To answer the two questions, we need to know two important equations involving centripetal movement:
v = ωr (ω represents angular velocity in radians)
a = [tex]\frac{v^{2}}{r}[/tex]
Let's apply the first equation to question a:
v = ωr
v = ((1800*2π) / 60) * 0.26
Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.
Back to the equation:
v = ((1800*2π)/60) * 0.26
v = (1800*2(3.14)/60) * 0.26
v = (11304/60) * 0.26
v = 188.4 * 0.26
v = 48.984
v = 49 (m/s)
Now that we know the linear velocity, we can find the centripetal acceleration:
a = [tex]\frac{v^{2}}{r}[/tex]
a = [tex]\frac{49^{2}}{0.26}[/tex]
a = 9234.6 (m/[tex]s^{2}[/tex])
Wow! That's fast!
We now have our answers for a and b:
a. 49 (m/s)
b. 9.2 * [tex]10^{3}[/tex] (m/[tex]s^{2}[/tex])
If you have any questions on how I got to these answers, just ask!
- breezyツ
49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as the drawing shows. The centripetal acceleration measured at corner A is n times as great as that measured at corner B. What is the ratio L1/L2 of the lengths of the sides of the rectangle when n
Answer:
[tex]\frac{L_1}{L_2} = \sqrt{(n^2 - 1)}[/tex]
Explanation:
For this interesting problem, we use the definition of centripetal acceleration
a = v² / r
angular and linear velocity are related
v = w r
we substitute
a = w² r
the rectangular body rotates at an angular velocity w
We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A
the distance OB = L₂
the distance AB = L₁
the sides of the rectangle
It is indicated that the acceleration in in A and B are related
[tex]a_A = n \ a_B[/tex]
we substitute the value of the acceleration
w² r_A = n r_B
the distance from the each corner is
r_B = L₂
r_A = [tex]\sqrt{L_1^2 + L_2^2}[/tex]
we substitute
\sqrt{L_1^2 + L_2^2} = n L₂
L₁² + L₂² = n² L₂²
L₁² = (n²-1) L₂²
Which diagram best represents the gravitational forces, F, be-
tween a satellite, S, and Earth?
Answer:
Diagram (3).
Explanation:
N3L states that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A ([tex]F_{A} = -F_{B}[/tex]).
The diagram which best represents the gravitational forces, F, between a satellite, S, and Earth is; Choice (3).
The Newton's law of gravitation states that the Force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.The law clearly states a Force of attraction; the two objects come towards each other.
Consequently, Choice (3) best represents the gravitational forces, F, between a satellite, S, and Earth.
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A 69.5-kg person throws a 0.0475-kg snowball forward with a ground speed of 31.5 m/s. A second person, with a mass of 57.5 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.35 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged
Answer:
- After throwing the snow, velocity of the thrower is 2.33 m/s
- the velocity of the receiver is 0.026 m/s
Explanation:
Given the data in the question;
Using conservation of momentum,
Initial thrower has a momentum of mv; [tex]m_{total[/tex]v
(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s
Now, When he throws it at 31.5 m/s, these constitutes a momentum of;
(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s
hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s
To get his velocity, we say;
161.94035 = mv
{ he lost weight of the snow ball so, m = 69.5 kg )
161.94035 = 69.5 × v
v = 161.94035 / 69.5
v = 2.33 m/s
Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s
Next is the Receiver;
the receiver will gain momentum of 1.49625 kg.m/s
he has no momentum initially and after he catches the snow ball;
1.49625 kg.m/s = mv
1.49625 kg.m/s = ( 57.5 kg + 0.0475 kg ) × v
1.49625 kg.m/s = 57.5475 kg × v
v = ( 1.49625 kg.m/s ) / 57.5475 kg
v = 0.026 m/s
Therefore, the velocity of the receiver is 0.026 m/s
A mass of slug, when attached to a spring, stretches it feet and then comes to rest in the equilibrium position. Starting at , an external force equal to is applied to the system. Find the equation of motion if the surrounding medium offers a damping force that is numerically equal to times the instantaneous velocity.
Answer:
Equation of motion is x(t) = [tex]-te^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t)
Explanation:
P.S - The exact question is -
Given - A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at t = 0, an external force equal to [tex]f(t) = 8 cos(4t)[/tex] is applied to the system.
To find - Find the equation of motion if the surrounding medium offers a damping force that is numerically equal to 8 times the instantaneous velocity.
Proof -
Given that,
Mass = 1 slug
We know that, 1 slug = 32 lb
Now,
Force, f = kx
⇒32 = k(2)
⇒k = 16
Now,
Given that, C = 8 ( 8 times the instantaneous velocity)
Now,
The differential equation of motion is equals to
mx'' + Cx' + kx = 8 cos(4t)
⇒x'' + 8x' + 16x = 8 cos(4t) ...........(1)
Let the General solution of equation (1) be
x(t) = x(c) + x(p)
Now,
The auxiliary equation is
m² + 8m + 16 = 0
m² + 4m + 4m + 16 = 0
m (m+4) + 4 (m+4) = 0
⇒(m+4)(m+4) = 0
⇒m = -4, -4
So,
The Complimentary equation becomes
x(c) = [tex]Ae^{-4t} + Bte^{-4t}[/tex] ...........(2)
Now,
Let the particular solution be
x(p) = C cos(4t) + D sin(4t)
x'(p) = -4C sin(4t) + 4D cos(4t)
x''(p) = -16C cos(4t) - 16D sin(4t)
It also satisfy equation (1)
Equation (1) becomes
-16C cos(4t) - 16D sin(4t) + 8 [ -4C sin(4t) + 4D cos(4t) ] + 16 [ C cos(4t) + D sin(4t) ] = 8 cos(4t)
⇒-16C cos(4t) - 16D sin(4t) - 32C sin(4t) + 32D cos(4t) ] + 16C cos(4t) + 16D sin(4t) ] = 8 cos(4t)
⇒-4C sin(4t) + 4B cos(4t) = cos(4t)
By comparing, we get
4B = 1 , A = 0
⇒ B = [tex]\frac{1}{4}[/tex] , A = 0
So, The particular solution becomes
x(p) = [tex]\frac{1}{4}[/tex] sin(4t)
Now,
The General solution becomes
x(t) = [tex]Ae^{-4t} + Bte^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t) .......(3)
Now,
Given that, At t = 0, initial velocity is zero and the system starts equilibrium
⇒x(0) = 0, x'(0) = 0
By putting t = 0 in equation (3) , we get
A = 0
Now,
Differentiate equation (3), we get
x'(t) = [tex]-4Ae^{-4t} + Be^{-4t} - 4Bte^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] *4 cos(4t)
Put t = 0, we get
0 = -4A + B + 1
⇒B = -1
∴ we get
The general solution becomes
x(t) = [tex]-te^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t)
Equation of motion is x(t) = [tex]-te^{-4t}[/tex] + [tex]\frac{1}{4}[/tex] sin(4t)
Can someone help me here
Answer:
1. grass Can survive by itself because it is not dependent on food, it is self-sustaining. and since grass it's not dependent on food it is a producer.
2. and 3. they would all decrease, since the prairie dogs are herbivores they eat basically plants such as grass and if half of the grass is gone then they would have less to eat and therefore decrease.
since ferrets eat prairie dogs, if the prairie dogs population decreases the ferrets would have less to eat and therefore decrease as well.
one of the predators of ferrets are foxes and if the ferrets decrease then the fox would also decrease since it will have less to eat.
4. if we would continue advancing years soon the ferret, prairie dogs and foxes would be no more because if there is less grass they would keep on decreasing over the years until they would be extinct.
Plz help
What factors determine
how the speed of the marbles changes in a
collision?
Answer:
Force,friction,inertia and momentum
Explanation:
The speed that the marble is moving at can be determined by the amount of force used when pushed or pulled and what kind of surface it's on.Momentum is also a factor because of the mass of the marbles.
Which of the following is true of the
thermocline layer of the ocean?
A. rapidly decreases in temperature
B. warmest and least dense of the ocean layers
C. is the bottom layer of the ocean
D. is the top layer of the ocean
Answer:
d is trueeeeeeeeeeeeeeeee
Which statement is true about how early humans met their needs?
Answer:
they were hunter gatherers
Explanation:
can someone please help me !!!!
Answer:
it's A subduction, deep water trench
PLEASE HELP! I'LL GIVE BRAINLEST
Does latitude has an effect on weight? PLEASE HELP!
Answer:
yes
it does you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.
Chemical energy may be released during a chemical reaction in the form of heat ..... True or false?
Answer:
heat and gas sometimes so true
In a movie production, a stunt person must leap from a balcony of one building to a balcony 3.0 m lower on another building. If the buildings are 2.0 m apart, what is the minimum horizontal velocity the stunt person must have to accomplish the jump? Assume no air resistance and that ay = −g = −9.81 m/s2 . (Ans. 2.6m/s) PLS SHOW WORK
This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.
The minimum horizontal velocity required is "2.6 m/s".
First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.
[tex]h=v_it+\frac{1}{2}gt^2[/tex]
where,
h = height = 3 m
vi = initial vertical speed = 0 m/s
t = time interval = ?
g = acceleration due to gravity = 9.81 m/s²
therefore,
[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]
t = 0.78 s
Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:
[tex]s = vt\\\\v = \frac{s}{t}[/tex]
where,
s = horizontal distance = 2 m
t =0.78 s
v = minimum horizontal velocity = ?
Therefore,
[tex]v = \frac{2\ m}{0.78\ s}[/tex]
v = 2.6 m/s
Learn more about equations of motion here:
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The attached picture shows the equations of motion in the horizontal and vertical directions.
A student hangs a block from a light string that is attached to a massive pulley of unknown radius R, as shown in the figure. The student allows the block to fall from rest to the floor. Which two of the following sets of data that could be measured or determined should the student use together to determine the final angular velocity of the pulley just before the block hits the floor? Select two answers. Justify your selections.
Answer:
The mass of the block, the distance of the block above the floor, and the time it takes the block to reach the floor, because these quantities can be used to determine the acceleration of the block.
The radius and the mass of the pulley, because these quantities can be used together to determine the rotational inertia of the pulley.
Explanation:
If the motion starts from rest, the initial angular velocity will be zero and the final angular velocity can be determined with the product of angular acceleration and time of motion of the pulley.
Angular velocity is defined as the change in the angular displacement per change in time of motion. This can be expressed mathematically as follows;
[tex]\omega = \frac{\Delta \theta}{\Delta t} = vr[/tex]
where;
Ф is the angular displacementt is the time of the motionv is the linear velocityr is the radius of the circular path.In a circular motion that starts from rest and ends with final velocity, the equation is given as;
[tex]\omega_f =\omega_i + \alpha t[/tex]
Where;
[tex]\omega_f[/tex] is the final angular velocity[tex]\omega_i[/tex] is the initial angular velocity[tex]\alpha[/tex] is the angular accelerationThus, if the motion starts from rest, the initial angular velocity will be zero and the final angular velocity can be determined with the product of angular acceleration and time of motion of the pulley.
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Fluid mechanics questions and answers
Answer:
Fluid mechanics is considered one of the toughest subdisciplines within mechanical and aerospace engineering. It is unique from almost any other field an undergraduate engineer will encounter. It requires viewing physics in a new light, and that's not always an easy jump to make.
A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of 0.20 m and released from rest. Find the speed with which the object passes through its original position on the way up.
Answer:
e = 0.0898m
v = 2.07m/s
Explanation:
a) According to Hooke's law
F = ke
e is the extension
k is the spring constant
Since F = mg
mg = ke
e = mg/k
Substitute the given value
e = 1.1(9.8)/120
e = 10.78/120
e = 0.0898m
Hence it is stretched by 0.0898m from its unstrained length
2) Total Energy = PE+KE+Elastic potential
Total Energy = mgh +1/2mv²+1/2ke²
Substitute the given value
5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²
Solve for v
5.0 = 2.156+0.55v²+0.48338
5.0-2.156-0.48338= 0.55v²
2.36 =0.55v²
v² = 2.36/0.55
v² = 4.29
v ,= √4.29
v = 2.07m/s
Hence the required velocity is 9.28m/s
A standard 1kilogram weight is a cylinder 50.5mm in height and 52.0mm in diameter. What is the density of the meterial?(kg/m^3)
Answer:
The correct answer is - 93.24×10^4 kg/m^3.
Explanation:
Given:
height of cylinder: 50.5 mm
diameter = 52.0
then radius will be diameter/2 = 52/2 = 26
Formula:
Density = mass/ volume
Volume = πr^2h
solution:
Now the volume of a cylinder is v = (22/7)×r^2×h
= 22/7×26×26×50.5
= 107261.59 mm^3
Now volume in cubic meter V =10.7261 ×10^(-5) m^3
So density d = m/V = 1/(10.7261 ×10^(-5))
Or d = 93.24×10^4 kg/m^3
The indices of refraction for her contact lens, cornea, and the fluid behind her cornea are 1.6, 1.4, and 1.3, respectively. Light is incident from air onto her contact lens at an angle of 30 ∘∘ from the normal of the surface. At what angle is the light traveling in the fluid behind her cornea?
Answer:
[tex]23^{\circ}[/tex]
Explanation:
n = Refractive index of air = 1
[tex]n_1[/tex] = Refractive index of contact lens = 1.6
[tex]n_2[/tex] = Refractive index of cornea = 1.4
[tex]n_3[/tex] = Refractive index of fluid = 1.3
From Snell's law
[tex]n\sin30^{\circ}=n_1\sin\theta\\\Rightarrow \theta=\sin^{-1}\dfrac{1\sin30^{\circ}}{1.6}\\\Rightarrow \theta=18.21^{\circ}[/tex]
[tex]n_1\sin\theta=n_2\sin\theta_1\\\Rightarrow \theta_{1}=\sin^{-1}\dfrac{1.6\times \sin18.21^{\circ}}{1.4}\\\Rightarrow \theta_1=20.92^{\circ}[/tex]
[tex]n_2\sin\theta_1=n_3\sin\theta_3\\\Rightarrow \theta_3=\sin^{-1}\dfrac{1.4\sin20.92^{\circ}}{1.3}\\\Rightarrow \theta_3=22.62^{\circ}\approx 23^{\circ}[/tex]
The angle is the light traveling in the fluid behind her cornea is [tex]23^{\circ}[/tex].
The angle is the light traveling in the fluid will be 23⁰. Light is traveling in a particular direction with an angle.
What is snell law?"The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for a given set of media,
The given data in the problem is;
n is the refractive index of air = 1
n₁ is the refractive index of contact lens = 1.6
n₂ is the refractive index of cornea = 1.4
n₃ is the refractive index of fluid = 1.3
According to Snell's law. The formula for Snell's law is
[tex]\rm n sin30^0 = n_1 sin\theta \\\\ \theta = sin^{- 1}(\frac{1sin30^0}{1.6} )\\\\ \theta = 18.21 ^0[/tex]
For contact lenses;
[tex]\rm n_1sin\theta = n_2 sin\theta_1 \\\\ \theta_1 = sin^{-1}\frac{1.6 \times sin 18.21^0}{1.4} \\\\ \theta_1 =20.92 ^0[/tex]
For fluid;
[tex]n_2 sin\theta_1 = n_2 sin \theta_3\\\\ \theta_3 = sin^{-1}\frac{1.4 sin 20.92^0}{1.3} \\\\ \theta_3 = 22.62 ^ 0 =23^0[/tex]
Hence the angle is the light traveling in the fluid will be 23⁰.
To learn more about snell's law refer to the link;
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Coherent monochromatic light of wavelength l passes through a narrow slit of width a, and a diffraction pattern is observed on a screen that is a distance x from the slit. On the screen, the width w of the central diffraction maximum is twice the distance x. What is the ratio a>l of the width of the slit to the wavelength of the light
Answer:
λ = a
Explanation:
This is a diffraction exercise that is described by the expression
a sin θ = m λ
sin θ = m λ/ a
the first zero of the diffraction occurs for m = 1
sin θ = λ / a
angles are generally very small and are measured in radians
sin θ = θ = y / x
we substitute
[tex]\frac{y}{x} = \frac{\lambda}{a}[/tex]
the width of the central maximum is twice the distance to zero
w = 2y
in the exercise indicate that this width is equal to twice the distance to the screen (2x)
W = 2x
2y = 2x
we substitute
1 = λ/ a
λ = a
we see that the width of the slit is equal to the wavelength used.
an object is acted by force of 22 newtons to the right and a force of 13 newtons to left
Answer:
Explanation:
I'm going to guess that you want the net force.
These two forces are acting in opposite directions. Therefore the forces are subtracted in effect.
F = F1 - F2
F1 = 22N to the right
F2 = 13N to the left
F = 22 - 13 = 9 N to the right.
Specifying the direction is very important. Forces do have directions and you must specify what that is. Otherwise, the question should be marked incorrect.
A 41.6-kg person, running horizontally with a velocity of +4.21 m/s, jumps onto a 14.6-kg sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?
Answer:
a) v = 3.116 m / s, b) μ = 1.65 10⁻²
Explanation:
a) to find the velocity of the set, let's define a system formed by the person and the sled, so that the forces during the collision are internal and the moment is conserved
initial instant. Before the crash
p₀ = M v₀
final instant. After the crash
p_f = (M + m) v
the moment is preserved
M v₀ = (M + m) v
v = [tex]\frac{M}{M+m} \ v_o[/tex]
let's calculate
v = [tex]\frac{41.6}{41.6 + 14.6} \ 4.21[/tex]
v = 3.116 m / s
b) for this part let's use the relationship between work and kinetic energy
W = ΔK
as the body has its final kinetic energy is zero
the work of the friction forces is
W = - fr x
the negative sign is because the friction forces always oppose the movement
let's write Newton's second law
Y axis
N - W_sled -W_person = 0
N = mg + M g
N = (m + M) g
X axis
fr = ma
the friction force has the expression
fr = μ N
fr = μ g (m + M)
we substitute
- μg (m + M) x = 0- ½ (m + M) v²
μ = [tex]\frac{1}{2} \ \frac{v^2 }{g \ x }[/tex]
let's calculate
μ = [tex]\frac{1}{2} \ \frac{3.116^2}{9.8 \ 30.0}[/tex]
μ = 0.0165
μ = 1.65 10⁻²
1.
is the rate that velocity changes
O Acceleration
O
Time
O
Distance
O
Mass
Answer:
a) acceleration
Explanation:
Acceleration is, by definition, the change of an object's velocity.
Which waves can travel through space?
a. Electromagnetic waves only
b. Mechanical waves only
c. Electromagnetic and mechanical waves
d. Longitudinal and electromagnetic waves
Answer:
electromagnetic waves only
Explanation:
I just took the test, Hope it helps!
Answer:
A: Electromagnetic waves only
Explanation:
A wave has a frequency of 2 Hz. Find its period
A 25.0kg girl pushes a 50.0kg boy so that he accelerates at 4.00m/s2. What is the force of the boy on the girl? A. 200N B. 100N C. 12.5 D. 400N
Answer:
a
Explanation:
so the answer is 200N
and I hope it is correct
Please help The position of masses 4kg, 6kg, 7kg, 10kg, 2kg, and 12kg are (-1,1), (4,2), (-3,-2), (5,-4), (-2,4) and (3,-5) respectively. Determine the position of the center of mass of this system?
Answer:
(1.9756, -2.1951)
Explanation:
The center of mass equation is: [tex]x_{cm}[/tex] = [tex]\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex], where m represents the masses and x represents the position.
In order to find the coordinates of the center of mass, we need to use this equation for both the x-values and the y-values.
x-values:
[tex]x_{cm}[/tex] = [tex]\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex] = [tex]\frac{4(-1)+6(4)+7(-3)+10(5)+2(-2)+12(3)}{4+6+7+10+2+12}[/tex] = [tex]\frac{(-4)+(24)+(-21)+(50)+(-4)+(36)}{41}[/tex] = [tex]\frac{81}{41}[/tex] = 1.9756
y-values:
[tex]y_{cm}[/tex] = [tex]\frac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3} + m_{4}y_{4} + m_{5}y_{5} + m_{6}y_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}[/tex] = [tex]\frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12}[/tex] = [tex]\frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41}[/tex] = [tex]\frac{-90}{41}[/tex] = -2.1951
center of mass:
(1.9756, -2.1951)
Question 8: Unspooling Thread (100 points) A 110 g spool of thread with a 4.2 cm radius is held up by a peg through its center and allowed to freely rotate. Assume the thread is ideal (i.e., it does not stretch or slip, and its mass is negligibly small). A 160 g needle is tied to the loose end of the thread. The needle is dropped, and it accelerates to the floor as the thread unwinds. Find the tension in the thread and the magnitude of the acceleration of the needle as it falls.
Answer:
a = 7.29 m / s², T = 0.40 N
Explanation:
To solve this exercise we must apply Newton's second law to each body
The needle
W -T = m a
mg - T = ma
The spool, which we will approach by a cylinder
Σ τ = I α
T R = I α
the moment of inertia of a cylinder with an axis through its center is
I = ½ M R²
angular and linear variables are related
a = α R
α = a / R
we substitute
T R = (½ M R²) a / R
T = ½ M a
we write our system of equations together
mg - T = m a
T = ½ M a
we solve
m g = (m + ½ M) a
a = [tex]\frac{m}{m + \frac{1}{2} M} \ g[/tex]
let's calculate
a = [tex]\frac{0.160}{0.160 + \frac{1}{2} 0.110} \ 9.8[/tex]
a = 7.29 m / s²
now we can look for the tension
T = ½ M a
T = ½ 0.110 7.29
T = 0.40 N
Explain in your
own words how
you would find
the density of a
regular-shaped
object.
Answer:
Use a ruler to measure the length (l), width (w) and height (h) of the object.Place the object on the top pan balance and measure its mass.Calculate the volume of the cube using (l*w*h).Use the measurements to calculate the density of the object.35 POINTSS!!! PLSSSS HELLPPP!!!
Work is the transfer of power from one object to another.
Please select the best answer from the choices provided
T
F
Answer:
T
beacuse:
Energy can be transferred from one object to another by doing work. ... When work is done, energy is transferred from the agent to the object, which results in a change in the object's motion (more specifically, a change in the object's kinetic energy).
g A velocity selector consists of crossed electric and magnetic fields. The electric field has a magnitude of 480 N/C and is in the negative z direction. What should the magnetic field (magnitude and direction) be to select a proton moving in the negative x direction with a velocity of 3.50 cross times 10 to the power of 5 m/s to go un-deflected
Answer:
B = 1.37 mT
Explanation:
Given that,
The magnitude of the electric field, E = 480 N/C
The speed of the proton, [tex]v=3.50 \times 10^5\ m/s[/tex]
We need to find the magnitude of the magnetic field. In a velocity selector, the electric field is balanced by the magnetic field. So,
[tex]qE=qvB[/tex]
Where
B is the magnetic field
[tex]B=\dfrac{E}{v}\\\\B=\dfrac{480}{3.5\times 10^5}\\\\B=1.37\times 10^{-3}\ T\\\\or\\\\B =1.37\ mT[/tex]
So, the magnetic field is equal to 1.37 mT.