per second, what was the magnitude of his average velocity during that time period?
Answer:
0.146 m/s
Explanation:
We can see it in the pic.
Is it possible for an object’s velocity to increase while its acceleration decreases? (A) No, because if acceleration is decreasing the object will be slowing down (B) No, because velocity and acceleration must always be in the same direction (C) Yes, an example would be a falling object near the surface of the moon (D) Yes, an example would be a falling object in the presence of air resistance
Answer: Ok, the acceleration decreases means if you have an object with acceleration equal 100 meters for second square, and it starts to decrease slowly to 90, 80... etc, there you still have positive acceleration, so your object's velocity keeps increasing, but more slower than before.
remmember, acceleration means the change in velocity, if you have positive acceleration, your velocity will increase.
the correct answer is D, the air resistance will fight against the gravity, but the object will keep accelerating (with less intensity) ence will go faster and faster.
An object's velocity can increase even as its acceleration decreases, such as in the case of an object falling with air resistance where the rate of speed increase slows down due to the opposing force of air resistance.
Explanation:It is indeed possible for an object's velocity to increase while its acceleration decreases. This can occur, for example, during a scenario where an object continues to speed up, but the rate at which it is speeding up is decreasing over time. A typical example of this is an object in free fall with air resistance. Initially, as the object begins to fall, it accelerates due to gravity. However, as the object's speed increases, air resistance begins to counteract some of the acceleration due to gravity, causing the acceleration to decrease. Despite the reduction in acceleration, the object's velocity can continue to increase up until it reaches terminal velocity, at which point the acceleration will become zero, but the velocity remains constant and positive.
It is also important to note that acceleration is not always in the direction of motion. When an object is moving and its acceleration is in the direction of motion, it speeds up, while if the acceleration is in the opposite direction of motion, it will slow down or decelerate.
Some plants disperse their seeds by having the fruit split and contract, propelling the seeds through the air. The trajectory of these seeds can be determined by using a high-speed camera. In one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.7 m/s . The launch angle is measured from the horizontal, with +90∘ corresponding to an initial velocity straight upward and −90∘ straight downward.
Answer: 23000 frames/s
Explanation:
The rest of the statement of the question is presented below:
The experiment is designed so that the seeds move no more than 0.20 mm between photographic frames. What minimum frame rate for the high-speed camera is needed to achieve this?
We know the maximum initial speed at which the seeds are dispersed is:
[tex]V_{i}=4,7 m/s[/tex]
In addition, we know the maximum distance at which the seeds move between photographic frames is:
[tex]d_{max}=0.20 mm \frac{1m}{1000 m}=0.0002 m[/tex]
And we need to find the minimum frame rate of the camera with these given conditions. This can be found by finding the time [tex]t[/tex] for each frame and then the frame rate:
Finding the time:
[tex]t=\frac{d_{max}}{V_{i}}[/tex]
[tex]t=\frac{0.0002 m}{4.6 m/s}[/tex]
[tex]t=0.00004347 s/frame[/tex] This is the time for each frame
Now we need to find the frame rate, which is the frequency at which the photos are taken.
In this sense, frequency [tex]f[/tex] is defined as:
[tex]f=\frac{1}{t}[/tex]
[tex]f=\frac{1}{0.00004347 s/frame}[/tex]
Finally:
[tex]f=23000 frames/s[/tex]
Hence, the minimum frame rate is 23000 frames per second.
A hockey puck is struck so that it slides at a constant speed and strikes the far side of the rink, 58.2 m away. The shooter hears the sound of the puck after 1.9 seconds. The speed of sound is 340 m/s.
How fast was the puck moving?
Answer:
[tex]v = 33.66 m/s[/tex]
Explanation:
Let hockey puck is moving at constant speed v
so here we have
[tex]d = vt[/tex]
so time taken by the puck to strike the wall is given as
[tex]t = \frac{58.2}{v}[/tex]
now time taken by sound to come back at the position of shooter is given as
[tex]t_2 = \frac{58.2}{340}[/tex]
[tex]t_2 = 0.17s[/tex]
so we know that total time is 1.9 s
[tex]1.9 = t + t_2[/tex]
[tex]1.9 = t + 0.17[/tex]
[tex]1.9 - 0.17 = t[/tex]
[tex]t = 1.73 s[/tex]
now we have
[tex]1.73 = \frac{58.2}{v}[/tex]
[tex]v = 33.66 m/s[/tex]
A coach tells his Little League players that hitting a 0.275 batting average, within 7% error, means that they had a really great season. Seven-year-old Tommy ended the season with a 0.258 batting average. According to his coach, did Tommy have a great season?
Yes. According to the coach's mathematical criteria, Tommy had a great season.
Sadly, Tommy doesn't even know it. His tone-deaf coach decided to describe success in terms that are absurd for 7-yr-olds, as well as for most of their parents.
A steel ball is dropped from a building's roof and passes a window, taking 0.14 s to fall from the top to the bottom of the window, a distance of 1.20 m. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.14 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 2.22 s. How tall is the building?
Answer:
The building is 20.68m high.
Explanation:
In order to solve this problem we must first do a drawing of what the situation looks like. (Look at the attached picture). Then, we determine what was the velocity of the ball when it reached the top of the window (let's call this [tex]V_{0A}[/tex] and when it reached the bottom of the window (let's call this [tex]V_{0B}[/tex]. We can start by finding the velocity the ball has at the top of the window [tex]V_{oA}[/tex] when it's falling.
In order to do so, we can use the following equation:
[tex]\updelta y=V_{0A}t+\frac{1}{2} at^{2}[/tex]
We can now solve this equation for [tex]V_{0A}[/tex] so we get:
[tex]v_{0A}=\frac{\updelta y-\frac{1}{2} at^{2}}{t}[/tex]
or when simplified we get:
[tex]v_{0A}=\frac{2\updelta y-at^{2}}{2t}[/tex]
now we can substitute values in the formula so we get:
[tex]v_{0A}=\frac{2(-1.20m)-(-9.8m/s^{2})(0.14s)^{2}}{2(0.14)}[/tex]
and solve:
[tex]v_{0A}=-7.886m/s[/tex]
Once we got the first velocity, we can proceed and find the second velocity.
We can find that by using the following formula:
[tex]a=\frac{V_{f}^{2}-V_{0}^{2}}{2y}[/tex]
So now we can solve for [tex]V_{f}[/tex]
When solving for [tex]V_{f}[/tex], we get the following formula:
[tex]V_f=\sqrt{2ya+V_{0}^{2}}[/tex]
So we can now substitute some values.
[tex]V_{0B}=\sqrt{2(-1.20m)(-9.8m/s\{2})+(-7.885m/s)^{2}}[/tex]
When solving this equation we get an answer of:
[tex]V_{0B}=-9.257m/s[/tex]
Once we got those velocities, we can use them to find the distance from the roof to the highest part of the window and the distance between the
lowest part of the window and the floor.
Let's start with the top portion of the window:
[tex]y=\frac{V_{f}^{2}-V_{0}^{2}}{2A}[/tex]
in this case, the initial velocity is 0 because we are dropping the ball from the roof. So we can now substitute values so we get:
[tex]y=\frac{(-7.885m/s)^{2}-(0)^{2}}{2(-9.8m/s^{2}}[/tex]
when solving this for y we get:
y=-3.17m
Now, we can find the height between the lowest part of the window and the floor, so we get:
[tex]\updelta y=V_{0B}t+\frac{1}{2} at^{2}[/tex]
In this case the time is 1.11s because it's half of the total time the ball is beneath the window when it's falling. So when substituting values we get:
[tex]\updelta y=(-9.257m/s^{2})(1.11s)+\frac{1}{2} (-9.8m/s^{2})(1.11s)^{2}[/tex]
When solving that expression we get that:
[tex]\updelta y=-16.31m[/tex]
so now we have enough information to solve the problem, notice that the heights appear as negative. This is because the bal is falling at this time. Since we only care about the magnitud, we can make them positive.
So in order to find the height of the building, we must ad the three lengths we just found, so we get:
h=3.17m+1.20m+16.31m=20.68m
So the building has a height of 20.68m
Scientists use different types of microscopes to observe objects that are not visible to the naked eye. A scientist is reviewing various samples of pond water to determine what species of microorganisms live in the pond. The scientist wishes to make drawings of the structure of each microorganism and study each one's method of movement. Which of the following microscopes would be best for the scientist to use?a. transmission electron microscope
b. scanning electron microscope
c. compound light microscope
d. dissecting microscope
Answer:
c. compound light microscope
Explanation:
Microorganisms (such as bacteria, fungi and protozoa) can be observed through a compound light microscope, which allows to increase the observed image since 400 times to 1000 times (in more advanced microscopes). This kind of organisms have sizes in the order of micrometers (μm, a million times lesser than a meter), so you cannot observe them with a dissecting microscope, which only increase the image 10-30 times. In fact, dissecting microscopes are employed to observe thin sections of tissues (of plants and animals) and, as the name say it, to "dissect" in very tiny parts.
The other two microscopes, transmission electron and scanning electron microscopes, are used to observe macromolecules, and physical changes in compounds and matter, in the order of nanometers (nm, a thousand million times lesser than a meter).
Final answer:
The scientist should use a compound light microscope to observe and draw living microorganisms from pond water, as it allows for the necessary magnification while preserving the organisms' life for movement study.
Explanation:
To observe and draw the structure of microorganisms in pond water and study their movement, the best type of microscope for a scientist to use would be a compound light microscope. This type of microscope uses visible light that passes through and is bent by the lens system, allowing the user to observe living organisms, which is essential for studying movement. Moreover, light microscopes can magnify cells up to approximately 400 times, which is typically sufficient for viewing microorganisms.
While electron microscopes like the transmission electron microscope and scanning electron microscope provide much higher magnification and resolution, they are not suitable for observing living specimens because the sample preparation process kills the organisms. Dissecting microscopes, on the other hand, provide a three-dimensional view of the specimen but have lower magnification and are more suitable for larger objects such as tissues, not microorganisms in pond water.
A chair of mass 15.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force of 60.0 N that is directed at an angle of 40.0° below the horizontal, and the chair slides along the floor. Use a free-body diagram and Newton’s laws to calculate the normal force that the floor exerts on the chair.
Answer:
[tex]F_n = 185.72 N[/tex]
Explanation:
As per FBD we can say that net downward force perpendicular to contact plane must be counterbalanced by the net upward Normal force
So here we will say
[tex]F_n = F sin40 + mg[/tex]
now we have
[tex]F = 60.0 N[/tex]
[tex]m = 15 kg[/tex]
now we have
[tex]F_n = 60 sin40 + 15(9.81)[/tex]
[tex]F_n = 185.72 N[/tex]
The normal force that the floor exerts on the chair is approximately 185.72 N.
First, let's draw a free-body diagram for the chair.
The forces acting on the chair are the force you apply (60.0 N) and the weight of the chair (mg), where m is the mass of the chair (15.0 kg) and g is the acceleration due to gravity (9.81 m/s^2).
Now, since the chair is sliding along the floor, there must be a frictional force opposing its motion. However, the problem doesn't provide the coefficient of friction or any information about it. Therefore, we cannot consider the frictional force in this calculation.
Next, let's break down the force you apply into its components. The force you apply is directed at an angle of 40.0° below the horizontal. We can resolve this force into two components: one perpendicular to the floor and one parallel to the floor.
The component perpendicular to the floor is given by F * sinθ, where F is the magnitude of the force (60.0 N) and θ is the angle (40.0°). This component contributes to the normal force.
The component parallel to the floor is given by F * cosθ. This component contributes to the frictional force, but since we are not considering friction in this calculation, we won't use this component.
Now, let's calculate the normal force. The normal force is the net force acting perpendicular to the contact plane. We can find it by adding the force component perpendicular to the floor (F * sinθ) and the weight of the chair (mg).
So, the normal force (Fn) is given by Fn = F * sinθ + mg.
Substituting the given values, we have:
Fn = 60.0 N * sin(40.0°) + 15.0 kg * 9.81 m/s².
Evaluating this expression, we find that the normal force is approximately 185.72 N.
Therefore, the normal force that the floor exerts on the chair is approximately 185.72 N.
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A plane is flying horizontally with speed 167 m/s at a height 4040 m above the ground, when a package is dropped from the plane. The acceleration of gravity is 9.8 m/s 2 . Neglecting air resistance, when the package hits the ground, the plane will be:
Answer:
The plane will be 4795.23 meters from where he threw the package.
Explanation:
Data:
v = 167 m/s
h = 4040 m
g = 9.8m/s²
package:
after it is dropped there is no horizontal force, so...
[tex]X - Xo = Vt[/tex]
t = [tex]\frac{X - Xo}{V}[/tex] I
in the vertical we have only [tex]P = mg[/tex]
[tex]Y - Yo = Vot -\frac{gt^{2} }{2}[/tex] but Vo = 0 because the package is dropped
[tex]Y - Yo =-\frac{gt^{2} }{2}[/tex] II
replacing I in II
[tex]Y - Yo =-\frac{g(X - Xo)^{2} }{2V^{2}}[/tex]
[tex]0 - 4040 =-\frac{9.8(X - Xo)^{2} }{2*167^{2}}[/tex]
X - Xo = 4795.23
The distance from where the plane drops the package and where it hits the ground is the same as the plane flies horizontally, as there is no acceleration at x.
A football is kicked vertically upward from the ground and a student gazing out of the window sees it moving upward past her at velocity U. The window is a height h above the ground. You may ignore air resistance. How high does the football go above ground?
Answer:
[tex]H = \frac{u^2}{2g} + h[/tex]
Explanation:
Let the football is kicked up vertically with some speed given as
[tex]v = v_o[/tex]
now its speed when it will reach to height "h" above the ground is given as "u"
so we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
here we have
[tex]u^2 - v_o^2 = 2(-g)h[/tex]
so we have
[tex]v_o^2 = u^2 + 2gh[/tex]
now we know that when football will reach to maximum height then it will have zero final velocity
So we will have
[tex]v_f^2 - v_i^2 = 2 a s[/tex]
[tex]0 - (u^2 + 2gh) = 2(-g)H[/tex]
so we have maximum height given as
[tex]H = \frac{u^2 + 2gh}{2g}[/tex]
[tex]H = \frac{u^2}{2g} + h[/tex]
The maximum height a football reaches after being kicked upward past a window can be computed by using the kinematic equation and setting the final velocity to zero at the peak. The formula H = h + (U²) / (2g) allows us to find the maximum height by adding the window's height above ground to the kinetic energy at the window level, divided by twice the acceleration due to gravity.
Explanation:Calculating the Maximum Height of a Football Kicked Upward
To find the maximum height a football reaches after being kicked upward past a window at velocity U and height h, we can use the principles of kinematics under constant acceleration due to gravity. Initially, when the ball is kicked, it possesses kinetic energy which gets converted into gravitational potential energy as it ascends. The football reaches its maximum height when its velocity becomes zero.
Using the kinematic equation:
v² = u² + 2gh
Where:
v is the final velocity (0 m/s at the maximum height)u is the initial velocity (U at the window height)g is the acceleration due to gravity (-9.8 m/s²)h is the change in height (maximum height - window height)Let H be the maximum height above ground and h the window height above ground, then:
0 = U² - 2g(H - h)
Solving for H gives us:
H = h + (U²) / (2g)
Plugging in the values for U and h as well as g, we can calculate the maximum height H.
Consider a deer that runs from point A to point B. The distance the deer runs can be greater than the magnitude of its displacement, but the magnitude of the displacement can never be greater than the distance it runs.True / False.
Answer:
True
Explanation:
Distance is defined as the length of the actual path traveled by the body.
Displacement is defined as the minimum distance between the two points.
the magnitude of displacement is always less than or equal to the distance traveled by the body.
As a deer runs from A to b , so it means the distance traveled by the deer is either equal to the magnitude of displacement or always greater than the magnitude of displacement of the deer.
Displacement can never be greater than the distance.
Thus, the option is true.
A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first rebound. (Ignore the small amount of time the ball is in contact with the floor.) (a) Determine the ball’s speed just before it hits the floor on the way down. (b) Determine the ball’s speed as it leaves the floor on its way up to its first rebound height. (c) How long is the ball in the air from the time it is dropped until the time it reaches its maximum height on the first rebound?
Answer:
a) [tex]V=14.01 m/s[/tex]
b) [tex]V=8.86 \, m/s[/tex]
c)[tex]t = 2.33s[/tex]
Explanation:
Our most valuable tool in solving this problem will be the conservation of mechanical energy:
[tex]E_m = E_k +E_p[/tex]
That is, mechanical energy is equal to the sum of potential and kinetic energy, and the value of this [tex]E_m[/tex] mechanical energy will remain constant. (as long as there is no dissipation)
For a point particle, we have that kinetic energy is:
[tex]E_k = \frac{1}{2} m \, V^2[/tex]
Where m is the mass, and V is the particle's velocity,
Potential energy on the other hand is:
[tex]E_p= m\, g\, h[/tex]
where g is the acceleration due to gravity ([tex]g=9.81 \, m/s^2[/tex]) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which [tex]h=0[/tex], a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.
Let's calculate the mechanical energy just before the ball is dropped:
As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:
[tex]E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m[/tex]
That's it, the actual value of m is not important now, as we will see.
Now, what's the potential energy at the bottom? Let's see:
At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set [tex]h=0[/tex]
[tex]E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2[/tex]
So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:
[tex]E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\[/tex]
And so we have found the velocity of the ball as it hits the floor.
[tex]V = \sqrt[]{2g\cdot 10m}=14.01\, m/s[/tex]
Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:
[tex]V = \sqrt[]{2g\cdot h}[/tex]
where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:
[tex]V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s[/tex]
Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.
As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average? It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:
[tex]V_{avg}\cdot t=h\\t=h/V_{avg}[/tex]
what's the average speed when the ball is descending?
[tex]V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s[/tex]
so the time it takes the ball to go down is:
[tex]t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\[/tex]
Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:
[tex]V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s[/tex]
and the time it takes to go up is:[tex]t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s[/tex]
When we add both times , we get:
[tex]t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s[/tex]
To find the speed of the tennis ball just before it hits the floor on its way down, we can use the equation y = 0 + voyt - 1/2gt^2. The graph of the ball's vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec and end at v = -0.98 m/s at t = 0.65 sec. The time the ball is in the air from the time it is dropped until it reaches its maximum height on the first rebound is 2.5 s.
Explanation:(a) While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We can find the time for this by using Equation 4.22: y = yo + voyt - 1/2gt^2. If we take the initial position yo to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:
(b) The graph of the ball's vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec (there should be a data point at t = 0.05 sec, v = 4.90 m/s). It should then have a slope of -9.8 m/s², crossing through v = 0 at t = 0.55 sec and ending at v = -0.98 m/s at t = 0.65 sec.
(c) This gives t = 2.5 s. Since the ball rises for 2.5 s, the time to fall is 2.5 s.
Learn more about Speed of a tennis ball during free fall here:https://brainly.com/question/33880642
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Soon after Earth was formed, heat released by the decay of radioactive elements raised the average internal temperature from 300 to 3000 K, at about which value it remains today. Assuming an average coefficient of volume expansion of 3.2 × 10–5 K–1, by how much has the radius of Earth increased since the planet was formed?
Answer:
97.03%
Explanation:
The equation for volumetric expansion due to thermal expansion is as follows
V/Vo=(1+γΔT)
V=final volume
Vo=initial volume
γ=coefficient of volume expansion=3.2 × 10–5 K–1
ΔT=
temperature difference
assuming that the earth is a sphere the volume is given by
V=(4/3)pi R^3
if we find the relationship between the initial and final volume we have the following
[tex]\frac{V}{Vo} =\frac{ \frac{4}{3} \pi r^{3} }{ \frac{4}{3} \pi ro^{3}}=\frac{r^{3} }{ro^{3}}[/tex]
taking into account the previous equation
r/ro=(1+γΔT)^(1/3)
r/r0=(1-3.2x10-5(3000-300))^(1/3)=
r/ro=0.9703=97.03%
find the x-component of this vector 18.4,0.250. Remember, angles are measured from the +x axis. Find the x-component and y-component
Answer:
x = 0.237
y = 0.0789
Explanation:
Vector with direction 18.4° and magnitude 0.250 has x and y components of:
x = 0.250 cos 18.4°
x = 0.237
y = 0.250 sin 18.4°
y = 0.0789
Answer: x = 0.237
y = -0.0789
Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of the particles have a negative charge: q1 = -8.2 nC and q2 = -16.4 nC . The remaining particle has a positive charge, q3 = 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?
The net electric force acting on particle 3 due to particle 1 and particle 2 is 4.55 N.
Explanation:To find the net electric force acting on particle 3 due to particle 1 and particle 2, we can use Coulomb's Law, which states that the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is: F = k × (|q1 × q3| / r^2)
Where:
F is the net electric forcek is the electrostatic constantr is the distance between the chargesq1 and q3 are the charges of particle 1 and particle 3 respectivelySubstituting the given values into the formula:
F = (9 × 10^9 Nm^2/C^2) × (|(-8.2 × 10^-9 C) × (8.0 × 10^-9 C)| / (3.3 × 10^-2 m)^2)
Simplifying the equation, we get:
F = 4.55 N
Therefore, the net electric force acting on particle 3 is 4.55 N.
The net electric force acting on particle 3 is approximately [tex]\( 1.62 \times 10^{-3} \, \text{N} \)[/tex] at an angle of [tex]\( 30^\circ \)[/tex] from the horizontal axis.
Given:
[tex]\( q_1 = -8.2 \, \text{nC} \)\\ \( q_2 = -16.4 \, \text{nC} \)\\ \( q_3 = 8.0 \, \text{nC} \)[/tex]
- Side length of the equilateral triangle, [tex]\( a = 3.3 \, \text{cm} = 0.033 \, \text{m} \)[/tex]
Calculate the Magnitude of the Forces
The magnitude of the force between two charges is given by Coulomb's law:
[tex]\[ F = k_e \frac{|q_1 q_2|}{r^2} \][/tex]
where:
- [tex]\( k_e \)[/tex] is Coulomb's constant, [tex]\( k_e = 8.99 \times 10^9 \, \text{N.m}^2/\text{C}^2 \)[/tex]
- r is the distance between the charges, [tex]\( r = 0.033 \, \text{m} \)[/tex]
Force between [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex]:
[tex]\[ F_{13} = k_e \frac{|q_1 q_3|}{a^2} = 8.99 \times 10^9 \frac{|-8.2 \times 10^{-9} \times 8.0 \times 10^{-9}|}{(0.033)^2} \]\[ F_{13} = 8.99 \times 10^9 \frac{65.6 \times 10^{-18}}{0.001089} \]\[ F_{13} = 8.99 \times 10^9 \times 6.025 \times 10^{-14} \]\[ F_{13} = 5.41 \times 10^{-4} \, \text{N} \][/tex]
Force between [tex]\( q_2 \)[/tex] and [tex]\( q_3 \)[/tex]:
[tex]\[ F_{23} = k_e \frac{|q_2 q_3|}{a^2} = 8.99 \times 10^9 \frac{|-16.4 \times 10^{-9} \times 8.0 \times 10^{-9}|}{(0.033)^2} \]\[ F_{23} = 8.99 \times 10^9 \frac{131.2 \times 10^{-18}}{0.001089} \]\[ F_{23} = 8.99 \times 10^9 \times 1.205 \times 10^{-13} \]\[ F_{23} = 1.08 \times 10^{-3} \, \text{N} \][/tex]
Determine the Direction of the Forces
Since the triangle is equilateral, the angles between the forces are 60 degrees. The forces are directed along the lines connecting the charges.
- [tex]\( F_{13} \)[/tex] is directed from [tex]\( q_1 \)[/tex] to [tex]\( q_3 \)[/tex] (repulsive force, but [tex]\( q_1 \)[/tex] is negative, so the force direction is towards [tex]\( q_1 \))[/tex].
- [tex]\( F_{23} \)[/tex] is directed from [tex]\( q_2 \)[/tex] to [tex]\( q_3 \)[/tex] (repulsive force, but [tex]\( q_2 \)[/tex] is negative, so the force direction is towards [tex]\( q_2 \))[/tex].
Calculate the Net Force
We need to resolve the forces into components and sum them.
Resolving [tex]\( F_{13} \)[/tex]:
- Along x-axis: [tex]\( F_{13x} = F_{13} \cos(30^\circ) = 5.41 \times 10^{-4} \cos(30^\circ) \)[/tex]
- Along y-axis: [tex]\( F_{13y} = F_{13} \sin(30^\circ) = 5.41 \times 10^{-4} \sin(30^\circ) \)[/tex]
Resolving [tex]\( F_{23} \)[/tex]:
- Along x-axis: [tex]\( F_{23x} = F_{23} \cos(30^\circ) = 1.08 \times 10^{-3} \cos(30^\circ) \)[/tex]
- Along y-axis: [tex]\( F_{23y} = F_{23} \sin(30^\circ) = 1.08 \times 10^{-3} \sin(30^\circ) \)[/tex]
Since both forces have the same y-component direction:
[tex]\[ F_{13x} = 5.41 \times 10^{-4} \times \frac{\sqrt{3}}{2} \approx 4.68 \times 10^{-4} \, \text{N} \]\[ F_{13y} = 5.41 \times 10^{-4} \times \frac{1}{2} \approx 2.70 \times 10^{-4} \, \text{N} \]\[ F_{23x} = 1.08 \times 10^{-3} \times \frac{\sqrt{3}}{2} \approx 9.35 \times 10^{-4} \, \text{N} \]\[ F_{23y} = 1.08 \times 10^{-3} \times \frac{1}{2} \approx 5.40 \times 10^{-4} \, \text{N} \][/tex]
The net force components are:
[tex]\[ F_{net_x} = F_{13x} + F_{23x} = 4.68 \times 10^{-4} + 9.35 \times 10^{-4} \approx 1.40 \times 10^{-3} \, \text{N} \]\[ F_{net_y} = F_{13y} + F_{23y} = 2.70 \times 10^{-4} + 5.40 \times 10^{-4} \approx 8.10 \times 10^{-4} \, \text{N} \][/tex]
Calculate the Magnitude of the Net Force
[tex]\[ F_{net} = \sqrt{F_{net_x}^2 + F_{net_y}^2} = \sqrt{(1.40 \times 10^{-3})^2 + (8.10 \times 10^{-4})^2} \]\[ F_{net} = \sqrt{1.96 \times 10^{-6} + 6.56 \times 10^{-7}} \]\[ F_{net} = \sqrt{2.62 \times 10^{-6}} \]\[ F_{net} \approx 1.62 \times 10^{-3} \, \text{N} \][/tex]
Determine the Direction of the Net Force
The direction of the net force can be found using the angle:
[tex]\[ \theta = \tan^{-1} \left( \frac{F_{net_y}}{F_{net_x}} \right) = \tan^{-1} \left( \frac{8.10 \times 10^{-4}}{1.40 \times 10^{-3}} \right) \]\[ \theta = \tan^{-1} (0.579) \]\[ \theta \approx 30^\circ \][/tex]
Ship A is located 4.2 km north and 2.7 km east of ship B. Ship A has a velocity of 22 km/h toward the south and ship B has a velocity of 39 km/h in a direction 36 ° north of east. What are the (a) x-component and (b) y-component of the velocity of A relative to B?
Answer:
(a) The x-component of velocity is 31.55 km/h
(b) The y-component of velocity is 44.92 km/hr
Solution:
As per the solution:
The relative position of ship A relative to ship B is 4.2 km north and 2.7 km east.
Velocity of ship A, [tex]\vec{u_{A}}[/tex] = 22 km/h towards South = [tex]- 22\hat{j}[/tex]
Velocity of ship B, [tex]\vec{u_{B}}[/tex] = 39 km/h Towards North east at an angle of [tex]36^{\circ}[/tex] = [tex]\vec{u_{B}} = 39sin36^{\circ} \hat{j}[/tex]
Now, the velocity of ship A relative to ship B:
[tex]\vec{u_{AB}} = \vec{u_{A}} - \vec{u_{B}}[/tex]
[tex]\vec{u_{A}} = - 22\hat{j}[/tex]
[tex]\vec{u_{B}} = 39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}[/tex]
Now,
[tex]\vec{u_{AB}} = - 22\hat{j} +39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}[/tex]
[tex]\vec{u_{AB}} = 31.55\hat{i} - 44.92\hat{j}[/tex]
A 7.1 cm diameter horizontal pipe gradually narrows to 5.4 cm . When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.5 kPa and 20.6 kPa , respectively. What is the volume rate of flow?
Answer:
Q=22.6L/s
Explanation:
First you must consider the continuity equation at points 1 and 2, which indicates that both flows are of equal value, in this way you get an equation between the two flow rates.
Then you raise the Bernoulli equation taking into account that the height is the same, which makes the term h1-h2 zero.
Using the equations above to calculate one of the speeds.
Finally you find the flow by multiplying the speed by the area.
I attached procedure
Why is net force a vector sum A.All forces have direction and magnitude B.Forces can only be attractive C.All forces are contact forces D.There are four fundamental forces
Answer:
A
Explanation:
Forces are vector quantities because they have both magnitude and direction.
A is the right answer i just got it right for A P E X
The captain orders his starship to accelerate from rest at a rate of "1 g" (1 g = 9.8 m/s2). How many days does it take the starship to reach 5% the speed of light?
(Light travels at 3.0 × 10^8 m/s.)
Answer:
17.7 days
Explanation:
Since the ship accelerates from the rest, its initial velocity would be equal to 0.
So,
[tex]v_{i}=0[/tex]
Acceleration of the star-ship = a = 1 g = 9.8 m/s²
We need to find how many days will it take the ship to reach 5% of the speed of light. Speed of light is [tex]3 \times 10^{8}[/tex] m/s.
5% of the speed of light = [tex]0.05 \times 3 \times 10^{8}=1.5\times 10^{7}[/tex] m/s
This means, the final velocity of the star-ship will be:
[tex]v_{f}=1.5\times 10^{7}[/tex]
We have the initial velocity, final velocity and the acceleration. We need to find the time(t). First equation of motion relates these quantities as:
[tex]v_{f}=v_{i}+at[/tex]
Using the values in this equation, we get:
[tex]1.5 \times 10^{7}=0+9.8(t)\\\\ t=\frac{1.5\times10^{7}}{9.8}\\\\ t=1,530,612.245[/tex]
Thus, the star-ship will take 1,530,612.245 seconds to reach to 5% the speed of light. Now we need to convert this time to days.
Since, there are 60 seconds in a minute:
1,530,612.245 seconds = [tex]\frac{1,530,612.245}{60}=25510.20[/tex] minutes
Since, there are 60 minutes in an hour:
25,510.20 minutes = [tex]\frac{25,510.20}{60}=425.17[/tex] hours
Since, there are 24 hours in a day:
425.17 hours = [tex]\frac{425.17}{24}=17.7[/tex] days
Thus, it will take approximately 17.7 days (approximately 17 days and 17 hours) to reach to 5% the speed of light
An airplane flying horizontally at a constant speed of 350 km/h over level ground releases a bundle of food supplies. Ignore the effect of the air on the bundle. What is the bundle's initial vertical velocity?
Answer:
0 km/h vertically, horizontally yet 350 km/h
Explanation:
The bundle was inside the plane, so it start a free fall movement from rest, i.e., 0km/h, but of course the speed increases with gravity, at 9,8m/s^2 over time, but since here it is asked the vertical speed at the beginning, it is 0 km/h.
The taillights of a vehicle are dimly illuminated whenever the brake lights are on. Technician A says this could be caused by a blown fuse. Technician B says an open ground circuit could cause this problem.
Who is correct?
Answer: Technician B
Explanation: A taillights, and several kind of ilumation devices, works by converting a current flow trought a material in Light. This mean that the Current Flow (Amperes) which goes through a lights is proportional to the light it gives (Lumens).
This means, that if the tailights has a dimly light the current that goes through is less than the one it is suppose to be.
A fuse is a protection device for over-current, if the current at any given time goes beyond a limit (designed on the fuse) the fuse will melt and cutting all the current to the circuit. A blown fuse will cut all the current from the circuit and the tailights will be completely off.
However a open ground circuit is a differente kind of failure. In this cases, there is a pact from where the current on the circuit "escapes" from it. This could be by several reason, unprotected wires the most usual. While the current escapes from his intended course, not all of the energy goes away from the load. This explain why the tailight still has enough energy through it to light dimly.
Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one?
a) 16F
b) 8F
c) 4F
d) F
e)F/4
Answer:
New force, F' = F
Explanation:
Given that, two small balls, A and B, attract each other gravitationally with a force of magnitude F. It is given by :
[tex]F=G\dfrac{m_Am_B}{r^2}[/tex]
If we now double both masses and the separation of the balls, the new force is given by :
[tex]F'=G\dfrac{2m_A\times 2m_B}{(2r)^2}[/tex]
F' = F
So, the new force remains the same as previous one. Hence, the correct option is (d) "F"
Answer:
After studying the law of gravitational attraction, students constructed a model to illustrate the relationship between gravitational attraction (F) and distance. If the distance between two objects of equal mass is increased by 2, then the gravitational attraction (F) is 1/4F or F/4. How would this model, situation A, change if the mass of the spheres is doubled?
A) A
B) B
C) C
D) D
If you came here from usa test prep it is:
Actually A
But for the question given right now is D.
Explanation: