Answer:
P = P1 + P2 where P is a vector quantity
P1 = P2 = .10 kg * 30.0 m/s = 3 kg-m/sec
P = (p1^2 + p2^2)^1/2 add vectors and find magnitude of resultant
p = (3.0^2 + 3.0^2)^1/2 = 4.24 length of resultant vector
Since vectors are at right angles p is the length of the hypotenuse
tan θ = 3.0 / 3.0 = 1 where theta is measured with respect to the positive x-axis and will be below that axis
θ = tan^-1 1 = 45 deg (45 degrees below positive x-axis)
One can find the value of tan θ from the length of the x-y vectors and from tan θ obtain θ
A 5 kg block is moved up a 30 degree incline by a force of 50 N, parallel to the incline. The coefficient of kinetic friction between the block and the incline is 0.25. What is the net work done on the block over this distance?
The net work done on the block over the given distance is 39.4d (joules)
What is the net work done on the block over this distance?The net work done on the block over the given distance is calculated by applying the following equation as shown below:
W(net) = F(net) x d where;
F(net) is the net force on the blockd is the distance moved by the blockF(net) = F - μmgcosθ where;
μ is the coefficient of kinetic frictionm is the mass of the blockg is acceleration due to gravityθ is the angle of inclination of the planeF(net) = 50 N - (0.25 x 5 x 9.8 x cos30)
NF(net) = 50 N - 10.6 NF(net) = 39.4 N
The net work done on the block over the given distance is calculated as: W = 39.4 N x dwhere;
d is the distance moved by the block = length of the incline
W = 39.4d (joules)
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a cat is moving at 2 m/s when it accelerates at 4 m/s for 2 seconds. his new velocity is
vf = vi + at = 2 m/s + (4 m/s²) (2 s) = 10 m/s
Monkey Joe and Money Jane are pulling a boat through water. Each exerts a force of 600N directed at a 30 angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the resistive force Fr, exerted in the boat by the water.
Answer:
1039 N
Explanation:
The 'x' components of their pulling forces = the force of friction:
2 x 600 cos 30 = 1039 N
Turner’s treadmill runs with a velocity of −1.2 m/s and speeds up at regular intervals during a half-hour workout. After 25 min, the treadmill has
a velocity of −6.5 m/s. What is the average acceleration of the treadmill
during this period?
0.212 m/s² is the average acceleration of the treadmill.
during this period
initial velocity, u = -1.2 m/s
time= 25 min
final velocity, v= −6.5 m/s
acceleration = (v-u)÷ t
acceleration =5.3/25
acceleration =0.212 m/s²
The rate at which an item changes its velocity is known as acceleration, a vector variable. If an object's velocity is changing, it is accelerating. A moving object can occasionally alter its velocity by the same amount every second. a moving object that changes its speed by 10 m/s per second. Since the velocity is changing by a fixed amount every second, this is known as a constant acceleration. It is important to distinguish between an item with a constant acceleration and one with a constant velocity. Be not deceived! An object is accelerating if its velocity is changing, whether by a fixed amount or a variable quantity. Additionally, a moving item with a constant speed is not accelerating.
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Kate was doing some resistance training where she would push against a brick wall
for 30 seconds. Which of the following statements best explains this situation in
terms of Work and Energy.
Kate did not do any work on the wall, because the wall did not move.
Kate used stored energy in her body which was likely converted to heat energy-
Kate did not change the Kinetic Energy of the wall; she did not change the
gravitational potential energy of the wall; and she did not change the elastic
potential energy of the wall.
All of the above statements are true.
Answer:
Kate did not do any work on the wall , because the wall did not move
Please help me with this looking to go over my work
0.99 joule must the work done by friction to stop the box.
m=87g
k=82N/m
x=11cm=0.11m
spring force=kx
spring force=82N/m×0.11m
spring force=9.02 N
work done= Fx
work done=9.02 N×=0.11m
work done=0.99 joule
When an item is moved across a distance by an external force, at least a portion of that force must be applied in the direction of the displacement. This is known as work done in physics. By multiplying the length of the path by the component of the force operating along the path, work may be calculated if the force is constant. The work W is equal to the force f times the distance h, or W = mgh, to mathematically describe this idea. The work is W = mas if the force is applied at an angle of to the displacement.
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Two 10¢ coins (dimes) carrying identical charges are lying 2.5 m apart on a table. If each of these coins experiences an electrostatic force of magnitude 2.0 N due to the other coin, how large is the charge on each coin?
I know the answer is 52 micro coulombs, but I keep getting 37.26...
Answer:
Charge q = 37.27 micro coulombs
Explanation:
Given:
q₁ = q₂ = q
r = 2.5 m
F = 2.0 N
________
q - ?
Coulomb's law:
F =k·q₁·q₂ / r² = k·q·q / r² = k·q² / r²
Charge:
q = √ (F·r² / k)
q = √ (2·2.5² / (9·10⁹) ≈ 37.27·10⁻⁹ C
or q = 37.27 micro coulombs
The correct answer is not 52, but !!! :))
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!
Answer:
The force of friction and air resistance slow down the movement of the bus.
c) A forse is changing the motion of the bus
A weightlifter curls a 25 kg bar, raising it each time a distance of 0.50 m. How many times must he repeat this exercise to burn off the energy in one slice of pizza? Assume 25% efficiency. Energy content of one slice of pizza is 1260 kJ .
42 times must he repeat this exercise to burn off the energy in one slice of pizza.
What is energy?
Energy is the ability or capability to do tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.
Energy burn by the weightlifter = potential energy
Potential energy = mgh
Potential energy = 25.(9.8)(0.50)
Potential energy = 122.5 Joule.
Assume 25% of efficiency so energy burn = 122.5*25/100
energy burn = 30.625 joule
Number of times = 1260/30.625
Number of times = 42 times.
42 times must he repeat this exercise to burn off the energy in one slice of pizza.
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A spaceship is coasting in orbit around a planet.
A second spaceship sits motionless on the launch
pad. The two ships define frames that are accelerated with respect to each other, yet both might be
regarded as inertial frames. Explain.
As per Drag theory , one possible explanation is v=0 that means velocity of earth relative to ether is zero.
For t1=t2, as per equation
[tex]t1=2l/c [ 1 + v^{2} /c^{2} ]\\t2= 2l/c [ 1 + v^{2} /2c^{2} ][/tex]
Then there is no relative velocity between earth and ether. In other words the ether is dragged with the motion of earth with the same velocity as the earth. However if this explanation is acceptable for there should be no aberration of light. Even if the ether is considered to be dragged partially , the absence of fringes shift and value of aberration cannot be explained simultaneously.
It has been verified by experiments that newton's frame of reference taking stars to be fixed , is an inertial one , while the other one i.e. reference frame fixed to earth is not an inertial frame. Since the earth rotates about its axis as well as round the sun.
Even Galilean Transformation equations explain the structure of inertial frames.
For example - Length of a rod is invariant under Galilean transformations.
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a boy throws a ball of mass of 0.25kg straight upward with initial velocity of 20 m/s when the ball returns its speed it 17m/s. how much work does air resistance do on the ball
1.125 joule work done does air resistance do on the ball
work done= kinetic energy= mΔv²/2
m=0.25kg
Δv=3 m/s
work done= mΔv²/2
work done=-0.25×9÷2
work done=-1.125 joule
When an item is moved across a distance by an external force, at least a portion of that force must be applied in the direction of the displacement. This is known as work done in physics. By multiplying the length of the path by the component of the force operating along the path, work may be calculated if the force is constant. The work W is equal to the force f times the distance h, or W = mgh, to mathematically describe this idea. The work is W = mas if the force is applied at an angle of to the displacement.
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Starting from rest, a car accelerates at 2.2 m/s2 up a hill that is inclined 5.2 ∘ above the horizontal. How far horizontally has the car traveled in 13 s ?
The horizontal distance travelled by the car in the given time period of 13 seconds is determined as 168.3 m.
What is the horizontal distance travelled by the car?The horizontal distance travelled by the car is calculated as follows;
s = ut + ¹/₂acosθ(t²)
where;
v is the final horizontal speed of the caru is the initial horizontal speed of the cara is the acceleration of the cart is the time of motion of the carθ is the angle of inclination of the planes = 0 + ¹/₂(2) x cos(5.2) x (13)²
s = 168.3 m
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Write down in words the formula connecting acceleration, initial speed, final speed and time.
Formula connecting acceleration, initial speed, final speed and time.
v=u+at,
v2=u2+2as,
s=ut+1/2at2.
How many different acceleration formulas are there?For motion on a straight line with constant acceleration, there are five common formulas. The formulas are provided in terms of the initial velocity (u), final velocity (v), displacement (x), acceleration (a), and time (t).
What is the relationship between distance and acceleration?The following equation can be used to determine distance from acceleration: D = v*t + 1/2*a*t2. where an is the acceleration, t is the passage of time, and v is the velocity.
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Katie pulls her wagon by the handle. She exerts a force of 60 N 30 degrees above the
horizontal. The force of friction from the sidewalk on the wagon is 30 N. The wagon has
a mass of 20 kg. What is the normal force on the wagon?
a. 226 N
b. 196 N
c. 166 N
d. 136 N
Answer: D: 136 N
..
..
Pls help :)
An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion, as shown in this graph.
Answer the following questions:
What is the period and frequency of the pendulum’s motion?
How many seconds out of phase with the displacements shown would graphs of the velocity and acceleration be?
What is the acceleration due to gravity on the surface of the planet in ? Determine the number of g-forces.
Show any necessary calculations.
a. ) The period is 1/frequency = 5sec / 9 cycles = 0.555 seconds
b.) the seconds out of phase with the displacements shown would graphs of the velocity and acceleration be 5/18 (0.2777...) sec delayed with respect to displacement
c.) Gravity = (324 π²) (0.2) / 25 = 25.582 m/sec²
What is gravity?
Gravity is defined as a fundamental interaction which causes mutual attraction between all things with mass or energy.
Solving for the acceleration due to gravity on the surface of the planet, we have the following:
For small swing angles, the period of an ideal pendulum anywhere is
T = 2pi √(length / local gravity) .
The astronaut has already done the pendulum and transmitted the data to us, so
we can use his data and this formula to calculate the local gravity where he is located
P = 2pi √(length / local gravity)
5/9 sec = 2π √(0.2m / gravity)
√(0.2m / gravity) = 5/9sec / 2π
Take the reciprocal of each side: √(gravity) / √(0.2) = 18π / 5
Multiply each side by √(0.2): √(gravity) = 18π √(0.2) / 5
Square both sides: Gravity = (324 π²) (0.2) / 25 = 25.582 m/sec²
The value gotten is about 2.608 times the Earth's gravity.
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Excluding air pressure, there is how many force or force(s) acting on a book lying at rest on a tabletop?
Answer: 2 Forces
Explanation:
1. Normal force- object is on a surface
2. Weight
State the name given to reflected sound waves
Echo
Explanation:
It is termed as Echo. Echo is a reflection of sound that arrives at the listener with a delay after the direct sound.
A 70kg student, happy to see snow in early
November, rushes out to sled on a hill with a 10kg
sled. They start h=12m above a horizontal plane.
How fast are they moving by the time they reach
the bottom of the hill? (assume the coefficient of
friction is zero here) After a short distance the
reach a point where there are some dry leaves are
scattered producing an effective coefficient of
friction uk=0.4. How far do they travel through the leaves before they
come to rest?
Use the fundamental principle
(a) The speed of the student at the bottom of the hill is 15.34 m/s.
(b) The speed of the student at the presence of friction force is 11.88 m/s.
What is the speed of the student at the bottom of the hill?
The speed of the student at the bottom of the hill is calculated by applying the principle of conservation of energy as shown below.
Kinetic energy at the bottom hill = potential energy at maximum height
¹/₂mv² = mgh
where;
m is mass of the studentv is the speed of the speed of the student at bottom hillh is the height of the hillv² = 2gh
v = √2gh
v = √(2 x 9.8 x 12)
v = 15.34 m/s
The speed of the student at the presence of friction force is calculated as;
Kinetic energy at the bottom hill + work done against friction = potential energy at maximum height
¹/₂mv² + μmgh = mgh
¹/₂v² + μgh = gh
v² + 2μgh = 2gh
v² = 2gh - 2μgh
v² = (2 x 9.8 x 12) - (2 x 0.4 x 9.8 x 12)
v² = 141.12
v = √141.12
v = 11.88 m/s
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I am not sure how to approach this question.
Answer:
35.2 AU
Explanation:
Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
[tex]\boxed{T^2 \propto a^3 \implies T^2=ka^3}[/tex]
Kepler's Third Law
[tex]T^2=\dfrac{4 \pi^2}{GM}a^3[/tex]
where a is the semi-major axis of the ellipse.
If expressed in the following units:
T = Earth years.a = Astronomical units AU (a=1 AU for Earth).M = Solar masses.then:
[tex]\implies\dfrac{4 \pi^2}{GM}=1[/tex]
Therefore Kepler's Third Law can be expressed as:
[tex]\boxed{T^2=a^3}[/tex]
Given:
T = 75.6 yearsFrom inspection of the given diagram:
[tex]\implies 2a=x+0.57[/tex]
[tex]\implies a=\dfrac{x+0.57}{2}[/tex]
Substitute these values into the equation and solve for x:
[tex]\implies (75.6)^2=\left(\dfrac{x+0.57}{2}\right)^3[/tex]
[tex]\implies (75.6)^\frac{2}{3}=\dfrac{x+0.57}{2}[/tex]
[tex]\implies 2(75.6)^\frac{2}{3}=x+0.57[/tex]
[tex]\implies x=2(75.6)^\frac{2}{3}-0.57[/tex]
[tex]\implies x=35.1883819...[/tex]
[tex]\implies x=35.2\; \text{AU}[/tex]
10. If an evaporator has a pressure drop of 10 psig and the low side pressure is 15 psig, what is the pressure at the
inlet of the evaporator?
a.5 psig
b.10 psig
c.15 psig
d.25 psig
If an evaporator has a vapour pressure drop of 10 psig and the low side pressure is 15 psig, pressure at the inlet of the evaporator will be 25 psig
change pressure= low side pressure-pressure drop
change pressure=15-(-10)
change pressure=25 psig
The vapor pressure of a liquid, sometimes referred to as the equilibrium pressure of a vapor above its liquid, is the pressure of the vapor produced by the evaporation of a fluid (or solid) over a sample of the liquid (or solid) in such a closed container (or solid).
Vapor pressure is the term for the force that is produced as liquids evaporate. Three factors commonly have an effect on a vapor press: surface area, intermolecular forces, and temperature. The vapor pressures of molecules change with temperature.
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950 kg
3.
If the car has 30.400 J of kinetic
energy, how fast is it moving?
An elevator lifts 2400 kg a distance of 55 m in 18.9 s. How much power does the
elevator demonstrate?
The work done by the elevator over the 55 meters is easily calculable: W = mgh = (2400)(9.8)(55) = 1293600Joules. The total time of the trip can be calculated from the velocity of the elevator: t = x/v=55/18.9 = 2.91. Thus the average power is given by: P =W/t=1293600/2.91=444536 Watts
How massive would a black hole have to be in order for it to evaporate due to Hawking radiation in
only one year? How big is that mass compared to
some object with which you are familiar? (On the
surface of the Earth, 1 kg ≈ 2.2 lb.)
It takes a black hole an astonishingly long time to convert all of its mass into energy via Hawking radiation. A googol, or 10100 years, would be required for a supermassive black hole to completely vanish.
How are black holes able to dissipate energy?A black hole radiates, losing mass as it does so and beginning to release more radiation, which accelerates the evaporation of the black hole. It eventually contracts to a size close to the Planck mass, where its Schwarzschild radius and DeBroglie wavelength coincide.
How long does it take a black hole to be destroyed by Hawking radiation?Sagittarius A* 1087 years for Ton 618, the biggest black hole ever detected, and Sagittarius A* 1087 years for Sagittarius A* to evaporate from Hawking radiation It would take more than 10100 years for it to evaporate away (weighing a staggering 66 billion solar masses).
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A model of a helicopter rotor has four blades, each of length 3.4 m from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of 600 rev/min.
What is the radial acceleration of the blade tip expressed as a multiple of the acceleration due to gravity, g ?
Express your answer as a multiple of g .
The radial acceleration of the blade tip is 1370 g rad/s².
What is radial acceleration?Radial acceleration is defined as "the acceleration of the object is along the radius, directed towards the center" in uniform circular motion.
Given parameters:
Rotational speed of the helicopter rotor: ω = 600 rev/min
= 600 × 2π/60 rad/min.
= 20π rad/min.
Length of the blade: r = 3.4 m.
Then, the radial acceleration = ω²r = (20π)²× 3.4 rad/s².
= {(20π)²× 3.4}/9.8 g rad/s².
= 1370 g rad/s².
Hence, the radial acceleration of the blade tip is 1370 g rad/s².
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Compare the momenta of a 6300-kg elephant walking 0.11 m/s and a 50-kg dolphin swimming 10.4 m/s.
Answer:
Explanation:
p₁ = m₁·V₁ = 6300·0.11 = 693 kg·m/s
p₂ = m₂·V₂ = 50·10.4 = 520 kg·m/s
693 kg·m/s > 520 kg·m/s
A star loses 6.022×10²³ mass per second. The total mass of the star is 12.044×10⁴⁶ in how much time will the star lose all its mass.
please helpp!!!!
It would take the star 2 seconds to lose all it's mass.
I got this by dividing 12.044×10⁴⁶ by 6.022×10²³
Question 15 of 34
Which of the following would increase the mechanical advantage of a lever?
A. Moving the fulcrum closer to the output force
O B. Moving the fulcrum farther from the output force
C. Decreasing the input force without moving the fulcrum
D. Increasing the input force without moving the fulcrum
The option that would increase the mechanical advantage of a lever is option A. Moving the fulcrum closer to the output force.
What influences a lever's mechanical advantage?Mechanical advantage quantifies a lever's effectiveness (how easy it is to lift the load). - When compared to the distance between the load (resistance) and the fulcrum, the advantage relies on the distance between the effort and the latter (effort arm) (resistance arm).
Note that in Levers of Class 1:
The mechanical advantage will increase when the fulcrum is brought closer to the load.The mechanical advantage will increase when the effort is moved further from the fulcrum. You might need a longer lever for this.Learn more about mechanical advantage from
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Calculate weight if an object, given its mass
Answer:
Weight is a measure of the force of gravity pulling down on an object. It depends on the object's mass and the acceleration due to gravity
Explanation:
When an earthquake occurs, potential energy in the Earth's crust is released and waves of energy move across the Earth. What are
these waves called?
OA. seismic waves
OB. crust waves
OC. rock waves
OD. solid waves
Answer:
seismic waves
Explanation:
A set of lights consists of 20 lamps connected in series to the 230 V mains electricity supply. 1. (a) When the lights are switched on and working correctly, the current through each lamp is 0.25 A. (i) What is the total current drawn from the mains supply?
The total current that is drawn from the mains supply is 5 A.
What is current?The term current has to do with the flow of charges in a circuit. Now we have been told that A set of lights consists of 20 lamps connected in series to the 230 V mains electricity supply. It was also said in the question that when the lights are switched on and working correctly, the current through each lamp is 0.25 A.
Now we have the following information;
Current passing through each lamp = 0.25 A
Number of lamps = 20 lamps
Hence;
Total current = 0.25 A * 20
= 5 A
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