The mechanical advantage of ramp A is greater than others and it will require the least force to move the load to greater distance.
Let the height of the ramp = hLet the length of ramp B, C, D and E = LLet the length of the ramp A = 2LThe mechanical advantage of the ramp is calculated as follows;
[tex]M.A = \frac{L}{h}[/tex]
The mechanical advantage of the ramp B, C, D and E is calculated as;
[tex]M.A = \frac{L}{h} \\\\[/tex]
The mechanical advantage of the ramp A is calculated as follows;
[tex]M.A = \frac{2L}{h} \\\\M.A = 2(\frac{L}{h} )[/tex]
Since the length of the ramp A is greater than other ramps, the mechanical advantage will be greater and it will require the least force to move the load to greater distance.
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What is the difference between real and apparent weightlessness?
Answer:
In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.
Explanation:
Hope this helps:)
What is the voltage if the current is 4 A and the resistance is 10 Ω?
Answer:
40 volts
Explanation:
Use the equation [tex]V=IR[/tex]
[tex]V=(4)(10)[/tex]
[tex]V=40[/tex]
An object accelerates from rest, with a constant acceleration of 6.4 m/s2, what will its velocity be after 7s?
I also need the Formula
Hi there!
The formula for velocity given acceleration:
v = at
Plug in given values:
v = 6.4(7) = 44.8 m/s
why does air pressure decrease with increasing altitude?
Need help with dot product
[tex]\textbf{A}\cdot\textbf{B} = 11.5[/tex]
Explanation:
The dot product between two vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B}[/tex] is defined as
[tex]\textbf{A}\cdot\textbf{B} = AB\cos{\theta}[/tex]
where A and B are the magnitudes of the vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B},[/tex] respectively and [tex]\theta[/tex] is the angle between the two. Since A = 3, B = 5 and [tex]\theta = 40°,[/tex] the dot product [tex]\textbf{A}\cdot\textbf{B}[/tex] is
[tex]\textbf{A}\cdot\textbf{B} = (3)(5)(0.766) = 11.5[/tex]
Velocity and Acceleration Quick Check
Item 1
Use this graph of velocity vs. time for two objects to answer the question.
Item 2
Item 3
С
Item 4
Item 5
D
velocity
time
Which statement makes an accurate comparison of the motions for objects C and D?
(1 point)
lol
Answer:it’s C
Explanation:
by using graph of velocity vs. time for two objects, Item 4 and Item 5 statement makes an accurate comparison of the motions for objects. thus option C is correct.
What is velocity ?
velocity is the rate of change of the position of the object with respect to reference and it is complicated but velocity is basically speeding a particular object in a specific direction.
Velocity is a vector quantity which means both magnitude (speed) and direction are combinedly define define velocity. The SI unit of velocity is meter per second (ms-1) and the magnitude or the direction of velocity of a body changes leads to acceleration.
Speed and velocity are the two closest term but the major difference between speed and velocity is that speed gives us an idea that the object with the faster rate of movement r where as velocity speed up as well as tells us the direction of the body
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The 6 strings on a guitar all have about the same length and are stretched with about the same tension. The highest string vibrates with a frequency that is 4 times that of the lowest string. 1)If the strings are made of the same material, how would you expect the diameters of the lowest and highest strings to compare
Answer:
not sure i need points
Explanation:
What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1000 m/sm/s
Explanation:
Since the neutron is only moving at 1000 m/s, we are going to ignore the relativistic effects on its mass and energy. The mass of a neutron in [tex]m_n = 1.67×10^{-27}\:\text{kg}[/tex] so its kinetic energy KE is
[tex]KE = \frac{1}{2}m_nv^2[/tex]
[tex]\:\:\:\:\:\:\:\:= \frac{1}{2}(1.67×10^{-27}\:\text{kg})(10^3\:\text{m/s})^2[/tex]
[tex]\:\:\:\:\:\:\:\:= 8.35×10^{-22}\:\text{J}[/tex]
A photon's energy E is defined as
[tex]E = h\nu[/tex]
where [tex]\nu[/tex] is the photon's frequency and h is the Planck's constant. Solving for the frequency, we get
[tex]\nu = \dfrac{E}{h} = \dfrac{8.35×10^{-22}\:\text{J}}{6.63×10^{-34}\:\text{J-s}}[/tex]
[tex]\:\:\:\:\: = 1.26×10^{12}\:\text{Hz}[/tex]
which is right around the infrared radiation range.
Tectonic plate movement is the reason why northern California has a very different landscape than southern California. Two different tectonic plates, each moving in different directions, border the western side of the North American Plate. Use the map to identify the two tectonic plates that border the North American Plate to the west.
Answer:
Remember, NORTH ^, EAST >, SOUTH v, WEST <
Explanation:
It doesn't have to be a super complex answer. All you have to do is look to the left (west) of the North American plate. What are the 2 plates that you see? The Pacific and the Juan de Fuca, yeah? To the South, there is the Cocos amongst a few others.
I am not going to share the answer for sure as I haven't completed the test yet but that's how I'm solving it. You should write the answer in your own words anyways. Hope this helps! Have a good day :)
Answer:
The Juan de Fuca Plate and the Pacific Plate both border the west side of the North American Plate.
Explanation:
Edmentum
The car on this ramp starts from rest. When released, it
accelerates at a constant rate. It has an initial position of 12 cm
from the top of the ramp, and has an average velocity of 1.20 m/s
for a total of 1.80 s. Which is the correct final position of the car?
Answer:
Explanation:
s 0.12 + 1.20(1.80) = 2.28 m from the top.
A cyclist on a training ride records the distance she travels away from home. The data only shows the first150 minutes of the ride before her cycling computer ran out of battery.
Answer:
A) 58 km
B) 30 mins
Explanation:
In pic details
graph in pic
A flywheel, rotating about its axis at a rate of 4 rev/s is acted upon by a torque of 25 Nm for 10 sec. If the wheel has moment of inertia of 1.2 kgm^2, what would be the speed of the wheel at the end in rev/s?
The applied torque increases the angular speed by the application of an
angular acceleration.
The speed after 10 seconds is approximately 37.16 rev/s.Reasons:
The speed of the flywheel at the axis = 4 rev/s
The torque applied, T = 25 N·m
The time the torque is applied, t = 10 sec
Moment of inertia of the flywheel, I = 1.2 kg·m²
Required:
The speed at the end of the 10 seconds
Solution:
T = I·α
Where;
α = Angular acceleration
[tex]\displaystyle \alpha = \frac{T}{I}[/tex]
Therefore;
[tex]\displaystyle \alpha = \frac{24 \ N\cdot m}{1.2 \ kg \cdot m^2} = \mathbf{20\frac{5}{6} \ s^{-2}}[/tex]
The rotational speed, ω = ω₀ + α·t
Which gives;
[tex]\displaystyle \mathrm{The \ angular \ speed, } \ \omega = \frac{2 \cdot \pi \times 4 \ rad }{s} = \frac{8 \cdot \pi \ rad }{s}[/tex]
ω₀ = 8·π rad/s
Which gives;
[tex]\displaystyle \omega = \mathbf{8 \cdot \pi +2 \frac{5}{6} \times 10} = 233.47[/tex]
The speed of the wheel in revolution per second is therefore;
[tex]\displaystyle Speed \ in \ rev/s = \frac{8 \cdot \pi +2 \frac{5}{6} \times 10}{2\cdot \pi} \approx 37.16[/tex]
The speed after 10 seconds is approximately 37.16 rev/s.
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An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 5.60 mm before stopping. How far does the lighter fragment slide
Answer:
M1 + M2 = 0 total momentum before explosion = momentum after explosion where M1 is the lighter fragment
M1 V1 + 7 M1 V2 = 0
V2 = -V1 / 7
The lighter fragment will slide 7 times as far - 39.2 mm because it must have 7 times as much velocity - assuming the distance slid is proportional to the original velocity
Accelerations are produced by
A. Masses
B.accelerations
C. Velocities
D.unbalanced, net forces
A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s, how high is the cliff, if it took 12 s for the ball to reach the ground?
I need the Formula
Hi there!
We can use the equation:
d = v₀t + 1/2at²
Where:
v₀ = initial velocity downward
a = acceleration due to gravity
t = time
Plug in given values:
d = 4(12) + 1/2(9.8)(12²)
d = 48 + 705.6 = 753.6 m
A 1.1 kg ball drops vertically onto a floor, hitting with a speed of 23 m/s. It rebounds with an initial speed of 5.0 m/s. (a) What impulse acts on the ball during the contact
Hi there!
We know that:
I = Δp = m(vf - vi)
Plug in the given values. Remember to take into account direction ⇒ let the rebound velocity be positive and initial be negative.
I = 1.1(5 - (-23)) = 30.8 Ns
The figure shows a rear view of a loaded two-wheeled wheelbarrow on a horizontal surface. It has balloon tires and a weight W = 684 N, which is uniformly distributed. The left tire has a contact area with the ground of AL = 6.20 × 10-4 m2, whereas the right tire is underinflated and has a contact area of AR = 9.20 × 10-4 m2. Find (a) the force from the left tire, (b) the pressure from the left tire, (c) the force from the right tire, (d) the pressure from the right tire that each tire applies to the ground.
Answer:
Explanation:
Summing moments about the CG to zero will show that the two normal forces are equal and have a value of FL = FR = 684/2 = 342 N
Pressure on the left
PL = 342 / 6.20e-4 = 551,612.9032... = 5.5e5 Pa
Pressure on the right
PR = 342 / 9.20e-4 = 371,739.1304... = 3.7e5 Pa
A) The force from the left tire is; FL = 342 N
B) The pressure from the left tire is; PL = 551613 N/m²
C) The force from the right tire is; FR = 342 N
D) The pressure from the right tire is; PR = 371739 N/m²
We see that;
FL and FR are upward forces
W is the downward force.
We know that in equilibrium;
Sum of upward forces = sum of downward forces
Thus;
FL + FR = W
We are given W = 684 N
Since W is at the center, it means that FL = FR. Thus;
FL = FR = 684/2
FL = FR = 342 N
We are given;
Contact area of left tire; AL = 6.2 × 10⁻⁴ m²
Contact area of right tire; AR = 9.2 × 10⁻⁴ m²
Formula for pressure is;
Pressure = Force/Area
Pressure on the left tire;
PL = FL/AL
PL = 342/(6.2 × 10⁻⁴)
PL = 551613 N/m²
Pressure on the t right tire;
PR = FR/AR
PR = 342/(9.2 × 10⁻⁴)
PR = 371739 N/m²
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A 10kg crate sits at rest on a rough flat surface. Astudent decides to pull the crate by attaching a rope at a 37 degree angle. Although the student pulls the rope with a force of 600 newtons, the coefficient of kinetic friction is large and has the force the student applies remains constant, how much time after he begins pulling the crate will it take before the crate has traveled a distance of 1.0 meter?
Answer:
Explanation:
Normal force of the surface on the box will be
N = mg - Fsinθ
Ν = 10(9.8) - 600sin37
N = -263
As normal force cannot be less than zero, the applied force lifts the crate off the surface.
Now it's just a matter of finding the acceleration
In the horizontal direction, the acceleration is
a = F/m
a = (600cos37) / 10
a = 47.9181... m/s²
the crate weight is mg = 10(9.8) = 98 N.
In the vertical direction the acceleration is
a = ((600sin37 - 98) / 10)
a = 26.3089... m/s²
total acceleration is
a = √(47.9181² + 26.3089²)
a = 54.6653... m/s²
s = ½at²
t = √(2s/a)
t = √(2(1.0)/54.6653)
t = 0.19127...
t = 0.19 s
can anyone explain how to do it for me? i don't understand...
Answer:
15[m].
Explanation:
1) the required distance is AD, for more info see the attached picture.
2) [tex]AD=\sqrt{AA_1^2+A_1D^2} =\sqrt{81+144}=15[m].[/tex]
b) A satellite is in a circular orbit around the Earth at an altitude of 1600 km above the Earth's surface. Determine the orbital period of the satellite in hours. [3]
Explanation:
The orbiting period of a satellite at a height h from earth' surface is
T=2πr32gR2
where r=R+h.
Then, T=2π(R+h)R(R+hg)−−−−−−−−√
Here, R=6400km,h=1600km=R/4
T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√
Putting the given values,
T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=7092s=1.97h
Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is 24h. Let us find the height h of such a satellite above the earth's suface in terms of the earth,'s radius.
Let it be nR.Then
T=2π(R+nR)R(R+nRg)−−−−−−−−−−√
=2π(Rg)−−−−−√(1+n)32
=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32
(5075s)(1+n)32=(1.41h)(1+n)32
For T=24h, we have (24h)=(1.41h)(1+n)32
or (1+n)32=241.41=17
or 1+n(17)23=6.61
or n=5.61
The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.
A tennis player strikes the tennis ball with an initial velocity of 44.7 m/s horizontally. The ball is initially 1.28 m above the ground and 12.9 m from the 0.914 m tall net. Does the tennis ball make it over the net?
Hi there!
We can begin by finding the total time taken for the ball to reach the net using the equation:
dₓ = vₓt
12.9 = 44.7t
12.9/44.7 = t = 0.289 s
Now, we can use the following equation to solve for displacement in the Y direction:
d = y₀ + vit + 1/2at²
There is no initial vertical velocity, so:
d = y₀ + 1/2at²
Plug in known values:
d = 1.28 + 1/2(-9.8)(0.289²)
d = 0.87m
Thus, since 0.87 m < 0.914 m, the tennis ball does NOT make it over the net.
An object will begin moving from rest when acted upon by which forces?
A. Forces that are slightly less than the force of friction
B. Forces that result in a net force of zero
C. Forces that are equal and act in opposite directions
D. Forces that are greater in one direction than in any other direction
Answer:
D
Explanation:
Process of elimnination + it's the only one that makes sense
An object will begin moving from rest when acted upon by forces that are greater in one direction than in any other direction. Hence, Option (D) is correct.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
When forces that are greater in one direction than in any other direction, resultant will be unbalanced forces. Unbalanced forces are those acting on a body when the net force acting on the body is greater than zero. The body alters its state of motion when unbalanced forces act on it.
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An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
Answer:
The distance traveled before takeoff is 1720 m
Explanation:
Given:a = + 3.2 m/s²
t = 32.8 s
Vᵢ = 0 m/s
To Find:d = ?
Now,
d = Vᵢ × t + 0.5 × a × t²
d = (0 m/s) × (32.8 s) + 0.5 × (3.20 m/s²) × (32.8 s)²
d = 1720 m
Thus, The distance traveled before takeoff is 1720 m
-TheUnknownScientist 72
What are cyclotrons and how are they used?
which of the following statements might be used to defend the Act of 1848
A 0.60-kgkg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3 mm on a frictionless horizontal surface. Part A If the cord will break when the tension in it exceeds 60 NN , what is the maximum speed the ball can have
11.4 m/s
Explanation:
The cord will break when the centripetal force exerted on it meets or exceeds the maximum tension [tex]T_{max}[/tex] that it can handle.
[tex]T_{max} = m\dfrac{v_{max}^2}{r}[/tex]
Solving for [tex]v_{max},[/tex] we get
[tex]v_{max}^2 = \dfrac{rT_{max}}{m}[/tex]
or
[tex]v_{max} = \sqrt{\dfrac{rT_{max}}{m}} =\sqrt{\dfrac{(1.3\:\text{m})(60\:\text{N})}{(0.6\:\text{kg})}}[/tex]
[tex]\:\:\:\:\:= 11.4\:\text{m/s}[/tex]
Please Help
A car of mass 400 kais sped up from 10m/s to 30m/s in a time of 15 seconds
What is the starting KE?
what is the ending KE?
what is the work done to speed up the car?
what is the power of watts?
Explanation:
The initial kinetic energy [tex]KE_0[/tex] is
[tex]KE_0 = \frac{1}{2}mv_0^2 = \frac{1}{2}(400\:\text{kg})(10\:\text{m/s})^2[/tex]
[tex]\:\:\:\:\:\;\:= 2×10^4\:\text{J} = 20\:\text{kJ}[/tex]
The final kinetic energy [tex]KE[/tex] is
[tex]KE = \frac{1}{2}mv^2 = \frac{1}{2}(400\:\text{kg})(30\:\text{m/s})^2[/tex]
[tex]\:\:\:\:\:\:\:= 1.8×10^5\:\text{J} = 180\:\text{kJ}[/tex]
The work done W on the car is
[tex]W = \Delta{KE} = KE - KE_0[/tex]
[tex]\:\:\:\:\:\:\:= 180\:\text{kJ} - 20\:\text{kJ} = 1.6×10^5\:\text{J}[/tex]
The power expended P is
[tex]P = \dfrac{W}{t} = \dfrac{1.6×10^5\:\text{J}}{15\:\text{s}} = 10667\:\text{Watts}[/tex]
[tex]\:\:\:\:\:= 10.7\:\text{kW}[/tex]
What is the birth rate of a population of 3000 chipmunks if 200 chipmunks are born each year?
o 0.022 births per chipmunk per year.
0.067 births per chipmunk per year.
O 0.82 births per chipmunk per year.
o 15 births per chipmunk per year.
Answer:
0.067 births per chipmunk per year
Explanation:
I took the test
Two objects are a distance of 1.7 x 103 meters apart. One object has a mass of 3 x 107 kg and the other has a mass of 6 x 108. Determine the gravitational force between the objects.
Answer: You need to use Newton's law for the equation --->
Explanation: G × M × m / separation. Thats how youll get your answer !!
6) The modern atomic model is called the _____.
A) central atom model
B) plum pudding model
C) nuclear atomic model
D) electron growth model
The modern atomic model is called the nuclear atomic model.
Therefore the correct answer is option C.
What are atomic models?There are many types of atomic models proposed in past based on their individual assumptions and the experimentations
John Dalton's atomic model.
The Plum Pudding Model, developed by J.J. Thomson,
Rutherford's model provided an explanation for the presence of a nucleus inside the atom.
In accordance with the current atomic model, atoms have a nucleus made up of protons and neutrons and an ethereal gradient or cloud surrounding them that houses electrons;
The modern atomic model is called the nuclear atomic model.
Thus, the correct answer is option C.
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