Answer:
M.A = load/ effort
1200N/400N
= 3
velocity ratio= radius of wheel/radius of Axle
40cm/10cm
=4
efficiency= 3/4*100
75%
i. mechanical advantage = 3
ii. velocity ratio = 4
iii. efficiency = 75%
iv. output work = 4800 J
v. input work = 64 J
What is mechanical advantage?
Mechanical advantages allows humans to perform tasks much easier in terms of the force they need to apply, but must always obey the conservation of energy.
i) mechanical advantage = Resistance force / Effort force
= 1200N / 400 N
= 3
ii) velocity ratio = velocity of effort / velocity of load
= distance moved by effort / distance moved by load
= 2π R (radius of wheel) / 2π r ( radius of axle )
= R / r = 40 cm/ 10 cm = 4
iii) efficiency = ?
Mechanical advantage = efficiency * velocity ratio
efficiency =( Mechanical advantage / velocity ratio) *100
= (3/ 4) * 100
= 75%
iv) output work = Force to lift the object * distance
= 1200 * 4 = 4800 J
v) efficiency = output work / input work
75 = 4800 J / input work
input work = 64 J
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find the velocity of the object for all relevent times find the position of the object for all relevent times a softball is popped up vertically velocity of 32 m/s
Answer:
whats the formula
Explanation:
calculate the opportunity acost of the production of good x at each point .what law dose the ternd in those value exhibit ?
Answer:
a
Explanation:
Where does a body have more weight the poor at the eqator of the earth.
Answer:
Explanation:
Your body weighs more at the pole for two important reasons. Both have to do to the spin of the earth on its axis.
Because of its spin the earth is thicker around the equator than it is through the poles. This means that when you stand on the equator, you are farther away from the center of earth than you would be at the poles. As gravity decreases with the inverse of the square of distance, gravity will be weaker at the equator.
As you are also spinning with the earth, you will have a required centripetal acceleration and force to keep you attached to the ground, This force decreases the effect of gravity so again, you would weigh less at the equator.
The ratio of RMS of velocity , Probable Velocity and Average velocity is ??
Answer:
2 :π8 :3
The ratio between the most probable velocity,mean velocity and root mean square velocity is 2 :π8 :3. Vrms=M3RT.
Explanation:
pls mark brainliest
Two point charges of equal magnitude are 7.5cm apart. At the midpoint of the line connecting them, their combined electric field has a magnitude of 49N/CFind the magnitude of the charges (pC)
Answer:
q 1 = q 2 = Q
E = k * Q / r²
The combined electric field:
E = 2 * k * Q / r²
k = 9 · 10^9 Nm²/C²
r = 0.075 : 2 = 0.0375 m
45 N/C = 2 · 9 · 10^9 · Q / 0.0375²
45 = 18 · 10 ^9 Q / 0.0014062
Q = 45 · 0.0014062 / 18 · 10^9
Q = 0.003515 · 10^(-9) C
Q = 3.5155 · 10 ^ (-12) C = 3.5155 pC
What is a downside of nonprofit fitness centers?
They offer fewer luxury services than for-profit health centersThey offer fewer luxury services than for-profit health centers , ,
Very few people live near a nonprofit fitness center.Very few people live near a nonprofit fitness center. , ,
Members tend to be wealthy and lack diversity. Members tend to be wealthy and lack diversity. , ,
Membership costs more than it does in for-profit centers.Membership costs more than it does in for-profit centers. , ,
they offer fewer luxury services because they lack the money to provide them.
An electric motor and a generator are similar in that both
A. Produce mechanical energy.
B. Use electricity.
C. Transform energy into a different form.
D. Create energy.
C. transform energy into a different from.
I hope this helps
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. What is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.
We have that the spring constant is mathematically given as
[tex]k=2.37*10^{11}N/m[/tex]
Generally, the equation for angular velocity is mathematically given by
[tex]\omega=\sqrt{k}{m}[/tex]
Where
k=spring constant
And
[tex]\omega =\frac{2\pi}{T}[/tex]
Therefore
[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]
Hence giving spring constant k
[tex]k=m((\frac{2 \pi}{T})^2[/tex]
Generally
Mass of earth [tex]m=5.97*10^{24}[/tex]
Period for on complete resolution of Earth around the Sun
[tex]T=365 days[/tex]
[tex]T=365*24*3600[/tex]
Therefore
[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]
[tex]k=2.37*10^{11}N/m[/tex]
In conclusion
The effective spring constant of this simple harmonic motion is
[tex]k=2.37*10^{11}N/m[/tex]
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Select the correct answer. Which of Newton's laws explains why your hands get red when you press them hard against a wall? A. Newton's law of gravity B. Newton's first law of motion C. Newton's second law of motion D. Newton's third law of motion
Answer:
D newton third law
Explanation:
good luck
if you jog at a speed of 1.5m/s for 20 seconds how far di you travel
Answer: 30m
Explanation:
Given:
Speed: 1.5m/s
Time: 20 seconds
Distance = speed × time
Distance = 1.5 × 20
= 30m
Therefore you will travel 30m
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The car has a mass of 0·50 kg. The boy
now increases the speed of the car to 6·0
ms-1 . The total radial friction between
the car and the track has a maximum
value of 7.0 N. Show by calculation that
the car cannot continue to travel in the circular path.
Answer:
A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
What is meant by centripetal force ?Centripetal force is described as the force applied to a body that is travelling in a circular motion and is pointed in the direction towards the center of the circular path.
Here,
Mass of the car, m = 0.5 kg
Velocity of the car, v = 6 m/s
Radial friction between the car and the track, f = 7 N
The necessary centripetal force for the car to execute the circular motion is provided by the maximum radial frictional force between the car and the track.
So, the condition that the car cannot continue to travel in the circular path is that the centripetal force required is greater than the maximum radial friction.
So,
mv²/r > f
0.5 x 6²/r > 7
Therefore, the radius of the circular track,
r < 18/7
r < 2.57 m
Hence,
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
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Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 55.3 kg, down a theta= 79.6º slope at constant acceleration a=-4.3 m/s2, as shown in Figure (here we assume the positive direction is going down the slope. So the given acceleration is a negative value, it means its direction is going up the slope, slowing down as it moving downward). So, the coefficient of friction between the sled and the snow is 0.100. How many Joules of work is done by the tension in the rope as the sled moves 2.1 m along the hill? Use g= 10 m/s2.
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
In this case, we need to construct the Free Body Diagram of the sled-victim System in order to determine what Forces are doing Work. Then, we construct the respective Energy equation by Newton's Laws of Motion, Work-Energy Theorem and definition of Work.
Given that system experiments an uniform Acceleration, we must solve the resulting model for the work done by the Tension in the rope.
From the Free Body Diagram (see image attached), we see that both Weight of the sled and Friction between sled and snow are doing work in favor of gravity, whereas Tension forces is against gravity. Normal force is not doing work as its direction is perpendicular to the direction of motion. The energy equation of this system is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex] (1)
Where:
[tex]W_{T}[/tex] - Work done by tension, in joules.
[tex]m[/tex] - Mass of the sled-victim system, in kilograms.
[tex]\mu[/tex] - Coefficient of kinetic friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex] - Travelled distance, in meters.
[tex]\theta[/tex] - Slope angle, in sexagesimal degrees.
[tex]a[/tex] - Net acceleration of the sled-victim system, in meters per square second.
If we know that [tex]\mu = 0.100[/tex], [tex]m = 55.3\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]s = 2.1\,m[/tex], [tex]\theta = 79.6^{\circ}[/tex] and [tex]a = -4.3\,\frac{m}{s^{2}}[/tex], then the work done by the tension in the rope is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex]
[tex]W_{T} = \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta -m\cdot a\cdot s[/tex]
[tex]W_{T} = (0.100)\cdot \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \cos 79.6^{\circ} + \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \sin 79.6^{\circ} - (55.3\,kg)\cdot \left(-4.3\,\frac{m}{s^{2}} \right) \cdot (2.1\,m)[/tex]
[tex]W_{T} = 1662.544\,J[/tex]
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
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a coach is travelling east wards at 12.6 m/s after 12 second its velocity is 9.5 m/s in the same direction. what is the acceleration and direction of its acceleration?
pls do it with the formula
thx mates :)
[tex]\\ \rm\longmapsto a=\dfrac{v-u}{t}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{12.6-9.5}{12}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{3.1}{12}[/tex]
[tex]\\ \rm\longmapsto \overrightarrow{a}=0.25m/s^2[/tex]
A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?
Answer:
A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?
Explanation:
The image shows a facility that converts the energy of moving water into
electrical energy. What is one advantage of using this technology in place of a
coal-burning power plant?
A. It is fueled by a nonrenewable resource.
B. It causes no harm to ecosystems.
оооо
C. It produces more water pollution.
D. It does not emit greenhouse gases.
Answer:
D. It does not emit greenhouse gases.
Explanation:
D. It does not emit greenhouse gases.
Please help! Can give brainliest too.
Explanation:
[tex]v = \sqrt{2ax}[/tex]
Take the square of both sides:
[tex]v^2 = \left(\sqrt{2ax}\right)^2 = 2ax[/tex]
Divide both sides by 2a and you will get
[tex]x = \dfrac{v^2}{2a}[/tex]
If the horizontal range and the max height of a body projected at an angle titre to the horizontal is K and Q respectively. Show that the muzzle velocity Vo is given by: Vo=√[2gQ + K²/8Q]
Explanation:
Recall that
[tex]K = \dfrac{v_0^2\sin2\theta}{g}\:\:\:\:\:\:\:\:\:(1)[/tex]
and
[tex]Q = \dfrac{v_0^2\sin^2\theta}{2g}\:\:\:\:\:\:\:\:\:(2)[/tex]
From Eqn(2), we can write
[tex]\sin\theta = \sqrt{\dfrac{2gQ}{v_0^2}}\:\:\:\:\:\:\:\:\:(3)[/tex]
Using the identity [tex]\sin\theta = 2\sin\theta \cos\theta[/tex], we can rewrite Eqn(1) as
[tex]\dfrac{gK}{2v_0^2} = \sin\theta \cos\theta[/tex]
Squaring the above equation, we get
[tex]\dfrac{g^2K^2}{4v_0^4} = \sin^2\theta \cos^2\theta[/tex]
[tex]\:\:\:\:\:\:\:\:\:=\sin^2\theta(1 - \sin^2\theta)\:\:\:\:\:\:\:(4)[/tex]
Use Eqn(3) on Eqn(4) and we will get the following:
[tex]\dfrac{g^2K^2}{4v_0^4} = \dfrac{2gQ}{v_0^2}(1 - \dfrac{2gQ}{v_0^2})[/tex]
This simplifies to
[tex]\dfrac{gK^2}{8v_0^2Q} = 1 - \dfrac{2gQ}{v_0^2}[/tex]
Rearranging this further, we get
[tex]1 = \dfrac{2gQ}{v_0^2} + \dfrac{gK^2}{8v_0^2Q}[/tex]
Putting [tex]v_0^2[/tex] to the left side, we get
[tex]v_0^2 = 2qQ + \dfrac{gK^2}{8Q}[/tex]
Finally, taking the square root of the equation above, we get the expression for the muzzle velocity [tex]v_0[/tex] as
[tex]v_0 = \sqrt{2gQ + \dfrac{gK^2}{8Q}}[/tex]
A man on top of a tall building 60m high projects a body vertically upwards with an initial velocity of 20m/s
(1) find the maximum height above the ground reached by the body.
(2) the velocity with which the body hits the ground.
Answer:
1) 65 m
2) 40 m/s downward
Explanation:
Using for both questions the kinematic equation
v² = u² + 2as
and ignoring air resistance.
1) h = 60 + √(20²/(2(9.8))) = 64.51753...
2) v = √(20² + 2(9.8)(60)) = 39.69886...
The average human walks at a speed of 5km per hour if your PE teacher asks you to walk for 30 minutes in gym class how far would you walk(km)?
Answer:
2.5 km
Explanation:
Answer:
2.5 km
Explanation:
Distance = speed x time
So =5 x 0.5
If there is no slipping, a frictional force must exist between the wheels and the ground. In what direction does the frictional force from the ground on the wheels act
Answer:
tire advances to the right, the friction force must be directed to the left, that is, opposing the possible movement of the tire.
Explanation:
For the movement of the wheel to be composed of a rotating part and a translational part, it is necessary that there be a static friction force between the floor and the tire.
As the tire advances to the right, the friction force must be directed to the left, that is, opposing the possible movement of the tire.
Calculate the magnetic field and its direction at point P, which is 2.0 cm away from the top wire and 4.0 cm from the bottom wire. Assume both wires are infinitely long and each carries a current of 1.5 A.
Answer:
Explanation : B = μ₀i / 2πr
B1 = (2×10^-7) × 1.5/0.02
= 15 μT
B2 = (2×10^-7 ) × 1.5/0.04
= 7.5 μT
total field = 15 μT + 7.5 μT
= 22.5 μT
8. A mass of 10 Kg is accelerating at 3 m/s2. What is the applied net force?
Answer:
Explanation:
F = ma
F = (10)(3)
F = 30 N
Answer:
[tex]\boxed {\boxed {\sf 30 \ Newtons}}[/tex]
Explanation:
We are asked to find the applied net force. According to Newton's Second of Law, force is the product of mass and acceleration.
[tex]F= m \times a[/tex]
The object has a mass of 10 kilograms and it is accelerating at 3 meters per second squared.
m= 10 kg a= 3 m/s²Substitute the known values into the formula.
[tex]F= 10 \ kg \times 3 \ m/s ^2[/tex]
Multiply.
[tex]F= 30 \ kg \times m/s^2[/tex]
1 kilogram meter per second squared is equal to 1 Newton, so our answer of 30 kg × m/s² is equal to 30 N.
[tex]F= 30 \ N[/tex]
The applied net force is 30 Newtons.
an aluminum atom has an atomic number of 13 and a mass number of 27,how many
a)protons
b) electrons
pls write the formula too
Element is
[tex]\boxed{\sf {}^{27}Al_{13}}[/tex]
Atomic number=13Mass number=27[tex]\\ \sf\longmapsto No\:of\:Protons=Atomic \:Number=13[/tex]
And[tex]\\ \sf\longmapsto No\:of\:Neutrons=Mass\:number-Atomic\:Number[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=27-13[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=14[/tex]
And
[tex]\\ \sf\longmapsto No\:of\:electrons=No\:of\:Protons=13[/tex]
Can you please help me please?
Explanation:
CH3CH2OH
That is the answer I hope this helps
An electron is moving at speed of 6.3 x 10^4 m/s in a circular path of radius of 1.7 cm inside a solenoid the magnetic field of the solenoid is perpendicular to the plane of the electron's path. Find its relevatn motion.
Answer:
Here, m=9×10
−31
kg,
q=1.6×10
−19
C,v=3×10
7
ms
−1
,
b=6×10
−4
T
r=
qB
mv
=
(1.6×10
−19
)(6×10
−4
)
(9×10
−31
)×(3×10
7
)
=0.28m
v=
2πr
v
=
2πm
Bq
=
2×(22/7)×9×10
−31
(6×10
−4
)×(1.6×10
−19
)
=1.7×10
7
Hz
Ek=
2
1
mv
2
=
2
1
×(9×10
−31
)×(3×10
7
)
2
J
=40.5×10
−17
J=
1.6×10
−16
40.5×10
−17
keV
=2.53keV
Explain why liquid particles at a high pressure would need more
energy to change to a gas than liquid particles at a low pressure.
Answer:
Liquids evaporate faster as they heat up and more particles have enough energy to break away. The particles need energy to overcome the attractions between them. ... At this point the liquid is boiling and turning to gas. The particles in the gas are the same as they were in the liquid they just have more energy.
Please help wil give brainiest & 40p.
Answer:
This is the answer
Explanation:
You can find the ans in the photo I attached.
a model car moves round a circular path of radius 0.3m at 2 revolutions per secs what is its angular speed, the period of the car and the speed of the car
Answer:
a) T = 0.5 s
b) v = 1.2π m/s ≈ 3.77 m/s
Explanation:
It makes two revolutions in one second so makes one revolution in ½ second
circumference of the circle is
C = 2πr = 0.6π m
which it traverses in one time period
0.6π m / 0.5 s = 1.2π m/s
To solve this, we must be knowing each and every concept related to speed and its calculations. Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
What is speed?Speed may be defined as the distance traveled by an item in the amount of time it requires to travel that distance. In other words, it measures how rapidly an item travels but does not provide direction.
Speed may be calculated in Science. The speed equation is a scientific formula that is used to calculate various types of speed.
Mathematically, the formula for speed can be given as
speed= distance/time
Values that are given
Time period= 0.5 s
Circumference = 2πr = 0.6π m
substituting all the given values in the above equation, we get
speed =0.6π m / 0.5 s
On calculations, we get
= 1.2π m/s
=3.77 m/s
Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
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what happens to the weight of the body when it is falling freely under the action of gravity
Answer:
A freely falling object has weight W=mg, where W-weight, m-mass of the object and g-acceleration produced due to the earth's gravity. ... This happens because the normal reaction force exerted on the object in the lift is equal to zero, and normal force equals to mg, which in turn equals the weight of the object
Explanation:
plz mark me as brainliest
Answer:
Gradually increases until the maximum weight reaches the surface of the earth
Explanation:
A boat is able to move at 7.6 m/s in still water. If the boat is placed on the south shore of a river (water current of 3.4 m/s [SE]), and the captain wants to head straight across to the north shore:
a) In what direction should the captain point the boat?
b) Calculate the time it will take to cross (the river is 212.0 m from the south to the north shore).
Answer:
I don't get it rewrite please
