Answer:
The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
Explanation:
Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ([tex]\dot Q[/tex]), measured in watts, that is, joules per second, by the following formula:
[tex]\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}[/tex] (1)
Where:
[tex]m[/tex] - Mass of the sphere, measured in kilograms.
[tex]c[/tex] - Specific heat of the material, measured in joules per kilogram-degree Celsius.
[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the sphere, measured in degrees Celsius.
[tex]\Delta t[/tex] - Time, measured in seconds.
In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:
[tex]\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}[/tex] (2)
Where:
[tex]m_{I}[/tex], [tex]m_{X}[/tex] - Masses of the iron and unknown spheres, measured in kilograms.
[tex]\Delta t_{I}[/tex], [tex]\Delta t_{X}[/tex] - Times of the iron and unknown spheres, measured in seconds.
[tex]c_{I}[/tex], [tex]c_{X}[/tex] - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.
[tex]c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}[/tex]
If we know that [tex]\Delta t_{I} = 6.35\,s[/tex], [tex]\Delta t_{X} = 4.59\,s[/tex], [tex]m_{I} = 0.515\,kg[/tex], [tex]m_{X} = 1.263\,kg[/tex] and [tex]c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the specific heat of the unknown material is:
[tex]c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)[/tex]
[tex]c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}[/tex]
Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
Your friend, a world-class long jumper, is trapped on the roof of a burning building. His only escape route is to jump to the roof of the next building. Fortunately for him, he is in telephone contact with you, a Physics 161 student, for advice on how to proceed. He has two options. He can jump to the next building by using the long-jump technique where he jumps at 45o to the horizontal. Or, he can take his chances by staying where he is in the hopes that the fire department will rescue him. You learn from the building engineers that the next building is 10 m away horizontally and the roof is 3 m below the roof of the burning building. You also know that his best long-jump distance is 7.9 m. What do you advise him to do
Answer:
y = -2.69 m
the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.
Explanation:
his problem must be solved with the missile launch equations.
Let's start by looking for the jumper's initial velocity
R = v₀² sin 2θ / g
for the long jump the angle used is tea = 45º, in the exercise they indicate that the best record is R = 7.9m
v₀² = R g / sin 2te
v₀ = [tex]\sqrt{ \frac{7.9 \ 9.8}{1 }[/tex]
v₀ = 8.80 m / s
Now suppose you jump with this speed to get to the other building, let's use trigonometry for the components of the speed
sin 45 = [tex]v_{oy}[/tex] /v₀
cos 45 = v₀ₓ / v₀
v_{oy} = v₀ sin 45
v₀ₓ = v₀ cos 45
v_{oy} = 8.8 sin 45 = 6.22 m / s
v₀ₓ = 8.8 cos 45 = 6.22 m / s
now let's calculate the sato with these speeds
x = [tex]v_{ox}[/tex] t
the minimum jump is x = 10 m
t = x / v₀ₓ
t = 10 / 6.22
t = 1.61 s
let's find the vertical distance for this time
y = v_{oy} t - ½ g t²
where zero is placed on the jump building
y = 6.22 1.61 - ½ 9.8 1.61²
y = -2.69 m
Let's analyze this result, the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.
When the bowling ball has fallen halfway down the building (height = 20 m), it has a speed of 19.8 m/s.
How much potential energy does the bowling ball have?
How much kinetic energy does the bowling ball have?
How much total energy (potential + kinetic) does the bowling ball have?
Of the bowling ball’s total energy, is more in the form of potential or kinetic energy?
Answer:
I think the answer is 19.8 potential energy
Explanation:
NONE.
An airplane is heading due south at a speed of 560 km/h . If a wind begins blowing from the southwest at a speed of 80.0 km/h (average). Calculate magnitude of the plane's velocity, relative to the ground.
Answer:
the magnitude of Vpg = 493.711 km/h
Explanation:
given data
speed Vpg = 560 km/h
speed Vwg = 80 km/h
solution
we get here magnitude of the plane velocity w.r.t. ground is
we know that the Vpg = Vpw + Vwg .....................1
writing the component of the velocity that is
Vpw = (0 km/h î - 560 km/h j )
Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)
adding these
Vpg = (0+80 cos 45 km/h ) î + ( -560 + 80 sin 45 km/h j)i
Vpg = (42.025 ) î (-491.92 km/h)j
now we take magnitude
the magnitude of Vpg = [tex]\sqrt{(42.025^2+(-491.92)^2)} km/h[/tex]
the magnitude of Vpg = 493.711 km/h
A cars mass is 950kg and it travels at a speed of 35 m/s when it rounds a flat curve of radius 215 m.
a. Determine the value of the frictional force exerted on the car.
b. Determine the value of the coefficient of friction between the tires and the road.
(a) It's the force of (static) friction that keeps the car on the road and prevents it from skidding, and this friction is directed toward the center of the curve.
Recall that centripetal acceleration has a magnitude a of
a = v ² / R
where
v = tangential speed
R = radius of the curve
so that
a = (35 m/s)² / (215 m) ≈ 5.69767 m/s² ≈ 5.7 m/s²
Parallel to the road, the only force acting on the car is friction. So by Newton's second law, we have
∑ F = Fs = m a
where
Fs = magnitude of static friction
m = mass of the car
Then
Fs = (950 kg) (5.7 m/s²) ≈ 5412.79 N ≈ 5400 N
(b) Perpendicular to the road, the car is in equilbrium, so its weight and the normal force of the road on the car are equal in magnitude. By Newton's second law,
N - W = 0
where
N = magnitude of normal force
W = weight
so that
N = W = m g = (950 kg) (9.8 m/s²) = 9310 N
Friction is proportional to the normal force by a factor of µ, the coefficient of static friction:
Fs = µ N
Assuming 35 m/s is the maximum speed the car can travel without skidding, we find
µ = Fs / N = (5400 N) / (9310 N) ≈ 0.581395 ≈ 0.58
what is the acceleration of an object that has a mass of 10kg and is pushed with a force of 50 N. just answer this question with the number no unit
Answer:
a = 5 [m/s²]
Explanation:
We know from Newton's second law that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = total force = 50 [N]
m = mass = 10 [kg]
a = acceleration [m/s²]
Now replacing:
[tex]a = \frac{F}{m} \\a=50/10\\a = 5 [m/s^{2} ][/tex]
Which of the following is a concern about the use of wind energy?
A. Wind farms produce a large amount of air and water pollution.
B. Wind farms may pose a threat to birds and bats.
O
C. Wind farms produce radioactive waste.
D. Wind farms often change the local weather patterns.
SUBMIT
Use of wind energy may pose a threat to birds and bats.
What is wind energy?The power of wind is harnessed to generate renewable energy through the use of wind turbines. These turbines are designed to capture the kinetic energy of the wind and convert it into electricity by rotating their blades. This process involves the movement of turbines, which in turn drives a generator that produces electricity. Wind energy is an eco-friendly and sustainable form of energy that does not emit any harmful gases or pollutants into the atmosphere. As countries worldwide seek to reduce their dependence on fossil fuels and transition towards a low-carbon economy, wind energy is gaining popularity as a clean source of electricity.
Which is a concern about the use of wind energy?The implementation of wind energy has raised concerns about its impact on wildlife, specifically birds and bats. The use of wind turbines can lead to collisions and habitat disruptions for these animals. Despite this, studies have shown that wind turbines cause fewer fatalities for birds and bats in comparison to other human-related sources such as buildings, power lines, and vehicles. To mitigate the impact on wildlife, measures like radar technology to detect animal movement and shutting down turbines during migration periods are being employed..
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According to Newton’s law of universal gravitation, in which of the following situations does the gravitational attraction between two bodies always increase
Answer:
When the mass increases or when distance between the bodies reduces
Explanation:
According to Newton's law of universal gravitation, the gravitational attraction between two bodies always increase if the mass increases and the distance between the bodies reduces.
The law of universal gravitation states that "the gravitational force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".
Mathematically;
Fg = [tex]\frac{G m1 m2}{r^{2} }[/tex]
G is the universal gravitation constant
m is the mass
r is the distance
__________ forces are not equal, and they always cause the motion of an object to change the speed and/or direction of an object.
Answer:
Unbalanced Forces.
Explanation:
Answer:
Unbalanced
Explanation:
Unbalanced forces are not equal and they always cause the motion of an object to change the speed and or direction of the object.
Balanced forces are equal and will not cause the motion of an object to change.
From Newton's law of inertia "an object will continue in its state of rest or of uniform motion unless acted upon by an external force".
The net force for a body with unbalanced forces is greater than zero.
For a body with balanced forces, the net force is zero
A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at some time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)
Answer:
Explanation:
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more .
Hence displacement is more in the downward slopping.
What is Displacement?Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more in the downward slopping.
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An astronaut with a mass of 75 kg stands on Mars (gravity = 3.7 m/s2). How much does the astronaut weigh?
Answer:
277.5 N.
Explanation:
From the question given above, the following data were obtained:
Mass (m) of astronaut = 75 kg
Acceleration due to gravity on Mar (gₘ) = 3.7 m/s²
Weight of astronaut on Mar (Wₘ) =.?
Weight of an object is simply defined by the following equation:
Weight (W) = mass × acceleration due to gravity (g)
W = m × g
With the above formula, we can obtain the weight of the astronaut on Mar as follow:
Mass (m) of astronaut = 75 kg
Acceleration due to gravity on Mar (gₘ) = 3.7 m/s²
Weight of astronaut on Mar (Wₘ) =.?
Wₘ = m × gₘ
Wₘ = 75 × 3.7
Wₘ = 277.5 N
Thus, the weight of the astronaut on Mar is 277.5 N
One disadvantage to experimental research is that experimental conditions do not always reflect reality.
Please select the best answer from the choices provided
T
F
Answer:
It's true I took the test on Edge.
Explanation:
Answer:
True
Explanation:
Got it right on edg
A Geiger–Müller tube can be used to detect ionising radiation. The tube is filled with low pressure gas. How does this detect ionising radiation
Explanation:
the gas will be hit by high energy particles as the radioactive decay happens and the decaying nuclei are breaking apart. these particles energize the gas and cause electrons to move around the orbitals. the flow of electrons is detected by the sensor.
The impulse given to a body of mass 1.5 kg, is 6.0 kg
• m•s-?
If the body was initially at rest, what
will its resulting kinetic energy be? Give your answer in J without units.
Show work
Answer:
12J
Explanation:
Given parameters:
Mass = 1.5kg
Impulse = 6kgm/s
let us start first by find the velocity with which this body moves;
Impulse = mass x velocity
Velocity = Impulse / mass = 6/ 1.5 = 4m/s
Initial velocity = 0m/s
Unknown:
Resulting kinetic energy = ?
Solution:
To solve this problem use the formula below:
K.E = [tex]\frac{1}{2}[/tex] m (v - u)²
m is the mass
v is the final velocity
u is the initial velocity
So;
K.E = [tex]\frac{1}{2}[/tex] x 1.5 x (4 - 0)²
K.E = 1.5 x 8 = 12J
How can force-time and force-displacement graphs be used to find the impulse or work done?
A. Area under force-time graph & Area under force-displacement graph
B. Area under force-displacement graph & Area under force-time graph
C. Gradient of force-time graph & Gradient of force-displacement graph
D. Gradient of force-displacement graph & Gradient of force-time graph
Answer:
A. Area under force-time graph & Area under force-displacement graph
Explanation:
To find the impulse or work done the area under force-time graph and area under force-displacement graph will give us these respective values.
Impulse = Force x time
Work done = Force x displacement
When we plot a graph of force and time, the area under it is the impulse.
When a graph of force and displacement is plotted, the area under is the work done.
Energy from the Sun is transferred from the Earth’s surface to the atmosphere, resulting in
atmospheric convection currents that produce winds. How do physical properties of the air
contribute to convection currents?
a -The warmer air sinks because it is more dense than cooler air.
b -The warmer air rises because it is more dense than cooler air.
c- The warmer air sinks because it is less dense than cooler air.
d -The warmer air rises because it is less dense than cooler air.
When an object with an electric charge of is from an object with an electric charge of , the force between them has a strength of . Calculate the strength of the force between the two objects if they are apart. Round your answer to significant digits.
The question is incomplete, the complete question is;
When an object with an electric charge of −7.0μC is 5.0cm from an object with an electric charge of 4.0μC, the force between them has a strength of 100.7N. Calculate the strength of the force between the two objects if they are 1.7cm apart. Round your answer to 2 significant digits
Answer:
865.1 N
Explanation:
F1 = Kq1q2/r1^2 ---------1
F2 = Kq1q2/r2^2 -------2
We have that;
r1 = 5cm
r2 =1.7 cm
F1 = 100.7 N
Comparing equations 1 and 2
F2 = F1r1^2/r2^2
F2 = 100.7N[(5cm)^2/(1.7cm)^2]
F2= 865.1 N
a 14n force is applied for 0.33 seconds, calculate the impulse
Answer:
4.62 N-s
Explanation:
recall that the formula for impulse is given by
Impulse = Force x change in time
in our case, we are given
Force = 14 N
change in time = 0.33s
Simply substituting the above into the equation for impulse, we get
Impulse = Force x change in time
Impulse = 14 x 0.33
= 4.62 N-s
[tex]\\ \sf\longmapsto Impulse=Force(Time)[/tex]
[tex]\\ \sf\longmapsto Impulse=14(0.33)[/tex]
[tex]\\ \sf\longmapsto Impulse=4.62Ns[/tex]
How does the force of friction for a sliding object vary with the area of contact?
Answer:
The force and texture of an object matter a lot.
Explanation:
If you were to try and run up a glass hill, could you? You maybe could, but it would be harder to than up carpet.
A tank, in the shape of a cube with each side measuring 2 meters, is half-full of water. Above the water is air that is pressurized to 5 kPa. Calculate the resultant force on one side of the tank due to the air and water (ignore atmospheric pressure) and determine the height of this force above the bottom of the tank.
Answer:
Explanation:
The tank is half full so height of water level is at 1 m . Centre of gravity of water will be at height of .5 m . Pressure will act at this point .
Pressure = h d g where h is height of centre of mass of water column , d is density of water and g is acceleration due to gravity .
Pressure of water column = .5 x 10³ x 9.8 = 4.9 k Pa .
Air is pressurized to 5 kPa so
resultant pressure on one side of the tank due to the air and water
= 4.9 + 5 kPa = 9.9 kPa .
Total force on one face = pressure x area of one face under water
= 9.9 x 10³ x .5 x 2²
= 19.8 kN .
A person pushes down on a lever with a force of 100 N. At the other end of the lever, a force of 200 N lifts a heavy object. What is the mechanical advantage of the lever?
A. 1/2, because the object will be lifted half the distance
B. -1, because the direction changes
C. 2, because the output force is twice the input force
D. 1, because the same amount of work is done
Answer:
Explanation:
C 200÷100=2
Output ÷ Input= MA
Your __________ environment includes the people you spend time with.
Answer:
social
Explanation:
Your social environment includes the people you spend time with – your family, your friends, classmates and other people in your community. Your social environment is healthier when you choose friends who show concern for their own health and yours.
When a rattlesnake strikes, its head accelerates from rest to a speed of 22 m/s in 0.48 seconds. Assume for simplicity that the only moving part of the snake is its head of mass 170 g. How much (average) power does the rattlesnake need to accelerate its head that fast? Answer in units of W.
Answer:
P = 85.72 W
Explanation:
Given that,
Initial speed, u = 0
Final speed, v = 22 m/s
Time, t = 0.48 s
Mass, m = 170 g = 0.17 kg
Let a be the acceleration of the rattlesnake.
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{22-0}{0.48}\\\\a=45.84\ m/s^2[/tex]
Let x is the displacement of a rattlesnake. It can be given by :
[tex]x=ut+\dfrac{1}{2}at^2\\\\x=0+\dfrac{1}{2}\times 45.84\times (0.48)^2\\\\x=5.28\ m[/tex]
The power of the rattlesnake is given by :
[tex]P=\dfrac{W}{t}\\\\P=\dfrac{m\times a\times x}{t}\\\\P=\dfrac{0.17\times 45.84\times 5.28}{0.48}\\\\P=85.72\ W[/tex]
So, the power of the rattlesnake is 85.72 W.
Which of these represent approaches to psychological science? (Choose every correct answer.)
Behavioral
Chemical
Investigative
Metaphysical
Sociocultural
Cognitive
Humanistic
Answer:
cognitive, humanistic, behavioral, sociocultural
Explanation:
Behavioral, sociocultural, cognitive, and humanistic are approaches to psychological science.
Psychology is a term to refer to the discipline that focuses on the study of various topics related to human thought such as:
The conductMental processes of individuals and human groups in different situations,Human experienceDue to the above, several subdisciplines have emerged that focus on the study of each of the topics. For example:
Behavioral psychology: focused on the study of human behavior.
Sociocultural psychology: focused on the study of human behavior and thought in different social situations.
Cognitive psychology: focused on mental processes related to learning.
Humanistic psychology: focused on the study of human thought from a comprehensive approach.
According to the above, options A, E, F, and G are correct because they mention different sub-disciplines of psychology while the other options mention terms that are not related to sub-disciplines or psychological sciences.
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what is energy? list the three mechanical forms of energy and their associated equations.
Answer:
mm hope that helps
Explanation:
What are the 3 forms of energy?
Forms of energy
Potential energy.
Kinetic energy.
Which has the most mass?
O The Moon
O A Pencil
O Your teacher.
O Earth
Answer:
Earth
lol... ....
A strong man and a weak man are trying to carry a ladder . How should they carry it in such a way that the weak person feels less weight of the ladder
Answer:
The strong person should carry the ladder at the front end and the weak person should carry it at the back end.
Explanation:
this is because in such a case the strong person has to pull the ladder whereas the weak person at the back end have to push the ladder. In such case it is easier to push because the weak person can use the force of gravity of his own body for pushing the ladde.
However in case of pulling the ladder one has to overcome his own gravity to pull the heavy object
I Hope it help you
Answer:
weak person: near the end of the ladder
strong person: nearer the center of the ladder
Explanation:
The closer the strong person is to the center of mass of the ladder, the greater the fraction of the ladder's weight that person will carry. The weak person will feel less weight if the strong person is nearer to the center of mass.
For example, if the strong person is 3/4 of the ladder's length from the end where the weak person is, then the strong person will carry twice the weight the weak person is having to carry. (Assuming the mass is uniformly distributed.)
A 1200 kg truck accelerates from 25 m/s to 53 m/s over a distance of 350 m.
What is the average net force on the truck?
Answer:
the average net force on the truck is 3744 N.
Explanation:
Given;
mass of the truck, m = 1200 kg
initial velocity of the truck, u = 25 m/s
final velocity of the truck, v = 53 m/s
distance traveled by the truck, d = 350 m
The acceleration of the truck is calculated as;
v² = u² + 2ad
53² = 25² + (2 x 350)a
700a = 53² - 25²
700a = 2184
a = 2184 / 700
a = 3.12 m/s²
The average net force on the truck is calculated as;
F = ma
F = 1200 x 3.12
F = 3744 N
Therefore, the average net force on the truck is 3744 N.
As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock. As the waves recede, they carry the sediment away. In this scenario, which process represents weathering, and which process represents erosion?
Answer:
WEATHERING is represented by the scenario (As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock).
Erosion is represented by the scenario (As the waves recede, they carry the sediment away).
Explanation:
A wave is a disturbance which travels through a medium and transfers energy from one point to another. When wind blows over a water body like the ocean, ocean waves are formed. As the generated energy from the wind is transported through the water by the waves, the can hit against rocks on the shores leading to its break down with time. WEATHERING occurs when tiny bit of sediments from rocks are loosened due to the impact of ocean waves.
Erosion can be described as the wearing away of the earth's surface due to the impact of wind, rainfall ( water) or waves. There are different types of erosion which is classified according it's cause of formation.
Wave erosion occurs when sediments such as sand, shell and other materials are carried to the shoreline by ocean waves. This erodes the shore over time as the sediments act like sandpapers.
A salmon jumps up a waterfall 2.4 m high. With what minimum speed did the salmon leave the water below to reach the top?
Answer:
6.86 m/s
Explanation:
The minimal velocity needed is when we have only vertical motion, then i will think in the problem only in one axis.
I suppose that the only force, in this case, is the gravitational force acting on the fish.
Then the gravitational equation of the fish will be:
a(t) = -9.8m/s^2
For the velocity equation we need to integrate over time to get:
v(t) = (-9.8m/s^2)*t + v0
Where v0 is the initial velocity of the fish and is what we want to find.
For the position equation we need to integrate over time again to get:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + v0*t + p0
p0 is the initial position of the fish, and because he starts one the water, the initial position is p0 = 0 m
Then the equation is:
p(t) = (1/2)*(-9.8 m /s^2)*t^2 + v0*t
p(t) = (-4.9 m/s^2)*t^2 + v0*t
We know that the maximum height is 2.4m
The value of time at which the fish gets his maximum height is when the velocity of the fish is equal to zero, then we first need to solve:
v(t) = (-9.8m/s^2)*t + v0 = 0
t = v0/9.8m/s^2
Now we replace this in the position equation to get the maxmimum height, which is equal to 2.4m
2.4m = p( v0/9.8m/s^2) = (1/2)*(-9.8 m /s^2)*(v0/9.8m/s^2)^2 + v0*(v0/9.8m/s^2)
2.4m = (1/2)(-v0)^2(-9.8 m /s^2) + v0^2/(9.8m/s^2))
2.4m = (1 - 1/2)*v0^2/(9.8m/s^2)
2.4m = 0.5*v0^2/(9.8m/s^2)
2.4m/0.5 = v0^2/(9.8m/s^2)
4.8m*(9.8m/s^2) = v0^2
√(4.8m*(9.8m/s^2)) = v0 = 6.86 m/s
16+109=................
Answer:
=125 tysm for points
Explanation:
Answer: 125
Explanation: I good at addition