A 1.8 kg book has been dropped from the top of the football stadium. Its speed is 4.8 m/s when it is 2.9 meters above the ground.
A). What is its total mechanical energy? (mechanical energy includes kinetic and gravitational potential)
B). What was the total mechanical energy of the book at the instant it was released?
C). How high is the stadium?
Answers
Answer:
Answer:
72.936 Joule
Explanation:
Mechanical Energy = Potential Energy + Kinetic Energy
Kinetic Energy = (1/2) x m x V² = (1/2) x 1.8 x 4.8² = 20.736 J
Potential Energy = m x g x h = 1.8 x 10 x 2.9 = 52.2 J
Total Mechanical Energy = 20.736 + 52.2 = 72.936 Joule
Explanation:
Related Questions
3. a ball is projected 125 m straight upward and then falls the same distance back to its starting point. neglecting air resistance, what is its total time in the air?
Answers
The total time in the air during uniform motion is 10.10 seconds.
We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule
vt = vo + a . t
vt² = vo² + 2a . s
s = vo . t + 1/2 . a . t²
where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.
From the question above, we know that
vo = 0 m/s
s = 125 m
a = g = 9.8 m/s²
By using the third equation, we can find the time taken to fall to the ground
s = vo . t + 1/2 . a . t²
125 = 0.t + 1/2 . g . t²
1/2 . 9.8 . t² = 125
t² = 250/9.8
t = 5.05 seconds
The total time should be twice
t total = 2t
t total = 2 . 5.05
t total = 10.10 seconds
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fluid mechanics
a) How does one generally describe friction in flows?
What is the difference between laminar and turbulent flow?
b) They should model the gas flow through the trachea!
1. What basic assumptions should you make?
2. The frictional force FR = 2πrlη (dvz/dr) ez acts on its lateral surface, acting on the top surface
the compressive force Fp = πr2( p1 − p2) ez. What is the flow for r = 0.5 * R? (Derivation!) Why is this model critical for colds? Please explain methodically!
c) A thought experiment on the blood vessels: Which factors determine the vascular resistance
with the flow? Why is this model critical for blood? Why is this simple model allowed
not be used for the renal vessels or the arterioles in the lungs?
Answers
This simple model is inadequate for the renal vessels or the arterioles in the lungs since it does not include all of the complexities related to the flow of blood in these vessels.
a) Friction in flows is generally described in fluid mechanics as a force that arises in the direction opposite to the fluid motion caused by a viscosity present in fluids. The viscosity determines the resistance of the fluid to shear or flow. In fluid dynamics, the movement of fluid is categorized into two types: laminar flow and turbulent flow.b)Assumptions for the model of gas flow through trachea: The following are the assumptions for the model of gas flow through trachea: The gas is considered as a continuum in the modeling of the gas flow through the trachea. The flow is steady, incompressible, and unidirectional. The flow of gas is considered to be laminar. There is no friction between the air and the walls of the trachea.The flow for r = 0.5 * R is: FR = 2πrlη (dvz/dr) ezThe frictional force FR = 2πrlη (dvz/dr) ez acts on its lateral surface. The compressive force Fp = πr2( p1 − p2) ez acts on its top surface. Vascular resistance is determined by a variety of factors, including the vessel's radius, length, and viscosity of the fluid. This model is crucial in blood because it aids in the comprehension of the dynamics of fluid flow through the blood vessels. This model is insufficient for the renal vessels or the arterioles in the lungs because it does not include all of the complexities associated with the flow of blood in these vessels.c) Vascular resistance is determined by the vessel's radius, length, and viscosity of the fluid. This model is crucial for blood because it aids in the comprehension of the dynamics of fluid flow through the blood vessels.
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A shark is cruising at 4 m/s when it sees a fish straight
ahead. The shark increases its speed to 9 m/s in 2
seconds. How far does the shark travel in this time?
Answers
Answer:
Distance, S = 13m
Explanation:
Given the following data;
Initial velocity, u = 4m/s
Final velocity, v = 9m/s
Time, t = 2 seconds
To find the distance, S;
First of all, we would determine the acceleration of the shark.
Acceleration = (v - u)/t
Acceleration = (9 - 4)/2
Acceleration = 5/2
Acceleration = 2.5m/s²
Now, to find the distance we would use the second equation of motion
S = ut + ½at²
Substituting into the equation, we have
S = 4(2) + ½*2.5*2²
S = 8 + 1.25*4
S = 8 + 5
Distance, S = 13m
acceleration; F = - F
A force of 21 N is applied to a 10 kg sled which experiences a small force of
friction of 3 N. What is the net force on the sled?
Your answer
Answers
Answer:
18N is the net force
Explanation:
Since friction always acts in the opposite direction of an objects motion we subtract the applied force from the frictional force to get our resultant net force of 18N.
Define 1 kg of mass as international standard
Answers
Answer:
1 kg is defined as the mass of a cylindrical Platinum iridium alloy whose height is equal to its diameter
in which part of science pressure belongs to
Answers
Answer: thermodynamic property
Explanation:
The pressure belongs to the thermodynamic property. The pressure is thus a scalar quantity. It also relates the vector area element (a vector normal to the surface) with the normal force acting on it.
In young's double slit experiment by using a source of light of wavelength 5000×10^-10 m, the fringe width obtained is 0.6 cm. If the distance between the screen and the slit is reduced to half. What should be the wavelength of the source to get fringes 0.003m wide?
Answers
\({ \purple{ \tt{Here \: is \: your \: answer...}}}\)
Umm it's quiet difficult to understand. But let us note down the given values.
\({ \green{ \tt{given : - }}}\)
\({ \red{ \tt{ λ = 5000 \times {10}^{ - 10} m}}}\)
\({ \red{ \tt{ λ = 5 \times {10}^{ - 7}m}}} \)
\({ \red{ \tt{ \beta = 0.6cm = 6 \times {10}^{ - 3} m}}}\)
\({ \red{ \tt{D' = \frac{D}{2}}}}\)
\({ \red{ \tt{ \beta' = 0.003m = 3 \times {10}^{ - 3}m}}}\)
ACTIVITY 4
Applying the equation learned, answer the following problems:
1. A bowling ball whose mass is 4.0 kg is rolling at a rate of 2.5 m/s. What is its momentum? p = m/s. What Is Its Momentum?
Given:
Find:
Formula:
Solution:
2. A skateboard is rolling at a velocity of 3.0 m/s with a momentum of 6.0 kg-m/s. What is its mass?
Given:
Find:
Formula:
Solution:
3. A pitcher throws a baseball with a mass of 0.5 kg and a momentum of 10 kg-m/s. What is its velocity?
Given:
Find:
Formula:
Solution:
Subject Is Science
Good Perfect Complete=Brainlist
Copy Wrong Incomplete=Report
Good Luck Answer Brainly Users:-)
Answers
Answer:
1) 10 kg-m/s
2) 2 kg
3) 20 m/s
Explanation:
The momentum of an object can be calculated using the equation:
\(\large\boxed{p=mv}\)
where:
p is momentum (measured in kilogram meters per second).m is mass (measured in kilograms).v is the velocity (measured in meters per second).\(\hrulefill\)
Question 1For this question we need to find the momentum of a bowling ball whose mass is 4.0 kg is rolling at a rate of 2.5 m/s.
Given values:
m = 4.0 kgv = 2.5 m/sSubstitute the given values into the momentum formula and solve for p:
\(p=4.0\;\text{kg} \cdot 2.5\;\text{m/s}\)
\(p=10\;\text{kg m/s}\)
Therefore, the momentum of the bowling ball is 10 kg-m/s.
\(\hrulefill\)
Question 2For this question we need to find the mass of a skateboard rolling at a velocity of 3.0 m/s with a momentum of 6.0 kg-m/s.
Given values:
p = 6.0 kg-m/sv = 3.0 m/sAs we want to find mass, rearrange the momentum formula to isolate m:
\(\large\boxed{m=\dfrac{p}{v}}\)
Substitute the given values into the formula and solve for m:
\(m=\dfrac{6.0\; \text{kg m/s}}{3.0\; \text{m/s}}\)
\(m=2\;\text{kg}\)
Therefore, the mass of the skateboard is 2 kg.
\(\hrulefill\)
Question 3For this question we need to find the velocity of a baseball with a mass of 0.5 kg and a momentum of 10 kg-m/s.
Given values:
p = 10 kg-m/sm = 0.5 kgAs we want to find velocity, rearrange the momentum formula to isolate v:
\(\large\boxed{v=\dfrac{p}{m}}\)
Substitute the given values into the formula and solve for v:
\(v=\dfrac{10\; \text{kg m/s}}{0.5\; \text{kg}}\)
\(v=20\;\text{m/s}\)
Therefore, the velocity of the baseball is 20 m/s.
The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the center of the wire's cross section as J(r) = Br, where r is in meters, J is in amperes per square meter, and B = 1. 95 ✕ 105 A/m3. This function applies out to the wire's radius of 2. 00 mm. How much current is contained within the width of a thin ring concentric with the wire if the ring has a radial width of 14. 0 μm and is at a radial distance of 1. 20 mm?
Answers
The amount of current contained within the width of a thin ring concentric with the wire is 1.257 × 10-6 A.Answer: 1.257 × 10-6 A.
Given data:
The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the center of the wire's cross-section as J(r) = Br, where r is in meters, J is in amperes per square meter, and B = 1.95 × 105 A/m³. This function applies out to the wire's radius of 2.00 mm.
Radial width of the thin ring = 14.0 μm
Radial distance of the ring = 1.20 mm
Formula used:
The current enclosed in a thin ring can be found by multiplying the current density and the area of the ring. Thus, I = JA. We get the area of the ring from its dimensions, i.e., A = 2πrΔr. Where Δr is the radial width of the ring, which is given to be 14.0 μm. We can use the radius of the ring as (r₁ + r₂)/2, where r₁ is the inner radius and r₂ is the outer radius of the ring.
Calculation:
Radius of the inner side of the ring, r₁ = 1.20 mm
Radius of the outer side of the ring, r₂ = (1.20 + 14.0) × 10⁻⁶ m = 1.214 × 10⁻³ m
Average radius of the ring, r = (r₁ + r₂)/2 = (1.20 × 10⁻³ m + 1.214 × 10⁻³ m)/2 = 1.207 × 10⁻³ m
The area of the thin ring, A = 2πrΔr = 2 × π × 1.207 × 10⁻³ m × 14.0 × 10⁻⁶ m = 5.343 × 10⁻⁹ m²
The current density is J(r) = Br
So, the current density at the average radius of the ring is
J = B × r
= 1.95 × 10⁵ A/m³ × 1.207 × 10⁻³ m
= 235.365 A/m²
The current enclosed in the thin ring is
I = JA = 235.365 A/m² × 5.343 × 10⁻⁹ m²
= 1.257 × 10⁻⁶ A
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A force of 2.0 × 102 newtons is applied to a lever to lift a crate. If the mechanical advantage of the lever is 3.43, what is the weight of the crate? A. 2.0 × 102 newtons B. 58 newtons C. 6.9 × 102 newtons D. 7.1 × 102 newtons
Answers
Answer:
C. \(6.9\times 10^2\ \text{N}\)
Explanation:
\(\text{Supplied force}=2\times 10^2\ \text{N}\)
M = Mechanical advantage = 3.43
Mechanical advantage is given by
\(M=\dfrac{\text{Weigh of crate}}{\text{Supplied force}}\\\Rightarrow \text{Weigh of crate}=M\times \text{Supplied force}\\\Rightarrow \text{Weigh of crate}=3.43\times 2\times 10^{2}\\\Rightarrow \text{Weigh of crate}=6.86\times 10^2\ \text{N}\approx 6.9\times 10^2\ \text{N}\)
The weight of the crate is \(6.9\times 10^2\ \text{N}\).
4. Compare reflecting telescopes and refracting
telescopes.
Answers
A hollow sphere is rolling along a horizontal floor at 5.0 m/s when it comes to a 30 o incline. How far up the incline does it roll before reversing direction?
Answers
To determine how far a hollow sphere will roll up a 30-degree incline before reversing direction, we need to consider the conservation of energy between kinetic and potential energy.
At the bottom of the incline, the sphere has kinetic energy but no potential energy. As it rolls up the incline, its kinetic energy decreases and its potential energy increases until it reaches the maximum height, where its kinetic energy is zero and its potential energy is at a maximum. At this point, the sphere will reverse direction and begin rolling back down the incline, with its potential energy decreasing and its kinetic energy increasing.
Since the sphere is hollow, we can use the parallel-axis theorem to find its moment of inertia, which is I = (2/3)MR^2, where M is the mass of the sphere and R is its radius.
Using conservation of energy, we can set the initial kinetic energy equal to the final potential energy at the maximum height. We can then solve for the height using the equation h = (v^2*sin^2(θ))/(2g(1+(I/(MR^2)))).
Plugging in the given values of v=5.0 m/s and θ=30 degrees, and using g=9.81 m/s^2, M=2.0 kg, and R=0.10 m, we get h = 0.33 meters.
Therefore, the hollow sphere will roll up the 30-degree incline for 0.33 meters before reversing direction.
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if the force of friction opposing the motion is 18 n, what force f (in n) is the person exerting on the mower? (enter the magnitude.)
Answers
The force f (in n) the person is exerting on the mower is 18 n.
The force of friction opposing the motion refers to the force that acts in the opposite direction to the motion of an object and makes it harder to move.
According to the given statement, the force of friction opposing the motion is equal to the force f (in N) the person is exerting on the mower.
This indicates that the person's pushing force must be equal to the force of friction opposing the motion for the mower to move at a constant speed.
Using the above information, we can calculate the force f (in N) that the person is exerting on the mower as 18 n since it is equal to the force of friction opposing the motion.
Therefore, the person must push with a force f (in N) of 18 n to overcome the friction and maintain a constant speed for the mower.
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A 66.0 kg diver is 3.10 m above the water, falling at speed of 6.10 m/s. Calculate her kinetic energy as she hits the water. (Neglect air friction)
Answers
Kinetic energy as she hits the water is 3300 joule.
To find the answer, we need to know about the Newton's equation of motion.
What's the Newton's equation of motion to determine the final velocity?The final velocity is determined asV²=U²+2aS
V= final velocity, U= initial velocity, a= acceleration and S= distanceWhat's the final velocity of the driver falling from 3.10m with initial velocity of 6.10m/s?Here, a= 9.8m/s², U= 6.10m/s and S= 3.10mSo, V²= 6.1²+2×9.8×3.10= 98 V= √98= 10m/sWhat's the kinetic energy of the driver when touches the water?Kinetic energy= 1/2×mass×velocity²
= 1/2 × 66 × 10²
= 3300J
Thus, we can conclude that the kinetic energy of the driver is 3300 Joule.
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What is the science behind Infrared thermometer ?
Answers
Answer:
Infrared thermometers employ a lens to focus the infrared light emitting from the object onto a detector known as a thermopile. The thermopile is nothing but thermocouples connected in series or parallel. When the infrared radiation falls on the thermopile surface, it gets absorbed and converts into heat.
The wavefunction of an electron is given by 1≤x≤ 16 y(x) = c = 0 x< 1 and x > 16 where c is a positive real number. (a) Calculate the probability of finding the electron in the range 2 ≤x≤ 11. (b) Where is the electron most likely to be found in the range 1 ≤ x ≤ 16? Explain your answer briefly, with one or two sentences only.
Answers
The given wave function of an electron is as follows:y(x) = c = 0 {x < 1} 1 ≤ x ≤ 16 {1 ≤ x ≤ 16} {x > 16}Where c is a positive real number.(a) The probability of finding the electron in the range 2 ≤ x ≤ 11 is calculated as follows.
Here, we use the formula for finding probability: P = ∫|y(x)|²dx for 2 ≤ x ≤ 11= ∫|c|²dx for 2 ≤ x ≤ 11= ∫c²dx for 2 ≤ x ≤ 11= c² (x) |₂ ᵢ=₁₁= c² (11) - c² (2) Hence, the required probability is c² (11) - c² (2). (b) The electron is most likely to be found at x = 8.5. ExplanationThe probability of finding the electron in the range 2 ≤ x ≤ 11 is calculated using the formula for probability, P = ∫|y(x)|²dx for 2 ≤ x ≤ 11.
Hence, the probability of finding the electron in the range 2 ≤ x ≤ 11 is c² (11) - c² (2). The electron is most likely to be found in the range where the wave function is the highest. The wave function has a maximum value at x = 8.5, so the electron is most likely to be found at x = 8.5.
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with what force will Annabella’s car hit a tree if her car has a mass of 5000 kg and an acceleration of 2 m/s2
Answers
Answer:
F=10,000N
Explanation:
F=ma
m=5000kg
a=2m/s^2
\(F=(5000kg)*(2m/s^{2} )\\\\\\\\F=10000N\)
Answer:
\(\boxed {\boxed {\sf 10,000 \ Newtons}}\)
Explanation:
We are asked to find the force of Annabella's car when it hits a tree.
According to Newton's Second Law of Motion, force is the product of mass and acceleration. Therefore, it is calculated using the following formula:
\(F=ma\)
The mass of the car is 5000 kilograms and the acceleration is 2 meters per second squared.
m= 5000 kg a= 2 m/s²Substitute the values into the formula.
\(F= (5000 \ kg)(2 \ m/s^2)\)
Multiply.
\(F= 10,000 \ kg*m/s^2\)
Convert the units. 1 kilogram meter per second squared is equal to 1 Newton, so our answer of 10,000 kg*m/s² is equal to 10,000 Newtons.
\(F= 10,000 \ N\)
Annabella's car hits the tree with a force of 10,000 Newtons.
Which change increases the electric forcWhich change increases the electric force between objects?e between objects?
Answers
Answer:
b
Explanation:
A 15 kg bowling ball is moving at 5m/s down the bowling lane, calculate its
momentum
Answers
The tendency to ignite or burn in air is known as what
Answers
Answer:
Combustible/combustion.
Explanation:
Answer:
Spontaneous Combustion
Explanation:
data reduction and error analysis for the physical sciences solution manual
Answers
There are a number of techniques that can be used in the physical sciences to decrease data and analyze faults.
Here are some often employed methods:
1. Averaging: Random mistakes can be minimized by taking numerous measurements of the same amount and averaging the results.
2. Error bars: Using error bars to represent data points gives each measurement's level of uncertainty a visual representation.
3. Error propagation: It's crucial to propagate errors when combining different measurements or employing observed quantities in calculations.
4. Instrument and equipment calibration helps to cut down on systematic mistakes. You may account for and correct any differences in your measurements by comparing them against recognized reference standards.
5. Control experiments: By performing control experiments, the sources of mistake can be found and measured since the variable of interest is consciously left out or maintained constant.
6. Replication: By carrying out experiments more than once, researchers can gauge how reliable and consistent their findings are.
7. Regression analysis and hypothesis testing are two statistical techniques that can be used to discover and quantify error sources, as well as to assess the relevance of the results and the dependability of the data.
8. Conducting an exhaustive uncertainty analysis is essential for comprehending the constraints and potential flaws in the data.
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The complete question is:
What methods can be used to reduce data and analyze errors in the physical sciences?
A whale travels 112 meters at 14 m/s How long does this take?
Answers
Answer:
8
Explanation:
112meters divided by 14
what is the kinetic energy in joules of a 1500kg car that is moving at a speed of 55 miles per hour (24.6m/s)
Answers
Answer:
453,870 J
Explanation:
KE = 1/2mv² = 1/2(1500 kg)(24.6 m/s)² = 453,870 J
Sally is riding her skateboard at 12.0 ft/s and passes under an apple tree. As she
moves past, she grabs a 2.0 kg branch with five apples right off the tree. Sally's mass
is 34 kg. What is her speed after grabbing the branch and apples?
Answers
Answer:
11.3
Explanation:
The period of a pendulum of length 0.500 m is
14.2 s
7.02 s
1.42 s
0.702 s
0.450 s
Answers
Answer:
C. 1.42 s
Explanation:
Givens:
length = 0.5m
gravity = 9.807 m/s
Solve:
T = 2π√(L/g)
T = 2π√(0.5/9.807)
T = 2π√(0.0509)
T = 2π * 0.22579
T = 6.28318 * 0.22579
T = 1.418 ≈ 1.42s
.
A sample of an ideal gas is contained at initial volume 34 L, temperature 450 K, and pressure 7.7 × 104 Pa. It then is allowed to expand to a volume of 40 L. To calculate the work involved in the expansion, which additional information, if any, must be known about the gas? Assume that known information includes the values of physical constants, such as the ideal-gas constant R.
A.the final pressure of the gas
B.The path of the change from initial to final conditions
C. the final temperature of the gas
Answers
Answer:
it have to be a
Explanation:
Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball to hit the ground.(a)Define your coordinate system, be thorough.(b)Write down the given infor-mation, be sure to include hidden information.(c)State what physics principleis at play here. How do you know this?(d)Select an equation.(e)Solve forthe time it takes for the ball to hit the ground.
Answers
Answer:
t = 5.05 s
Explanation:
This is a kinetic problem.
a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m
b) in this system the equations of motion are
y = v₀ t + ½ g t²
where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth
e) y = 0 + ½ g t²
t = √ (2y / g)
t = √(2 125 / 9.8)
t = 5.05 s
The time it takes for the Celestial sphere to rotate once relative to the Earth is called A. a solar day B. a sidereal day C. an astronomical day. D. a Celestial day E. a polar day
Answers
celestial day. the period of time required for one full rotation of the sky by a star here on celestial sphere. It can also be thought of as the length of time it takes for the earth to complete one rotation around the sun.
What are sidereal and solar days?The length of time it takes is for Earth to revolve around its axis that the Sun always appears at the same location in the sky is known as a solar day. The solar day is about 4 minutes longer than the sidereal day. The relative to the sun day is the length of time it takes the Earth, in relation to the "fixed" stars, to complete a rotation around its axis.
What makes it a celestial sphere?The celestial sphere, which is a fictional sphere centered on the Earth and onto all of which celestial bodies could be projected, is utilized to describe the positions of objects in the sky. An observer just on surface of the Earth may only see 50% of celestial sphere at any given time because.
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A vat of nitrogen at its boiling point (-196 degrees C) absorbs 384000 J of heat. How much mass of nitrogen burns off? (Unit = kg)
Answers
Answer:
1.92 kg of nitrogen.
Explanation:
The following data were obtained from the question:
Heat absorbed (Q) = 384000 J
Note: Heat of vaporisation (ΔHv) of nitrogen = 5600 J/mol
Next, we shall determine the number of mole of nitrogen that absorbed 384000 J.
This is illustrated below:
Q = mol·ΔHv
384000 = mole of N2 x 5600
Divide both side by 5600
Mole of N2 = 384000/5600
Mole of N2 = 68.57 moles
Next, we shall convert 68.57 moles of nitrogen, N2 to grams.
This can be obtained as follow:
Molar mass of N2 = 2 x 14 = 28 g/mol.
Mole of N2 = 68.57 moles.
Mass of N2 =..?
Mole = mass /molar mass
68.57 = mass of N2 /28
Cross multiply
Mass of N2 = 68.57 x 28
Mass of N2 = 1919.96 g
Finally, we shall convert 1919.96 g to kilograms.
This can be achieved as shown below:
1000g = 1 kg
Therefore,
1919.96 g = 1919.96/1000 = 1.92 kg.
Therefore, 1.92 kg of nitrogen were burned off.
Answer:
Explanation:
i cheat on Acellus cause I dont care about school. If i did I would be going to public school... Im never gonna use this info in my life so I dont bother learning it.
A thermometer has a scale which starts at -10 Celcius and ends at 110 Celcius. What is the value of the lower fixed point and of the upper fixed point of the scale?
Answers
Explanation:
The Kelvin, Celsius, and Fahrenheit scales are the three temperature scales that are most widely used. In most physical experiments the Celsius scale is used to take common measurements, but the Kelvin scale is used for most calculations.
