Answers
Answer:
True
Explanation:
This is because of the point where the forces are applied by our muscles and
the angle they have about the bones. Take for example the diagram I uploaded.
If we do a free body diagram and a sum of torques, we would get that:
[tex]F_{muscle}sin \theta r1 - mg r2 = 0[/tex]
In this case, mg is the same in magnitude as the force made by the hand to hold the ball, so:
[tex]F_{muscle}sin \theta r_{1} - F_{hand} r_{2} = 0[/tex]
If we solve the equation for the force of the muscle we would get that:
[tex]F_{muscle}=\frac{F_{hand}r_{2}}{r_{1}sin \theta}[/tex]
Since r2 is greater than r1 and the sin function can only return values that are less than 1, this means that the force of the muscle is much greater than the force used by the hand to hold the weight.
Let's use some standard values to prove this, let's say that r1=10cm, r2=35cm and theta=60 degrees. When inputing the values into the equation we get:
[tex]F_{muscle}=\frac{F_{hand}(35cm)}{(10cm)sin (60^{o})}[/tex]
which yields:
[tex]F_{muscle}=4.04 F_{hand}[/tex]
so in this example, the force made by the muscle is 4 times as big as the force exerted by the hand.
Related Questions
Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward. How fast does Bill see Carlos moving and in what direction?
a. 10 mis eastward
b. 5 m/s eastward
c. 15 m/s westward
d. 20 m/s westward
e. 10 m/s westward
Answers
Explanation:
Given that,
Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.
Taking eastward as positive direction, we have:
[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)
[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is
[tex]v_B=+5\ m/s[/tex]
Therefore, Carlos velocity in Bill's reference frame will be
[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]
So, the magnitude is 20 m/s and the direction is westward (negative sign).
How much work does the electric field do in moving a proton from a point with a potential of 170 V to a point where it is -50 V
Answers
Physics
Two identical point charges are fixed to diagonally opposite corners of a square that is 1.5 m on a side. Each charge is +2.0 µC. How much work is done by the electric force as one of the charges moves to an empty corner? I have W(fe)= -EPE=-q[V(f)-V(i)].
physics
Two point charges are placed on the x axis. (Figure 1) The first charge, q1 = 8.00nC , is placed a distance 16.0m from the origin along the positive x axis; the second charge, q2 = 6.00nC , is placed a distance 9.00m from the origin along the negative x
A car travelling at 14.0 m/s approaches a traffic light. The driver applies the brakes and is able to come to halt in 5.6 s. Determine the average acceleration of the car during this time interval.
Answers
Answer:
[tex]a=2.5\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the car, u = 14 m/s
Finally, it comes to rest, v = 0
Time, t = 5.6 s
We need to find the average acceleration of the car during this time interval. We know that,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0-14}{5.6}\\\\a=-2.5\ m/s^2[/tex]
So, the acceleration of the car is [tex]2.5\ m/s^2[/tex] in the opposite direction of motion.
An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 and a dielectric strength of 2 x 108 V/m. The separation between the plates of the capacitor is 5 mm. What is the maximum electric charge (in nC) that can be stored in the capacitor before dielectric breakdown
Answers
Answer: [tex]283.2\times 10^{-9}\ nC[/tex]
Explanation:
Given
Cross-sectional area [tex]A=0.4\ cm^2[/tex]
Dielectric constant [tex]k=4[/tex]
Dielectric strength [tex]E=2\times 10^8\ V/m[/tex]
Distance between capacitors [tex]d=5\ mm[/tex]
Maximum charge that can be stored before dielectric breakdown is given by
[tex]\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC[/tex]
Answer:
The maximum charge is 7.08 x 10^-8 C.
Explanation:
Area, A = 0.4 cm^2
K = 4
Electric field, E = 2 x 10^8 V/m
separation, d = 5 mm = 0.005 m
Let the capacitance is C and the charge is q.
[tex]q = CV\\\\q=\frac{\varepsilon o A}{d}\times E d\\\\q = \varepsilon o A E\\\\q = 8.85\times 10^{-12}\times0.4\times 10^{-4}\times 2\times 10^8\\\\q = 7.08\times 10^{-8}C[/tex]
The pan flute is a musical instrument consisting of a number of closed-end tubes of different lengths. When the musician blows over the open ends, each tube plays a different note. The longest pipe is 0.31 m long.
What is the frequency of the note it plays? Assume room temperature of 20∘C.
Answers
Answer:
f = 276.6 Hz
Explanation:
This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.
In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is
L = λ/ 4
speed is related to wavelength and frequency
v = λ f
λ = v / f
we substitute
L = v / 4f
f = v / 4L
the speed of sound at 20ºC is
v = 343 m / s
let's calculate
f = [tex]\frac{343 }{4 \ 0.31}[/tex]
f = 276.6 Hz
Write a short note of the following
a) Reflection
b) Refraction
c) Diffraction
Answers
Answer:
a) Light that passes through the floor to reveal yourself (not shadow).
b) 2 rays of light that bounce between 2 transparent media.
c) I don't know what is Diffraction?
A satellite measures a spectral radiance of 8 Watts/m2/um/ster at a wavelength of 10 microns. Assuming a surface emissivity of 0.90, what would be the estimated temperature
Answers
Two identical cars, each traveling at 16 m>s, slam into a concrete wall and come to rest. In car A the air bag does not deploy and the driver hits the steering wheel; in car B the driver contacts the deployed air bag. (a) Is the impulse delivered by the steering wheel to driver A greater than, less than, or equal to the impulse delivered by the air bag to driver B
Answers
Answer:
I = - m 16 the two impulses are the same,
Explanation:
The impulse is given by the relationship
I = Δp
I = p_f - p₀
in this case the final velocity is zero therefore p_f = 0
I = -p₀
For driver A the steering wheel impulse is
I = - m v₀
I = - m 16
For driver B, the airbag gives an impulse
I = - m 16
We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less
1) Consider an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper cable used to transmit this current has a diameter o = 2.00 cm and a length L = 150 km. If there are 8.43x10^28 free electrons per cubic meter (m^3 ) in the cable, calculate how long it would take for an electron to cross the entire length of the transmitter line.
Answers
Answer:
t = 1.27 x 10⁹ s
Explanation:
First, we will find the volume of the wire:
Volume = V = AL
where,
A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²
L = Length of wire = 150 km = 150000 m
Therefore,
V = 47.12 m³
Now, we will find the number of electrons in the wire:
No. of electrons = n = (Electrons per unit Volume)(V)
n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)
n = 3.97 x 10³⁰ electrons
Now, we will use the formula of current to find out the time taken by each electron to cross the wire:
[tex]I =\frac{q}{t}[/tex]
where,
t = time = ?
I = current = 500 A
q = total charge = (n)(chareg on one electron)
q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)
q = 6.36 x 10¹¹ C
[tex]500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}[/tex]
Therefore,
t = 1.27 x 10⁹ s
Example 2.13 The acceleration a of a particle in a time t is given by the equation a = 2+ 5t^2. Find the instantaneous velocity after 3s. Solution
Answers
Answer:
the instantaneous velocity is 51 m/s
Explanation:
Given;
acceleration, a = 2 + 5t²
Acceleration is the change in velocity with time.
[tex]a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s[/tex]
Therefore, the instantaneous velocity is 51 m/s
A bird has a kinetic energy of 3 J and a potential energy of 25 J. What is the mechanical energy of the bird?
Answers
Answer:
28 j
Explanation:
because when you add you get 28
A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a magnetic force of 0.16 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of 72.0° with respect to the wire.
Answers
Answer:
7.28×10⁻⁵ T
Explanation:
Applying,
F = BILsin∅............. Equation 1
Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°
Substitute these values into equation 2
B = 0.16/(68×34×sin72°)
B = 0.16/(68×34×0.95)
B = 0.16/2196.4
B = 7.28×10⁻⁵ T
Why must scientists be careful when studying
nanotechnology?
Answers
Answer:
When studying nanotechnology, scientists must be aware that their ideas may not work out. Their work could be very time consuming and cost a lot of money. Finally, scientists do not yet know all of the effects of nanotechnology on human health.
Hope it helps u:)
A 1030 kg car has four 12.0 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles
Answers
Answer:
The required fraction is 0.023.
Explanation:
Given that
Mass of a car, m = 1030 kg
Mass of 4 wheels = 12 kg
We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.
The rotational kinetic energy due to four wheel is
[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]
Linear kinetic Energy of the car is:
[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]
Fraction,
[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]
So, the required fraction is 0.023.
A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy … (use g = 10 m/s 2)
A) at the moment of impact.
B) 2 seconds after the stone is released.
C) after the stone has fallen 40 m.
D) when the stone is moving at 20 m/s.
At the moment of impact both Kinetic Energy and Potential Energy should be 0, right? So it can't be A), right? Or is this wrong? Is it indeed A)? Please show work and explain it well.
Answers
Answer:
Explanation:
The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.
From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.
What is the name of the invisible line that runs
down the center of the axial region?
Answers
Answer:
An axis is an invisible line around which an object rotates, or spins. The points where an axis intersects with an object's surface are the object's North and South Poles.
Explanation:
The Earth's axis is represented by the red line. The white circle represents axial precission, the slow "wobble" of the axis.
A wire carrying a 23.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's field, and there is a 2.45 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet?
Answers
Answer:
3.55 T
Explanation:
Applying,
F = BILsin∅.............. Equation 1
Where F = Force, B = magnetic Field, I = current, L = Length of the wire, ∅ = Angle between the wire and the magnetic field
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 2.45 N, L = 3.00 cm = 0.03 m, I = 23.0 A, ∅ = 90° (Perpendicular)
Substitute these values into equation 2
B = 2.45/(0.03×23×sin90)
B = 2.45/0.69
B = 3.55 T
In a similar rolling race (no slipping), the two objects are a solid cylinder and hollow cylinder of the same radius and mass. Which reaches the bottom first
Answers
Answer:
solid cylinder
Explanation:
the object that arrives first is the object that has more speed, let's use the concepts of energy
starting point. Highest point
Em₀ = U = m g h
final point. Lowest point
Em_f = K = ½ mv² + ½ I w²
since the body has rotational and translational movement
how energy is conserved
m g h = ½ mv² + ½ I w²
linear and angular velocity are related
v = w r
w = v / r
we substitute
m g h = ½ mv² + ½ I (v/r) ²
mg h = ½ v² (m + I /r²)
v = [tex]\sqrt{2gh \ \frac{m}{m + \frac{I}{r^2} } }[/tex]
the tabulated moments of inertia for the bodies are
solid cylinder I = ½ m r²
hollow cylinder I = m r²
we look for the speed for each body
solid cylinder
v₁ = [tex]\sqrt {2gh} \ \sqrt{\frac{m}{m + m/2} }[/tex]
v₁ = [tex]\sqrt{2gh} \ \sqrt{2/3}[/tex]
let's call v₀ = [tex]\sqrt{2gh}[/tex]
v₁ = 0.816 v₀
hollow cylinder
v₂ = [tex]\sqrt{2gh } \ \sqrt{\frac{m}{m+ m} }[/tex]
v₂ = v₀ √½
v₂ = 0.707 v₀
Therefore, the body that has the highest speed is the solid cylinder and since time is the inverse of speed, this is the body that spends less time to reach the bottom, that is, it is the first to arrive
At what angle torque is half of the max
Answers
1. A turtle and a rabbit are to have a race. The turtle’s average speed is 0.9 m/s. The rabbit’s average speed is 9 m/s. The distance from the starting line to the finish line is 1500 m. The rabbit decides to let the turtle run before he starts running to give the turtle a head start. If the rabbit started to run 30 minutes after the turtle started, can he win the race? Explain.
Answers
Answer:no
Explanation:because 0.9*(30*60)=0.9*1800=1620
The turtle has already won the race
Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
What will be the speed of the rabbit and the turtle?It is given
[tex]V_{t} = 0.9 \frac{m}{s}[/tex]
[tex]V_{r} = 9 \frac{m}{s}[/tex]
[tex]D=1500 m[/tex]
Time taken by turtle
[tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]
[tex]T=1666 minutes= 27 hours[/tex]
Time taken by rabbit
[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]
[tex]T=166 minutes[/tex]
since rabbit started 30 minutes after turtle then
[tex]T= 136+30=196 minutes[/tex]
[tex]T= 3.2 hours[/tex]
Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
To know more about average velocity follow
https://brainly.com/question/6504879
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (at 76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F ?
Answers
Answer:
The correct answer would be - Low pitch.
Explanation:
As it is known that if frequency increases then pitch will be increase as well as pitch depends on frequency, Now for the question it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases. As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
What is the meant of by renewable energy and non-renewrable with example of each.
Answers
Answer:
Renewable energy is a type of energy that can be renewed easily, such as sunlight. By using Solar panels to collect the suns energy, we are not depleting it, so this source is renewable.
Non-renewable energy is something that cannot easily be replenished. An example would be oil because oil takes millions of years to form and cannot be renewed easily.
Calculate the buoyant force due to the surrounding air on a man weighing 600 N . Assume his average density is the same as that of water. Suppose that the density of air is 1.20 kg/m3.
Answers
Answer:
[tex]F_b= 0.720 N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=600N[/tex]
Average density [tex]\rho=1.20kg/m^3[/tex]
Mass
[tex]m=\frac{W}{g}[/tex]
[tex]m=\frac{600}{9.81}[/tex]
[tex]m=61.22kg[/tex]
Generally the equation for Volume is mathematically given by
[tex]V =\frac{ mass}{density}[/tex]
[tex]V= \frac{61.22}{1000}[/tex]
[tex]V=0.06122 m^3[/tex]
Therefore
Buoyant force [tex]F_b[/tex]
[tex]F_b=\rho*V*g[/tex]
[tex]F_b= rho_air*V*g[/tex]
[tex]F_b= 0.720 N[/tex]
Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.
52 cm4
72 cm4
32 cm4
24 cm4
2 cm4
Answers
Answer:
Minimum Area of rectangle = 24 centimeter²
Explanation:
Given:
Length of rectangle = 6 centimeter
Width of rectangle = 4 centimeter
Find:
Minimum Area of rectangle
Computation:
Minimum Area of rectangle = Length of rectangle x Width of rectangle
Minimum Area of rectangle = 6 x 4
Minimum Area of rectangle = 24 centimeter²
distance of distinct vision.
is placed at a distance less than the distance of near point, its image o
will be blurred. Hence human eye can not see such object clearly.
ADDITIONAL INFORMATION
distance of distinct vision for a normal eye of different age groups
Babies = 7 cm
Adults = 25 cm
erson of age 55 years and above = 100 cm
ever, in our discussion we are concerned with a normal eye of an adult so least
The foulart position of an ahiect from a human eve so that the sh
Answers
The least distance up to which we can see the objects clearly without any strain is called least distance of distinct vision. Least distance of distinct vision for a normal human being is 25cm. For young people, the least distance of distant vision will be within 25cm which however it varies with age.
Answer:
25 you said ? thats incorecct
Explanation:
Suppose that 2 J of work is needed to stretch a spring from its natural length of 34 cm to a length of 46 cm. (a) How much work is needed to stretch the spring from 36 cm to 41 cm
Answers
Answer:
0.83 J of work
Explanation:
2 J of work is required to stretch a spring from 34cm to 46cm
So that is 12cm stretched with 2 J of work
We can make that 6cm for 1 J of work
So, we need the find the work for stretching 36cm to 41cm
Which is 5cm
So, What is the work required to stretch 5cm?
1 J of work for 6cm
x work for 5cm
So, by proportion method
1 : 6 :: x : 5
6 * x = 1 * 5
6x = 5
x = 5/6
= 0.83
So to stretch 36cm to 41cm we need 0.83 J of work
derive expression for pressure exerted by gas
Answers
Urgent please help me
Answers
1433 km
Explanation:
Let g' = the gravitational field strength at an altitude h
[tex]g' = G\dfrac{M_E}{(R_E + h)^2}[/tex]
We also know that g at the earth's surface is
[tex]g = G\dfrac{M_E}{R_E^2}[/tex]
Since g' = (2/3)g, we can write
[tex]G\dfrac{M_E}{(R_E + h)^2} = \dfrac{2}{3}\left(G\dfrac{M_E}{R_E^2} \right)[/tex]
Simplifying the above expression by cancelling out common factors, we get
[tex](R_E + h)^2 = \dfrac{3}{2} R_E^2[/tex]
Taking the square root of both sides, this becomes
[tex]R_E + h = \left(\!\sqrt{\dfrac{3}{2}}\right) R_E[/tex]
Solving for h, we get
[tex]h = \left(\!\sqrt{\dfrac{3}{2}} - 1\right) R_E= 0.225(6.371×10^2\:\text{km})[/tex]
[tex]\:\:\:\:\:= 1433\:\text{km}[/tex]
A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the string makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant
Answers
Answer:
[tex]T=8.1N[/tex]
Explanation:
From the question we are told that:
Mass m=0.40
Radius r=1.8m
Angle Beneath the Horizontal \theta =40 \textdegree
Speed v=5.0m/s
The Tension Angle
[tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]
[tex]\alpha=50 \textdegree[/tex]
Generally the equation for Tension is is mathematically given by
[tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]
[tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]
[tex]T=8.1N[/tex]
What would you expect to happen to the velocity of the bobber if the mass of the washers in the cylinder remained the same and the radius was doubled?
Answers
Answer:
The velocity becomes [tex]v\sqrt 2[/tex].
Explanation:
The force acting on the bobber is centripetal force.
The centripetal force is given by
[tex]F =\frac{mv^2}{r}[/tex]
when mass remains same, radius is doubled and the force is same, so the velocity is v'.
[tex]F =\frac{mv^2}{r}=\frac{mv'^2}{2r}\\\\v'=v\sqrt 2[/tex]
