Answer:
81.1 m/s
Explanation:
The net force of Jason is T - f = ma where T = thrust = 200 N f = frictional force = μN = μmg where μ = coefficient of kinetic friction of water = 0.10, m = mass of Jason plus skis = 75 kg, g = acceleration due to gravity = 9.8 m/s² and a = Jason's acceleration
So, T - f = ma
T - μmg = ma
a = T/m - μg
susbstμituting the values of the varμiables into the equation, we have
a = 200 N/75 kg - 0.1 × 9.8 m/s²
a = 200 N/75 kg - 0.1 × 9.8 m/s²
a = 2.67 m/s² - 0.98 m/s²
a = 1.69 m/s²
Using v = u + at, we find Jason's velocity v where u = initial velocity = 0 m/s (since he starts from rest), a = 1.69 m/s² and t = time = 48 s
So, v = u + at
v = 0 m/s + 1.69 m/s² × 48 s
v = 0 m/s + 81.12 m/s
v = 81.12 m/s
v ≅ 81.1 m/s
So, Jason's top speed is 81.1 m/s
Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimumpossible kinetic energy in MeV. Treat this problem as one-dimensional, and use the relativistic relationbetweenEandp. Give your answer to 2 significant figures. (The large value you will find is a strongargument against the presence of electrons inside nuclei, since no known mechanism could contain anelectron with this much energy.)
Answer:
39.40 MeV
Explanation:
Determine the minimum possible Kinetic energy
width of region = 5 fm
From Heisenberg's uncertainty relation below
ΔxΔp ≥ h/2 , where : 2Δx = 5fm , Δpc = hc/2Δx = 39.4 MeV
when we apply this values using the relativistic energy-momentum relation
E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,
Also in a nuclear confinement ( E, P >> mc )
while The large value will portray a Non-relativistic limit as calculated below
K = h^2 / 2ma^2 = 1.52 GeV
which one of the following is not a simple machine
1 ladder
2 wheel Barrow
3 pulley
4 electric pole
A body initially at rest is accelerated at a constant rate for 5.0 seconds in the positive x direction. If the final speed of the body is 20.0 m/s, what was the body's acceleration?
Answer:
[tex]a=4\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of a body, u = 0
Final speed of the body, v = 20 m/s
Time, t = 5 s
We need to find the acceleration of the body. We know that the acceleration of an object is equal to the rate of change of velocity divided by time taken. So,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{20-0}{5}\\\\a=4\ m/s^2[/tex]
So, the body's acceleration is equal to [tex]4\ m/s^2[/tex].
Please helppppppp!!!!
Answer:
I think the particles transferred between brandi's body and the door knob causing the shock is the electrons.
MCQs. How does the heat from the Sun reach the Earth.(options)a: conduction b: convertion c: radiation d: combustion.
HELP ME PLEASE !!!!!!!!!!!!
Answer:
D
Explanation:
Because the y axis is meter. If it is straight line at time and meter graph then it velocity and speed is 0
An atom has 20 protons and 22 neutrons and 18 electrons. The charge of this atom is: ________
Answer:
the number of electrons should equal to the the number of protons in a neutral atom
if there is a inequality between the numbers it means the atom has a + or - charge
The charge of this atom=+(20-18)=+2If two dogs are pulling a bone with force-20Newtons in opposite direction, then the resultant force is
Answer:
Newtons third law of motion: Balanced forces
Every action has a corresponding and opposing response, according to Newton's third law of motion. As a result, forces always work in pairs. Once more, tug-of-war is a prime illustration.
What force in opposite direction follow newton law?The third law of motion by Newton states that equal, but diametrically opposed forces always act in pairs. There is an equal but opposite reaction to every action, to put it another way.
The forces are balanced if the pullers are exerting equal force but going in the opposite direction on either side of the rope. There is hence no motion.
Although equal and opposite in nature, action and reaction forces cannot be balanced since they act on separate things and do not cancel one another out.
Therefore, This means that when you push against a wall, the wall pushes back against you with an equal amount of force.
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An astronaut hits a golf ball of mass m on the Moon, where there is no atmosphere and the acceleration due to gravity is g 6 , where g is the acceleration due to gravity on Earth. Assume that the golf club is in contact with the ball for a time t. Just after losing contact with the club, the ball has an initial speed v directed at an angle T above the horizontal. 6. What is the magnitude of the average force exerted by the club on the ball during contact
Answer:
The magnitude of the average force exerted by the club on the ball during contact = mv/t
Explanation:
Impulse exerted on the ball = Momentum of the ball = mass * velocity = m*v
As we know,
m*v = Integration of F.dt with limits 0 to T
Ft = mv
F = mv/t
The magnitude of the average force exerted by the club on the ball during contact = mv/t
Plants get their food by which of the following ways?
A. Making it from sunlight
B.Eating other plants
C. Absorbing it from the soil
D.Taking nutrients from the air
Answer:
C
Explanation:
Use your understanding of heat loss to ESTIMATE the cost of the lost energy through one standard window during an average summer day in Maryland. Use $0.17 per kWh as your energy cost.
Answer:
The cost of energy is $ 0.34.
Explanation:
The energy is the capacity to do work.
The energy is a scalar quantity and its SI unit is Joule.
The commercial unit of energy is kWh.
Cost of 1 kWh energy = $ 0.17
energy loss by standard window is 2 kWh .
So, the cost of lost of energy is
Cost = $ 0.17 x 2 = $ 0.34
tell me about Orion nebula
In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give
Answer:
The answer is " because it is in the opposite direction of the bullet".
Explanation:
Given:
[tex]m_1= 70 \ kg\\\\m_2 = 0.01\ kg\\\\v_2 = 500\ \frac{m}{s}\\\\v_1 = ?[/tex]
Using formula:
It recoils the velocity of the rifle [tex]= -0.07142 \ \ \frac{m}{s}[/tex]
Its negative sign displays the opposite direction of the rifle.
What is the speed acquired by a freely falling object 4 seconds after being dropped from a rest position? Use units of meter per second (m/s) and assume acceleration from gravity is 10 m/s2.
speed = 40 m/s
Explanation:
Since the object is dropped, V0y = 0.
Vy = V0y - gt
= -(10 m/s^2)(4 s)
= -40 m/s
This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.
The speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
What are the three equations of motion?There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
Keep in mind that these calculations only apply to uniform acceleration.
As given in the problem, we have to find the speed acquired by a freely falling object 4 seconds after being dropped from a rest position,
By using the first equation of motion,
v = u + at
initial velocity(u) = 0 m/s
acceleration(a) = 10 m/s²
v = 0 + 10×4
v = 40 meters/seconds
Thus, the speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
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A disk of Radius R with a uniform distibution of mass"m" rotater about an axis perpendicular to its place at the rim with angular speed "w" the moment of Inertia of the disc about an axis through the contre MR² What is the KE of the disk?
Answer:
[tex]\frac{1}{2}mR^2\omega^2[/tex]
Explanation:
The rotational kinetic energy of an object is given by [tex]KE_r=\frac{1}{2}I\omega^2[/tex], where [tex]I[/tex] is the object's moment of inertia/rotational inertia and [tex]\omega[/tex] is the object's angular speed.
What we're given:
The object's moment of inertia: [tex]I=MR^2[/tex] The object's radius, mass, and angular speed: [tex]R, m, \omega[/tex], respectivelySince no numerical value is given for any of these, it is implied the desired answer will be an equation in terms of the variables given.
Substituting [tex]I=MR^2[/tex]:
[tex]KE_r=\boxed{\frac{1}{2}mR^2\omega^2}[/tex]
A FBD of a rocket launching into space should include:
Answer:
Explanation:
the force of the rocket engine pushing it up, the force of gravity pulling it down, maybe some force of air resistance as the rocket goes fast, hmmm Free Body Diagrams (FBD) should have any and all forces on the model, unless they are negligible . or so slight they really make little difference in the total outcome.
Charissa says that bread goes through a chemical change when it is toasted. Which of the following observations would least support her conclusion?
A. a change in the smell of the bread
B. a change in the color of the bread
C. a change in the texture of the bread
D. a change in the shape of the bread
Answer:
A. the change in smell
Explanation:
sry if its wrong it's what I know
Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 ss to speed up from rest to its top speed of 1 rotation every 1.30 ss . The astronaut is strapped into a seat 5.90 mm from the axis. What is the astronaut's tangential acceleration during the first 40.0 s?
How many g's of acceleration does the astronaut experience when the device is rotating at top speed? Each 9.80 m/s^2 of acceleration is 1 g.
Answer:
speed = 0.9 mm/s
Explanation:
time, t = 40 s
initial angular speed, wo = 0 rad/s
final frequency, f = 1/1.03 rps = 0.97 rps
final angular speed, w = 2 x 3.14 x 0.97 = 6.1 rad/s
time, t = 40 s
distance, r = 5.9 mm
The angular acceleration is given y the first equation of motion.
[tex]w =wo + \alpha t\\6.1 = 0 +\alpha \times 40\\\alpha = 0.1525 rad/s^{2}[/tex]
The linear velocity is
[tex]v =5.9\times 10^{-3}\times 0.1525 = 9\times 10^{-4} m/s[/tex]
speed, v = 0.9 mm/s
A man who works for a moving company is loading a box onto a moving van. He pushes a 200N box up a 5m long ramp. If he pushes with a force of 60N and the ramp is 1m high, what is the efficiency of the inclined plane
Answer:
η = 0.667 = 66.7%
Explanation:
The efficiency of the man can be given by the following formula:
η = output/input
where,
η = efficiency of man = ?
output = potential energy gain of the box = Wh
input = work done by man = Fd
Therefore,
[tex]\eta = \frac{Wh}{Fd}[/tex]
where,
W = weight of box = 200 N
h = height gained by box = 1 m
F = force exerted by man = 60 N
d = length of ramp = 5 m
Therefore,
[tex]\eta = \frac{(200\ N)(1\ m)}{(60\ N)(5\ m)}[/tex]
η = 0.667 = 66.7%
Which image shows an example of potential energy?
Answer:
D
Explanation:
Potential energy involves the change of an object's position, which in this case a rocket is increasing its vertical displacement from the ground.
When a rocket is increasing its vertical displacement from the ground, it exhibits both potential and kinetic energy. Therefore option D is correct.
At the initial stage, when the rocket is on the ground and not moving, it possesses potential energy. This potential energy is in the form of stored energy due to its elevated position above the ground.
As the rocket launches and gains altitude, it continues to accumulate potential energy because it is moving higher against the force of gravity.
Simultaneously, as the rocket moves upward, it also gains kinetic energy. Kinetic energy is the energy associated with the rocket's motion.
The faster the rocket moves, the greater its kinetic energy becomes. As the rocket ascends, its speed increases, resulting in an increase in kinetic energy.
Therefore, in the context of a rocket increasing its vertical displacement from the ground, both potential energy (due to its height) and kinetic energy (due to its motion) are present.
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3. A skater moves across the ice a distance of 12 m before a constant frictional force of 15 N cause him to stop. His initial speed is 2.2 m/s . Calculate the skater's mass.
Explanation:
thankyou fo
r the poimts
Two balls are thrown from a cliff. One is thrown directly up, the other directly down. Both balls have the same initial speed, and both hit the ground below the cliff but at different times. Which ball hits the ground at the greater speed: (a) the ball thrown upward, (b) the ball thrown downward, or (c) both the same
Answer:
both the same
Explanation:
When a ball is thrown vertically upwards, it experiences that same acceleration due to gravity as an object thrown directly downwards.
This means that if we ignore the effects of air resistance, and the two balls have the same initial speed, they are expected both expected to hit the ground at the same speed as a result of the principle of conservation of energy.
Answer:
Both balls will hit the ground at the same speed because air resistance is ignored.
Explanation:
The electric field of a negative infinite line of charge: Group of answer choices Points perpendicularly away from the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and increases in strength at larger distances from the line charge Points perpendicularly away from the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge
Answer:
Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge
Explanation:
The electric field for a uniform line of charge is given by E = λ/2πε₀r where λ = charge density and r = distance from line of charge.
If λ is negative, E is negative so it points in the negative direction towards the line of charge.
Also, since for negative charges, electric field lines end up in them, the electric field for an infinitely long negative line of charge points towards the charge perpendicular to it.
Also as r increases, E decreases since E ∝ 1/r
So, the electric field decreases at larger distances from the line of charge.
So, the electric field of a negative infinite line of charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge.
After the big bang started the universe 13.78 billion years ago, the universe began to rapidly expand. This rapid expansion caused the universe to cool
down into a network of gaseous clouds composed mainly of which element?
O Oxygen
Hydrogen
O Neon
O Carbon
Answer:
hydrogen
Explanation:
hope it helps bye.
Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, the box (initially at rest) is traveling at a speed of 6 m/s. How much work (in Joules) did friction do in this process
Answer: 321 J
Explanation:
Given
Mass of the box [tex]m=3\ kg[/tex]
Force applied is [tex]F=25\ N[/tex]
Displacement of the box is [tex]s=15\ m[/tex]
Velocity acquired by the box is [tex]v=6\ m/s[/tex]
acceleration associated with it is [tex]a=\dfrac{F}{m}[/tex]
[tex]\Rightarrow a=\dfrac{25}{3}\ m/s^2[/tex]
Work done by force is [tex]W=F\cdot s[/tex]
[tex]W=25\times 15\\W=375\ J[/tex]
change in kinetic energy is [tex]\Delta K[/tex]
[tex]\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J[/tex]
According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy
[tex]\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J[/tex]
Therefore, the magnitude of work done by friction is [tex]321\ J[/tex]
The pair of forces described by Newton third law must be
Answer:
The answer is Newton's third law of motion states that every action has an equal and opposite reaction. This means that force always act in pairs
The pair of forces described by Newton third law must be in opposite direction.
What is Newton's third law of motion ?Every action have equal and opposite reaction. for example when we fire bullet from a gun, the gun will recoil back and bullet moves forward. In case of rocket, rocket is fired, thrust is reaction of force applied by the gas on the floor.
The motion of lift from an airfoil in which the air is diverted downward by the airfoil's action and the wing is pushed upward in response.
When a spinning ball moves, the air is deflected to one side, and the ball responds by travelling in the other direction.
A jet engine's motion generates thrust, and hot exhaust gases rush out the back of the engine, producing thrust in the opposite direction.
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POR FA ME PODRIAN AYUDAR. URGENTE!!!!
¿Cuál es la energía cinética de un 0,01kg bala viajando a una velocidad de 700Sra?
Answer:
K.E = 2450 Joules
Explanation:
Given the following data;
Mass = 0.01 kg
Velocity = 700 m/s
To find the kinetic energy;
La energía cinética (K.E) se puede definir como una energía que posee un objeto o cuerpo debido a su movimiento.
Matemáticamente, la energía cinética viene dada por la fórmula;
[tex] K.E = \frac{1}{2}mv^{2}[/tex]
Sustituyendo en la fórmula, tenemos;
K.E = ½ * 0.01 * 700²
K.E = 0.005 * 490000
K.E = 2450 Joules
A balloon contains 0.04m3 of air at a pressure of 1.20 x 105Pa. Calculate the pressure required to reduce its volume to 0.025m3 at constant temperature.
Answer:
you can simply answer p1v1=p2v2 p2 = p1v1÷ p2 then inter what given in the formula you get it pl=1.20x 105pa vl= 0.40m3 v2=0.025m3
what is the magnitude of an electric field (in 106 n/c) that balances the weight of a plastic sphere of mass 2.1 g that has been charged to 3.0 nc
Answer:
[tex]E=6.86\times 10^6\ N/C[/tex]
Explanation:
Given that,
Mass of the sphere, m = 2.1 g = 0.0021 kg
Charge, q = 3 nC
We need to find the magnitude of the electric field that balanced the weight of sphere. Let it is E. So,
qE = mg
[tex]E=\dfrac{mg}{q}[/tex]
Put all the values,
[tex]E=\dfrac{0.0021\times 9.8}{3\times 10^{-9}}\\\\E=6.86\times 10^6\ N/C[/tex]
So, the magnitude of the elecric field is [tex]6.86\times 10^6\ N/C[/tex].
Which statements are true of noble gases?
Check all that apply.
A. They are metalloids.
B. Their valence shells are full of electrons.
C. They are not very reactive.
D. All of the noble gases have at least two electron shielding layers.