Answer:
12: If the bird flies 10m in 2s, it's speed is then 10m/2s, which we can simplify giving us 5m/s
13: Flying 5000 km in eight hours means that it's flying at (5000 / 8)km/h, or 625km/h
14: A speed graph will not go below zero, an acceleration graph will.
If the mass of a moving object is doubled, which variable will also double? *
(7 points)
momentum
final velocity
acceleration
potential energy
Answer:
potential energy
Explanation:
this is because potential energy= mgh
so when m= 2m
p.e. = 2 times the previous value
Therefore , When mass is doubled , Potential energy gets doubled. When mass is halved , Potential energy decreases by half.
__________ and __________ both heavily relied on dream analysis in their treatment of patients. A. Alfred Adler . . . Albert Ellis B. Alfred Adler . . . Carl Jung C. Sigmund Freud . . . Carl Jung D. Sigmund Freud . . . Alfred Adler
Answer:
C. Sigmund Freud . . . Carl Jung
Explanation:
edge 2021
C.Sigmund Freud and Carl Jung both heavily relied on dream analysis in their treatment of patients.
What is Freud most famous for?Freud is well-known for inventing and developing the approach of psychoanalysis; for articulating the psychoanalytic idea of motivation, intellectual infection, and the structure of the unconscious; and for influencing medical and popular conceptions of human nature by using positing both everyday and strange thought.
Sigmund Freud was an Austrian neurologist who's perhaps maximum known as the founding father of psychoanalysis. Freud advanced a fixed of therapeutic strategies centered on communication therapy that worried the use of techniques that include transference, loose affiliation, and dream interpretation.
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A 64.0-kg person holding two 0.900-kg bricks stands on a 2.00-kg skateboard. Initially, the skateboard and the person are at rest. The person now throws the two bricks at the same time so that their speed relative to the person is 19.0 m/s. What is the recoil speed of the person and the skateboard relative to the ground, assuming the skateboard moves without friction?
Answer:
0.518 m/s
Explanation:
To solve this question, we would use the Law of conservation of momentum.
It is stated that the skateboard and the person were at rest initially, this means that their initial momentum is 0. Since they aren't moving.
And from the law if conservation of momentum, we know that initial momentum must be equal to final momentum.
The final momentum of the bricks will be 2* 0.9 * 19 = 34.2 kg m/s.
This means that the momentum of the person and that of the skateboard has to be 34.2 kg m/s also, although, it will be in the opposite direction.
Mass of the person plus that of the skateboard is
64 + 2 = 66 kg,
If we divide the momentum of the person and that of the skateboard by the mass of the both of them, we get the speed at which they are moving after the bricks are thrown
Recoil speed = (34.2 kg m/s)/(66 kg) = 0.518 m/s.
The recoil speed of the person is 0.518 m/s
Calculation of the recoil speed of the person:
As per the conservation of momentum, the initial momentum should be equivalent to the final momentum
So, here the final momentum should be
= 2* 0.9 * 19
= 34.2 kg m/s.
Now the mass of the person should be
= 64 + 2
= 66 kg,
Now the recoil speed should be
Recoil speed = (34.2 kg m/s)/(66 kg)
= 0.518 m/s.
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Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change
Answer:
No.
Explanation:
Given that Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change ?
Considering the formula
V = wr
Where
V = linear speed
W = angular speed
r = radius of the wheel.
But W = 2πrf
Where the the 2 and pi are constant. The radius of the first wheel will be small but counter balance with the larger frequency.
While the radius of the second wheel may be large but it will be of a small frequency.
We can therefore conclude that the reading on the speedometer will not change. Because speedometer will read the linear speed V.
Which of these is the BEST answer for why science is important?
Science can take us to other planets, even if it’s only through a telescope.
Science is part of human nature; it helps answer questions about how the world works.
Science helps us learn to think more critically and weigh evidence better.
Science gives us better tablet computers and games.
QQQUUUUCCCKKKK!!!!!!
A car with a mass of 1200 kg has a momentum of 15, 350 kg * m/s. What is its velocity?
Answer:
v = 12.79 m/s
Explanation:
Given that,
The mass of a car, m = 1200 kg
Momentum of the car, p = 15 350 kg-m/s
We need to find the velocity of the car. We know that, the formula for the momentum of an object is given by :
p = mv
Where
v is the velocity of the bject
So,
[tex]v=\dfrac{p}{m}\\\\v=\dfrac{15350}{1200}\\\\v=12.79\ m/s[/tex]
So, the velocity of the car is 12.79 m/s.
Assume your computer has a power rating of 50.0 W, and you use your computer for 7.0 hours a day for a normal school day. If the electric company charges $0.10 per kWh, how much does it cost you to use your computer each school day?
35$
$0.035
$0.005
$0.35
An atom of tin has an atomic number of 50 and a mass number of 119. How many protons, electrons, and neutrons are found in one neutral atom of tin?
O 50 protons, 69 electrons, 50 neutrons
O 50 protons, 50 electrons, 69 neutrons
69 protons, 50 electrons, 69 neutrons
69 protons, 69 electrons, 50 neutrons
Answer:
50 protons 50 electrons and 69 neutrons...
Explanation:
the number of protons is equal to number of electrons. then mass number is the sum of protons and neutrons in a nucleus so for we to get the number of neutrons we take the mass number subtract the protons number.
PLEASE HELP
Section 1 - Question 6
Wave Movement Through Media
What could be happening to the wave as it travels from left to right?
A
It's moving through a medium whose density stays the same
B
It's moving from a low density medium to a high density medium.
С
It's moving from a high density medium to a low density medium.
D
It's moving from a low density medium, to a high density medium, and then back to a low density medium
Answer: B
Explanation:
Carl works hard to get a grades on his report card because his mother pays him 25 dollars for each semester he earns straight as Carl’s behavior is being influenced by
On a cold winter day, a steel metal fence post feels colder than a wooden fence post of identical size because: a. The specific heat capacity of steel is higher than the specific heat capacity of wood. b. The specific heat capacity of steel is lower than the specific heat capacity of wood. c. Steel has the ability to resist a temperature change better than wood. d. The mass of steel is less than wood so it loses heat faster. Selected:e. Two of the above statements are true.
Answer:
The specific heat capacity of steel is lower than the specific heat capacity of wood
Explanation:
THERE IS ONLY 1 ON MY assignment i geot dis right please brainlyist
The specific heat capacity of steel is lower than the specific heat of a piece of wood. Therefore, option (2) is correct.
What is the specific heat capacity?Specific heat can be defined as the heat energy required to change the temperature of one unit mass of a substance of a constant volume by 1 °C. The S.I. unit of the specific heat capacity of a material is KJ/Kg.
The thermal capacity of a material is defined as a physical property of a substance. The amount of heat is given to a given mass to create a change in unit temperature.
The mathematical expression of specific heat capacity can be written as :
Q = m C ΔT Where C is the heat capacity.
The specific heat capacity is an intensive property of a substance as it does not depend upon the size of the material.
A steel metal fence post feels colder than a wooden fence post of similar size because the specific heat of steel is lower than the specific heat capacity of wood.
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what is amplitudes relationship between sound and amplitudes
Answer:
A sound wave's amplitude relates to changes in pressure. The sound is perceived as louder if the amplitude increases, and softer if the amplitude decreases. This is illustrated below. DOSITS short video on amplitude.
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One of the greatest dangers in a tornado is from flying objects. A 15 pound piece of lumber can turn into a flying missile that could severely damage walls and homes. A piece of steel with a mass of 200 pounds and travelling at the same velocity would cause even more damage. Select any evidence from the list below that you could use to explain why a 200 pound piece of steel would cause more damage than a 15 pound piece of wood travelling at the same velocity.
As the kinetic energy of an object increases, the force it can exert on another object decreases.
As the kinetic energy of an object increases, the force it can exert on another object increases.
Objects with more mass have less kinetic energy.
Objects with more mass have more kinetic energy.
As the velocity of an object increases, its kinetic energy decreases.
As the velocity of an object increases, its kinetic energy increases.
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu
Is there a way to see moon and the sun at once?
A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of 170 kgkg and is traveling east with a velocity of magnitude 4.60 m/sm/s. Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.
This question is not complete, the complete question is;
A railroad handcar is moving along straight, frictionless tracks with negligible air resistance.
In the following cases, the car initially has a total mass (car and contents) of 170 kg and is traveling east with a velocity of magnitude 4.60 m/s.
Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.
Part A
An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity.
Part B
An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.
Answer:
Part A) the final velocity of the car is 4.6 m/s
Part B) the final velocity of the car is 5.28 m/s
Explanation:
Given the data in the question;
Total mass (m₁+m₂) = 170 kg
velocity of magnitude Vx = 4.60 m/s
PART A)
An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity,
i.e
m₂ = 22.0 kg
so m₁ = 170 - 22 = 148 kg
so, we apply conservation of momentum
since the object thrown out of the car, it has nothing to do with the car's velocity.
(m₁+m₂)Vx = m₁Vx₁ + m₂Vx₂
we substitute
(170)4.60 = 148Vx₁ + 22(4.60)
782 = 148Vx₁ + 101.2
148Vx₁ = 782 - 101.2
148Vx₁ = 680.8
Vx₁ = 680.8 / 148
Vx₁ = 4.6 m/s
Therefore, the final velocity of the car is 4.6 m/s
Part B)
An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.
Vx = V(m₁+m₂) / ((m₁+m₂) - m₁)
we substitute
Vx = 4.60(170) / ((170) - 22)
Vx = 782 / 148
Vx = 5.28 m/s
Therefore, the final velocity of the car is 5.28 m/s
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the first light source has a wavelength of 640 nm. Two different interference patterns are observed. If the 10th order bright fringe from the first light source coincides with the 12th order bright fringe from the second light source, what is the wavelength of the light coming from the second monochromatic light source
Answer:
533.33 nm
Explanation:
Since dsinθ = mλ for each slit, where m = order of slit and λ = wavelength of light. Let m' = 10 th order fringe of the first slit of wavelength of light, λ = 640 nm and m"= 12 th order fringe of the second slight of wavelength of light, λ'.
Since the fringes coincide,
m'λ = m"λ'
λ' = m'λ/m"
= 10 × 640 nm/12
= 6400 nm/12
= 533.33 nm
A cylindrical body has 6 m height and its radius is 2 metre calculate its volume. Ans :75.428m3
Answer:
75.4
Explanation:
r= 2
h= 6
v= 22/7 *r*r*h
v= 75.42
Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.38 x 10-3 rad/s2 for 2.04 x 103 s. For the next 1.48 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.63 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.42 rad/s. Find the total angular displacement of the propeller.
Answer:
Δθ = 15747.37 rad.
Explanation:
The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:[tex]\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)[/tex]
Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:[tex]\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)[/tex]
Solving for Δθ in (2):[tex]\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)[/tex]
The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:[tex]\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)[/tex]
Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:[tex]\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)[/tex]
Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:[tex]\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)[/tex]
The total angular displacement is just the sum of (3), (4) and (6):Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad ⇒ Δθ = 15747.37 rad.Please help I need help I can’t fail .
Answer:
A
Explanation:
Gas particles move fastest.
Plasma lacks fixed volume and shape.
Liquid lacks fixed shape, but has fixed volume.
Solid has both fixed volume and shape.
It requires 3480 J to move a 675 N object how far?
If you blow across the open end of a soda bottle and produce a tone of 250 Hz, what will be the frequency of the next harmonic heard if you blow much harder?
___Hz
Answer:
Generally, the lowest overtone for a pipe open at one end and closed would be at y / 4 where y represents lambda, the wavelength.
Since F (frequency) = c / y Speed/wavelength
F2 / F1 = y1 / y2 because c is the same in both cases
F2 = y1/y2 * F1
F2 = 3 F1 = 750 /sec
Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe
and y1 = 3 y2
The second harmonic will be three times the first harmonic. The answer is 750 Hz
VIBRATION OF WAVES IN PIPESClosed pipes have odd multiples of frequencies or harmonics. That is,
If [tex]F_{0}[/tex] = fundamental frequency = first harmonic
[tex]F_{1}[/tex] = 3[tex]F_{0}[/tex] = second harmonic
[tex]F_{2}[/tex] = 5[tex]F_{0}[/tex] = third harmonic
[tex]F_{3}[/tex] = 7[tex]F_{0}[/tex] = fourth harmonic
Let assume that the first harmonic is 250 Hz, If you blow it much harder, second, third or fourth harmonic can be produced.
By using the formula above,
second harmonic will be 3 x 250 = 750Hz
Therefore, the frequency of the next harmonic heard if you blow much harder will be 750 Hz
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A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one with mass mA = 1.35 kg and the other with mass mB = 0.270 kg . In the explosion, 810 J of chemical energy is converted to kinetic energy of the two fragments.
Required:
a. What is the speed of each fragment just after the explosion?
b. It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.
Answer:
Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s
Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s
B. 475.3 m
Explanation:
A. Determination of the speed of each fragment.
I. Determination of the speed of the fragment with mass mA = 1.35 kg
Mass of fragment (m₁) = 1.35 kg
Kinetic energy (KE) = 810 J
Velocity of fragment (u₁) =?
KE = ½m₁u₁²
810 = ½ × 1.35 × u₁²
810 = 0.675 × u₁²
Divide both side by 0.675
u₁² = 810 / 0.675
u₁² = 1200
Take the square root of both side.
u₁ = √1200
u₁ = 34.64 m/s
Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s
II. I. Determination of the speed of the fragment with mass mB = 0.270 kg
Mass of fragment (m₂) = 0.270 kg
Kinetic energy (KE) = 810 J
Velocity of fragment (u₂) =?
KE = ½m₂u₂²
810 = ½ × 0.270 × u₂²
810 = 0.135 × u₂²
Divide both side by 0.135
u₂² = 810 / 0.135
u₂² = 6000
Take the square root of both side.
u₂ = √6000
u₂ = 77.46 m/s
Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s
B. Determination of the distance between the points on the ground where they land.
We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:
Maximum height (h) = 90.0 m
Acceleration due to gravity (g) = 10 m/s²
Time (t) =?
h = ½gt²
90 = ½ × 10 × t²
90 = 5 × t²
Divide both side by 5
t² = 90/5
t² = 18
Take the square root of both side
t = √18
t = 4.24 s
Thus, it will take 4.24 s for each fragments to get to the ground.
Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:
Velocity of fragment (u₁) = 34.64 m/s
Time (t) = 4.24 s
Horizontal distance travelled by the fragment (s₁) =?
s₁ = u₁t
s₁ = 34.64 × 4.24
s₁ = 146.87 m
Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:
Velocity of fragment (u₂) = 77.46 m/s
Time (t) = 4.24 s
Horizontal distance travelled by the fragment (s₂) =?
s₂ = u₂t
s₂ = 77.46 × 4.24
s₂ = 328.43 m
Finally, we shall determine the distance between the points on the ground where they land.
Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m
Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m
Distance apart (S) =?
S = s₁ + s₂
S = 146.87 + 328.43
S = 475.3 m
Therefore, the distance between the points on the ground where they land is 475.3 m
PLEASE HELP The United States spends over $20 billion a year on space exploration through NASA. Do you think that this has been worth the cost? In three to five sentences, provide two specific examples of things we have learned from space exploration, and explain how these examples influence your opinion.(4 points)
Answer: I think $20 billion a year it’s worth the cost. The reasoning behind that is because we can conduct research on various things that could help out humanity. Therefore we can conclude that’s spending billions of dollars every year is worth it.
Explanation:
1 A roller coaster starts from rest at A, rolls down the track to B, describes a circular loop of 12-m diameter, and travels up and down past point E. Determine the range of values of h for which the roller coaster will not leave the track at D or E. Assume no energy loss due to friction.
Answer:
15m
Explanation:
Given that a roller coaster starts from rest at A, rolls down the track to B, describes a circular loop of 12-m diameter, and travels up and down past point E. Determine the range of values of h for which the roller coaster will not leave the track at D or E. Assume no energy loss due to friction
Solution
At point A
The maximum potential energy = maximum K.E
At point A, the total energy = maximum P.E.
Down the track to point B, the P.E will be converted to maximum K.E.
Hence,
Mgh = 1/2mv^2
Also, the total energy at the roller coaster will be P.E + K.E
I.e mg2r + 1/2mv^2
Where 2r = height of the loop = diameter of the loop.
Since the energy is always conserved, hence
Mgh = mg2r + 1/2mv^2
Let also consider the centripetal acceleration to keep the object in the circle.
F = mV^2 / r = mg
Mass will cancel out
U^2 = rg
Substitute that in the last equation
Mgh = mg2r + 1/2mgr
mgh = mg ( 2r + 1/2r )
Mg will cancel out
h = 2.5r
Where r = 12/2 = 6
h = 2.5 × 6
h = 15m
Therefore, the values of h for which the roller coaster will not leave the track at D or E is 15m.
Which of the following changes when an unbalanced force acts on an object?
A. mass
B. motion
C. inertia
D. weight
The answer is Motion
Do blood cells have a nuclei? Describe the shape of blood platelets.
Answer: Mature red blood cells don't have nuclei.
Explanation: " Platelets are irregularly shaped, have no nucleus, and typically measure only 2–3 micrometers in diameter. "
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help me plz!!!!!!. what difference between center of gravity and center of mass
Answer:
I think that the centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field. Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field
A runner having mass of 55 kg moving at a speed of 9 m/s rounds a bend with a radius of 18 m. What is the centripetal
Answer:
247.5 N
Explanation:
From the question,
The expression of centriputal force is given as
F = mv²/r..................... Equation 1
Where m = mass of the runner, v = velocity of the runner, r = radius of the runner
Given: m = 55 kg, v = 9 m/s, r = 18 m
Substitute these values into equation 1
F = 55(9²)/18
F = 247.5 N.
Hence the centriputal force of the runner is
247.5 N
In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point 2.5 m from his foot at the edge of the pool.
Where does the spot of light hit the bottom of the pool, measured from the bottom of the wall beneath his foot, if the pool is 2.1 m deep pool?
Answer:
4.4 m
Explanation:
We are told the light from his flashlight, 1.3 m above the water level. Thus; h1 = 1.3m
Also,we are told that the light shone 2.5 m from his foot at the edge of the pool. Thus, L1 = 2.5 m
Angle of incidence θ1 is given by;
tan θ1 = L1/h1
tan θ1 = 2.5/1.3
tan θ1 = 1.9231
θ1 = tan^(-1) 1.9231
θ1 = 62.53°
Using Snell's law, we can find the angle of refraction from;
Sin θ2 = (η_air/η_water) Sin θ1
Where;
η_air is Refractive index of air = 1
η_water is Refractive index of water = 1.33
Thus;
Sin θ2 = (1/1.33) × sin 62.53°
Sin θ2 = 0.6671
θ2 = sin^(-1) 0.6671
θ2 = 41.84°
We want to find where the spot of light hit the bottom of the pool if the pool is 2.1 m deep. Thus, h2 = 2.1 m
Now, the spot can be found from;
L = L1 + L2
Where L2 = (h2) tan θ2
L = 2.5 + 2.1 tan 48.84
L = 2.5 + (2.1 × 0.8954)
L ≈ 4.4 m
12. Identify the Leader