To increase the mechanical advantage of incisors (front teeth) so that they act as wedges in the human body is through 2. an increase in the distance of the teeth from the fulcrum.
The decrease in the distance will decrease the mechanical advantage of incisors. The increase in the thickness at the widest part of the teeth will not necessarily support the incisors to become more effective wedges.
Thus, incisors will acquire some mechanical advantage by increasing the force that it exerts. This advantage can be achieved by increasing the distance of incisors from the fulcrum.
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Answer:
4 aka D
Explanation:
A baseball is hit so that it travels straight upward after being struck by the bat. If its initial velocity is 29 m/s , then what is the maximum height that it will reach?
Answer:
Explanation:
Use kinematic equation v² = u² + 2as
Rearrange for distance
s = (v² - u²) / 2a
Realize that at the top of its flight, the ball has zero velocity and gravity is acting downward in an assumed upward positive reference frame.
s = (0² - 29²) / (2(-9.8))
s = 42.90816...
s = 43 m
The seat on a carnival ride is fixed on the end of an 12.60-m-long beam, pivoted at the other end. If the beam sweeps through an angle of 141°, what is the distance through which the rider moves?
The distance through which the rider at the end of the beam moves is;
L = 15.5 m
We are told that the beam on which the carnival ride is fixed is 12.6m in length.
Since the seat is at the end of the beam with the other end pivoted and the beam sweeps through an angle of 141°, then we can say that the radius of this arc formed by the swing is;
Radius; r = 12.6 m
Also, θ = 141°
The distance through which the driver moves will be the length of the arc formed by the beam at angle of 141°.
Formula for length of arc is given as;
L = 2πr(θ/360)
Plugging in the relevant values gives;
L = 2π × 12.6 × 141/360
L = 15.5 m
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A car’s velocity as a function of time is given by Vx (t) = α.t + β.t 2 , where α= 3m/s and β= 0.1m/s 3 . Calculate the average acceleration for the time interval
b) t= 5 to t = 10 s
The definition of average acceleration allows to find the result for the average acceleration in the given time interval is:
[tex]a_{average}= 1.5 \ \frac{m}{s^2}[/tex]
Instantaneous acceleration is defined as the derivative of velocity with respect to time.
a = [tex]\frac{dv}{dt}[/tex]
Where a is the acceleration, v the velocity and t the time.
They indicate that the speed of the car is given by the relation.
v = α t + β t²
With α = 3 m / s and β = 0.1 m / s³
Let's make the derivative.
a = α + 2β t
Let's substitute
a = 3 + 2 0.1 t
Average acceleration is the change in velocity in the time interval.
[tex]a_{average} = \frac{\Delta v}{\Delta t }[/tex]
Let's find the velocity at the indicated time.
For t = 5 s
v₅ = 3 + 0.1 5²
v₅ = 5.5 m / s
For t = 10 s
v₁₀ = 3 + 0.1 10²
v₁₀ = 13 m / s
Let's calculate the average acceleration.
[tex]a_{average} = \frac{13 - 5.5 }{ 10 - 5 }\\[/tex]
[tex]a_{average}= 1.5 \ m/s^2[/tex]
In conclusion using the definition of mean acceleration we can find the result for the mean acceleration in the given time interval is:
[tex]a_{average} =[/tex] 1.5 m / s²
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The displacement of an object in SHM is described by the equation
[tex] x = cos\binom{2\pi}{3}t[/tex]
where x is in meters and t in seconds. Determine the velocity of the object at t = 0.6 s.
Answer:
[tex]-1.99\:\mathrm{m/s}[/tex]
Explanation:
Assuming that the equation is intended to be [tex]\displaystyle x=\cos\left(\frac{2\pi}{3}t\right)[/tex], we can find the velocity vs. time equation by taking the first derivative with respect to [tex]t[/tex]:
[tex]\displaystyle \frac{dx}{dt}=\frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)[/tex]
Recall the chain rule:
[tex]\displaystyle f(g(x))'=f'(g(x))\cdot g'(x)[/tex]
Therefore,
[tex]\displaystyle \frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex]
Therefore, the velocity vs. time equation of the object is [tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex].
Substitute [tex]t=0.6\text{ s}[/tex] into this equation to find the velocity at that given time:
[tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}(0.6)\right)\cdot \frac{2\pi}{3}\approx \boxed{-1.99\text{ m/s}}[/tex]
A diffraction grating, ruled with 300 lines per mm, is illuminated with a white light source at normal incidence.
(i) What is the angular separation, in the third-order spectrum, between the 400 nm and 600 nm lines? [5]
(ii) Water (of refractive index 1.33) now fills the whole space between the grating and the screen. What is the angular separation, in the first-order spectrum, between the 400 nm and 600 nm lines? [5]
the expression for diffraction grating allows to find the results for the questions for the angular separation are:
i) The third order is Δθ = 0.203 rad.
ii) The first order with water is Δθ = 0.046 rad.
The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.
d sin θ = m λ
Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.
i) Let's start by looking for the separation between two lines
Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.
d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm
d = 3.333 10⁻⁶ m
Let's find the angle of diffraction for the third order (m = 3) for each wavelength.
λ₁ = 400 nm = 400 10⁻⁹ m
sin θ₁ = [tex]\frac{m \ \lambda }{d}[/tex]m λ/ d
sin θ₁ = [tex]\frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]
θ₁ = sin⁻¹ 0.3600
θ₁ = 0.368 rad
λ₂ = 600 nm = 600 10⁻⁹ m
sin θ₂ = [tex]\frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]
θ₂ = sin⁻¹ 0.5401
θ₂ = 0.571 rad
The angular separation is
Δθ = θ₂ - θ₁
Δθ = 0.571 - 0.368
Δθ = 0.203 rad
ii) In this case, the separation between the network and the observation screen is filled with water.
When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.
[tex]n_i \ sin \theta_1 = n_r \ sin \theta_r[/tex]
The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.
Let's start looking for the incident angles for the first order of diffraction.
m = 1
λ₁ = 400 nm
θ₁ = sin⁻¹ [tex]\frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}[/tex]
θ₁ = 0.120 rad
λ₂ = 600 nm
θ₂ = sin⁻¹¹ [tex]\frac{1 \ 600 \ 10^{-9} }{3.33 \ 10^{-6}}[/tex]
θ₂ = 0.181 rad
we use the equation of refraction.
[tex]\theta_r[/tex] = sin⁻¹ ([tex]\frac{n_i}{n_r} \ sin \ \theta_i[/tex] )
λ₁ = 400 nm
θ₁ = sin¹ ([tex]\frac{1 sin 0.120}{1.33}[/tex]
θ₁ = 0.090 rad
λ₂ = 600 nm
θ₂ =sin⁻¹ [tex]\frac{1 sin 0.181}{1.33}[/tex]
θ₂ = 0.1358 rad
The angular separation is
Δθ = 0.1358 - 0.090
Δθ = 0.046 rad.
In conclusion using the relation for the diffraction grating we can find the results for the questions about angular separation are:
i) The third order is Δθ = 0.203 rad.
ii) The first order with water is Δθ = 0.046 rad.
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in order to keep heat in or out, you need a(n)
Answer:
What is instillation
Explanation:
Which of the following happens when a substance melts?
Answer:
hola como estas hablas español
Explanation:
.5A and 4.5V what’s the resistance
Explanation:
[tex]V = IR \Rightarrow R = \dfrac{V}{I}[/tex]
Plugging in the numbers, we get
[tex]R=\dfrac{4.5\:\text{V}}{0.5\:\text{A}} = 9.0[/tex] ohms
Which of the following statements about electromagnetic radiation it true? A.electromagnetic waves with long wavelength are more energetic then electromagnetic waves with short wavelength. B.all electromagnetic radiation carries the same amount of energy. C.electromagnetic radiation in a vacuum can change frequently to become more or less energetic. D.electromagnetic waves with high frequency are more energetic then electromagnetic waves with low frequency
Given what we know, we can confirm that option D, Electromagnetic waves with high frequency are more energetic than electromagnetic waves with low frequency is true.
Why are high-frequency waves more energetic?High-frequency waves are synonymous with short wavelengths. This means that the waves are oscillating much quicker and therefore carry more kinetic energy within them. This is transformed and released as electromagnetic radiation, which is the reason why high-frequency waves are more energetic than low-frequency electromagnetic waves.
Therefore, we can confirm that the statement "Electromagnetic waves with high frequency are more energetic than electromagnetic waves with low frequency" is true.
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what causes a solid air fresher to lose mass and volume
Answer:
A solid air freshener loses mass and volume as a result of sublimation, where solid particles skip the liquid state and change directly from a solid to a gas. Sublimation requires that the particles in the solid state gain enough energy to immediately become a gas.
Explanation:
Hope this helps.
A normal atom is electronically ** (positive/negative/neutral) because the number of ** (protons/neutrons/electrons), each with a positive charge, equals the number of ** (protons/neutrons/electrons), each with a negative charge.
The answers to multiple choice questions are in parentheses
Answer:
(neutral)
(protons)
(electrons)
Explanation:
Electrons have a negative charge (-) while protons have a positive (+) charge.
Atoms will usually be neutral, which means that there will be no charge.
For an atom to have a neutral charge, protons and electrons must "cancel" each other out. For this to happen, you need to have the same amount of each.
Positive will "cancel" out the negative.
1.Which term describes the sum of the number of protons and neutrons in an atom?
Your answer is → Mass number
A mover pushes a 45 kg crate across a level floor with a force of 300 N, but the crate accelerates at a rate of only 4.44 m/s2 because a friction force opposes the crate's motion. What is the magnitude of this force of friction? 300 N Force of Friction O A. 50 N N OB. 25 N C. 15 N 0 D. 100 N
By Newton's second law, the net force on the crate is
∑ F = 300 N - f = (45 kg) (4.44 m/s²)
where f is the magnitude of friction. Solve for f :
300 N - f = (45 kg) (4.44 m/s²)
f = 300 N - (45 kg) (4.44 m/s²)
f = 100.2 N ≈ 100 N (D)
What shape is the graph produced by a force vs acceleration graph?
A. Linear
B. Quadratic
C. Circular
D. Triangular
The answer to the question What shape is the graph produced by a force vs acceleration graph is A. Linear
Since Force, F = ma where m = mass and a = acceleration. For constant mass, F ∝ a. That is, F is directly proportional to acceleration, a.
Since this is a linear relationship, the graph of force vs acceleration will be linear.
The answer to the question What shape is the graph produced by a force vs acceleration graph is A. Linear
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A string is wrapped around a solid cylinder with mass M and radius R. The free end of the string is held in place, allowing the cylinder to fall. Recall that the moment of inertia of a solid cylinder rotated about its center is given by MR2/2. All answers to this problem should be symbolic, purely in terms of the variables M, R, and g. (a) Find the linear acceleration (in m/s2) of the cylinder and the tension in the string (in Newtons) as the cylinder falls. (b) Now suppose the cylinder is hollow instead of solid. The moment of inertia of a hollow cylinder rotated about its center is given by MR2. What is the acceleration and tension in this case?
Answer:
I will use (a / R) for alpha the angular acceleration
T R = I a / R torque equals angular acceleration for cylinder
M g - T = M a linear acceleration of center of mass
T = M (g - a) = I a / R^2 from first equation
If I = 1/2 M R^2 then M ( g - a) = M a / 2 from above
or g = 3 a / 2 and a = 2 g / 3
Also we have T = M (g - a) = M (g - 2 g / 3) = g / 3
Substitute I = M R^2 for the hollow cylinder
Looks like a = g/2 for hollow cylinder
1. As the angle of the ramp is increased, the normal force increases /decreases / remains the same and the friction-force increases /decreases / remains the same. [1 Point] 2. As the angle of the ramp is increased, the force parallel increases /decreases / remains the same. [1 Point] 3. The angle at which the force down the plane was equal to the force of friction (for the cabinet) was _____________. [1 Point] 4. Consider a very low (~ zero) friction, 5.0 kg skateboard on a ramp at an angle of 15o to the horizontal. What would be the net force that would cause acceleration when the skateboard is allowed to move
(1) As the angle of the ramp is increased, the normal force decreases.
(2) As the angle of the ramp is increased, the parallel force increases.
(3) The angle at which the force down the plane was equal to the force of friction is zero degree.
(4) The net force that would cause acceleration is 47.33 N.
Let the angle of inclination of the ramp = θ
(1)
The normal force on an object on the ramp inclined to the ramp is calculated as follows;
[tex]F_n = mgcos (\theta)[/tex]
when θ is 0;
[tex]F_n = mgcos (0)\\\\F_n = mg[/tex]
when θ is 90;
[tex]F_n = mgcos(90)\\\\F_n = 0[/tex]
Thus, as the angle of the ramp is increased, the normal force decreases.
(2)
The parallel force on an object on the ramp inclined to the ramp is calculated as follows;
[tex]F_x = mgsin(\theta)\\\\[/tex]
when θ is 0;
[tex]F_x = mgsin(\theta)\\\\F_x = mgsin(0) \\\\F_x = 0[/tex]
when θ is 90;
[tex]F_x = mgsin(90)\\\\F_x = mg[/tex]
Thus, as the angle of the ramp is increased, the parallel force increases.
(3)
The force of friction is calculated as follows;
[tex]F_n = \mu F_n[/tex]
[tex]F_k = \mu mgcos(\theta)[/tex]
[tex]F_k = \mu mg cos(0)\\\\F_k = \mu mg[/tex]
Thus, the angle is zero degree
(4)
The net force that would cause acceleration is calculated as follows;
[tex]F_k = Fn\\\\F_k = mg cos(\theta)\\\\F_k = 5 \times 9.8 \times cos(15)\\\\F_k = 47.33 \ N[/tex]
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i just need help plzzzz
Answer:
I think D it's the correct answer!! Let me know if i'm wrong
Explanation:
Part A click listen to a single source and select audio enabled and listener on the right control panel increase the frequency of the sound wave to about 600 hertz using he slider on the right. What happens to the sound generated
Answer:
When the frequency of the wave is increased, the pitch of the sound increases; that is, he sound becomes sharper or higher.
Explanation: Just completed on Edmentum
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A meteor falls from the sky to the Earth. The meteor already had an initial velocity downward when it was spotted. If it hit the Earth at 335 m/s after being seen for 30 seconds, then what was the initial velocity of the meteor?
Answer:
335 - 30 g
Explanation:
V = Vo + g*T = 335
Vo + 9.8*30 = 335
Vo =
Which statement is true
1) The phases of the moon are caused because sometimes the moon is in the Earth's
shadow
2) The phases of the moon is a result of the moon shinning different amounts of its own
light
3) We see different phases of the moon because the moon travels along its orbit around
the Earth and see different amounts of the illuminated half
4) We see different phases of the moon because different amounts of the moon is
illuminated by the sun
Answer:
I think the answer is option 3
Content from magnets
Answer:
magnets are magnetic :)
Explanation:
The first motor abilities a new born exhibits are
Answer:
here your answer
i am sorry if wrong
(c) Changing water into vapour is condensation true or false
Answer:
true
Explanation:
Find the speed required to throw a ball straight up and it return 6 seconds later. Neglect air resistance
Answer:
the ball will go up 3s and down 3s
v=gt
where t=3s and g=9.8m/s^2
distance=v0(t)+(1/2)gt^2
where initial velocity (v0)=0
Explanation:
The speed required to throw a ball straight up and returns 6 seconds later would be 29.43 meters/seconds.
What are the three equations of motion?There are three equations of motion given by Newton,
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
As given in the problem we have to find the speed required to throw a ball straight up and it returns 6 seconds later,
S = ut + 1/2*a*t²
0 = u×6 + 0.5×(-9.81)×6²
0 = 6u - 176.8
6u = 176.8
u = 176.8/6
u = 29.43 meters / seconds
Thus, the speed required to throw a ball straight up and returns 6 seconds later would be 29.43 meters/seconds.
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Explain why two acetate rods, both charged with silk repel
In a vacuum light travels at which speed?
Answer:
Light traveling through a vacuum moves at exactly 299,792,458 meters (983,571,056 feet) per second. That's about 186,282 miles per second
Explanation:
PLEASE I NEED HELP WITH 6 and 8 !!!! I will love u sm
Answer:
Explanation:
006
They are acting in opposite directions. Therefore the net force is found by subtraction. The sign is the same as the larger number.
Net Force = 99.6 - 52.8 = 46.8 N acting in the same direction as the 99.6 which is upward.
008
If the two forces act in the same direction, the net force is found by addition.
Net Force = 99.6 + 52.8 = 152.4 N downward.
Which of the following is correct about the time period of a pendulum, whose length (length of thread) is 2 m and mass of whose bob (metallic ball) is 2 kg, when it is set in motion?
Answer:
P = 2 * pi * (L / g)^1/2 period of simple pendulum
P = 6.28 * (2 / 9.8)^1/2 = 2.84 sec
Two spherically symmetric planets with no atmosphere have the same average density, but planet B has twice the radius of planet A. A small satellite of mass mA has period TA when it orbits planet A in a circular orbit that is just above the surface of the planet. A small satellite of mass mB has period TB when it orbits planet B in a circular orbit that is just above the surface of the planet.
A period of a satellite is the time taken by the satellite to travel round a
body.
The comparison between the periods [tex]T_B[/tex], and [tex]T_A[/tex] is [tex]\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}[/tex]
Reason:
The period, T, of a satellite is given as follows;
[tex]T = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \cdot M} }[/tex]
Volume of the planet A = [tex]\dfrac{4}{3} \cdot \pi \cdot r^3[/tex]
Mass of planet A, [tex]m_A[/tex] = [tex]\dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho[/tex]
Volume of the planet B = [tex]\dfrac{4}{3} \cdot \pi \cdot (2 \cdot r)^3 = \dfrac{32}{3} \cdot \pi \cdot r^3[/tex]
Mass of planet B, [tex]m_B[/tex] = [tex]\dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho[/tex]
Period of the satellite on planet A, [tex]T_A[/tex], is given as follows;
[tex]T_A = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{4}{3} \cdot \pi \times \rho} }[/tex]
Period of the satellite on planet B, [tex]T_B[/tex], is given as follows;
[tex]T_B = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{32}{3} \cdot \pi \times \rho} }[/tex]
Therefore, get;
[tex]\dfrac{T_A}{T_B} = \dfrac{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 4 \cdot \pi \times \rho} }}{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 32 \cdot \pi \times \rho} }} = \sqrt{\dfrac{32}{4} } = \sqrt{8} = 2 \cdot \sqrt{2}[/tex]
Therefore, [tex]T_A[/tex] = (2·√2)·[tex]T_B[/tex]
[tex]T_B = \dfrac{T_A}{2 \cdot \sqrt{2} } = \dfrac{\sqrt{2} \cdot T_A}{4 }[/tex]
The comparison between [tex]T_A[/tex] and [tex]T_B[/tex] is therefore;
[tex]\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}[/tex]Learn more here:
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When people do a bungee jump theyare asked how much they weigh. Why?
Answer:
The elasticity of the bungee cord reduces the gravitational forces applied on the body during bungee jumping. For example, if a 100-pound individual jumps from a building and encounters 900 pounds of deceleration force, they will feel 9 "G's" of force.
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