Answer:
0.055 kg
Explanation:
According to the given situation the solution of the mass of the string is shown below:-
Speed of the wave is
[tex]v = \sqrt{\frac{F_T\times Length\ of\ string}{Mass\ of\ string}}[/tex]
[tex]30.0 m/s = \sqrt{\frac{10 kg m/s^2\times 5.00 m}{Mass\ of\ string}[/tex]
Mass of string is
[tex]= \sqrt{\frac{10 kg m^2/s^2\times 5.00 m}{900 m^2 s^2}[/tex]
After solving the above equation we will get the result that is
= 0.055 kg
Therefore for calculating the mass of the string we simply applied the above formula.
The voltage across a membrane forming a cell wall is 72.7 mV and the membrane is 9.22 nm thick. What is the magnitude of the electric field strength? (The value is surprisingly large, but correct.) You may assume a uniform E-field.
Answer:
The magnitude of the electric field intensity is [tex]E = 7.89 *10^{6} \ V/m[/tex]
Explanation:
From the question we are told that
The voltage is [tex]\epsilon = 72.7 \ mV = 72.7 *10^{-3} V[/tex]
The thickness of the membrane is [tex]t = 9.22 \ nm = 9.22 *10^{-9} \ m[/tex]
Generally the electric field intensity is mathematically represented as
[tex]E = \frac{\epsilon }{t}[/tex]
substituting values
[tex]E = \frac{72.7 *10^{-3} }{9.22 *10^{-9}}[/tex]
[tex]E = 7.89 *10^{6} \ V/m[/tex]
The greater the frequency of the waves, the ____________ the pitch.
Answer:
Higher.
Explanation:
The greater the frequency the bigger the amplitude gets and the greater pitch gets.
Think - more energy, bigger waves, more waves, and higher sound
PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton box 0.8 meters. 2. A boy lifts a 5-newton box 0.8 meters. 3.A boy lifts a 8-newton box 0.2 meters. 4.A boy lifts a 10-newton box 0.2 meters.
Explanation:
Work done is given by the product of force and displacement.
Case 1,
1. A boy lifts a 2-newton box 0.8 meters.
W = 2 N × 0.8 m = 1.6 J
2. A boy lifts a 5-newton box 0.8 meters.
W = 5 N × 0.8 m = 4 J
3. A boy lifts a 8-newton box 0.2 meters.
W = 8 N × 0.2 m = 1.6 J
4. A boy lifts a 10-newton box 0.2 meters.
W = 10 N × 0.2 m = 2 J
Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.
Dr. Stein's hypothesis is that excess sugar causes hyperactivity. He is interested in doing research.
Which research method would be the best to use?
Answer:
The correct answer would be - dependent independent variable experiment.
Explanation:
Dr. Stein hypothesized that excess sugar causes hyperactivity, so sugar treatment /no sugar treatment would be independent variable. By giving some children sugar and others a sugar cookies he can manipulate the independent variable.
Similarly , the dependent variable is the result or outcome of independent variable, or what Dr. Stein hypothesize to be the result of excess sugar . In this sugar experiment, then, the dependent variable is the children's hyper activity level.
Thus, the correct answer would be - dependent independent variable experiment.
The best research method to use for the research of hyperactivity, would be dependent-independent variable experiment.
The given problem is based on the effect of sugar on hyperactivity. Hyper activity refers to the increased movement, impulse actions and a shorter attention span.
Dr. Stein hypothesized that excess sugar causes hyperactivity, so sugar treatment /no sugar treatment would be independent variable. By giving some children sugar and others a sugar cookies he can manipulate the independent variable.Similarly , the dependent variable is the result or outcome of independent variable, or what Dr. Stein hypothesize to be the result of excess sugar . In this sugar experiment, then, the dependent variable is the children's hyper activity level.
Thus, we can conclude that the best research method to use, would be - dependent-independent variable experiment.
Learn more about the hyperactivity here:
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In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with the Earth's moon, with mass Mm = 3.75 ✕ 1019 kg and radius Rm = 1.98 ✕ 105 m, giving it a free-fall acceleration of g = 0.0636 m/s2. One astronaut, being a baseball fan and having a strong arm, decides to see how high she can throw a ball in this reduced gravity. She throws the ball straight up from the surface of Mimas at a speed of 41 m/s (about 91.7 mph, the speed of a good major league fastball).
(a) Predict the maximum height of the ball assuming g is constant and using energy conservation. Mimas has no atmosphere, so there is no air resistance.
(b) Now calculate the maximum height using universal gravitation.
(c) How far off is your estimate of part (a)? Express your answer as a percent difference and indicate if the estimate is too high or too low.
Answer:
a) h = 13,205.4 m
b) r_f = 2.12 106 m
c) e% = 0.68%
Explanation:
a) This is an exercise we are asked to use energy conservation,
Starting point. On the surface of Mimas
Em₀ = K = ½ m v²
Final point. Where the ball stops
[tex]Em_{f}[/tex] = U = m g h
Em₀ = Em_{f}
½ m v² = m g h
h = ½ v² / g
let's calculate
h = ½ 41² / 0.0636
h = 13,205.4 m
b) For this part we are asked to use the law of universal gravitation, write the energy
starting point. Satellite surface
Em₀ = K + U = ½ m v² - GmM / r_o
final point. Where the ball stops
[tex]Em_{f}[/tex]= U = - G mM / r_f
Em₀ = Em_{f}
½ m v² - G m M / r_o = - G mM / r_f
In this case all distances are measured from the center of the satellite
1 / rf = 1 / GM (-½ v² + G M / r_o)
let's calculate
1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)
1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)
1 / r_f = 4,714 10⁻⁷
r_f = 1 / 4,715 10⁻⁷
r_f = 2.12 106 m
to measure this distance from the satellite surface
r_f ’= r_f - r_o
r_f ’= 2.12 106 - 1.98 105
r_f ’= 1,922 106 m
c) the percentage difference is
e% = 13 205.4 / 1,922 106 100
e% = 0.68%
The estimate of part a is a little low
A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done by gravity during the roll, in Joules
Answer:
The work done by gravity during the roll is 490.6 J
Explanation:
The work (W) is:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the displacement = 20 m
The force is equal to the weight (W) in the x component:
[tex]F = W_{x} = mgsin(\theta)[/tex]
Where:
m: is the mass of the bowling ball = 5 kg
g: is the gravity = 9.81 m/s²
θ: is the degree angle to the horizontal = 30°
[tex]F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N[/tex]
Now, we can find the work:
[tex]W = F*d = 24.53 N*20 m = 490.6 J[/tex]
Therefore, the work done by gravity during the roll is 490.6 J.
I hope it helps you!
A vertically polarized light wave of intensity 1000 mW/m2 is coming toward you, out of the screen. After passing through this polarizing filter, the wave's intensity is
Answer:
The intensity is [tex]I = 500 mW/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]
Generally the intensity of the light emerging from the polarizer is mathematically represented as
[tex]I = \frac{I_o}{2}[/tex]
substituting values
[tex]I = \frac{1000 *10^{-3}}{2}[/tex]
[tex]I = 500 *10^{-3} W/m^2[/tex]
[tex]I = 500 mW/m^2[/tex]
An ideal spring of negligible mass is 11.00cm long when nothing is attached to it. When you hang a 3.05-kg weight from it, you measure its length to be 12.40cm .
If you wanted to store 10.0J of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.
Express your answer numerically. If there is more than one answer, enter each answer, separated by a comma.
=
Answer
0.2067m or 0.2067m
Explanation;
Let lenght of spring= Lo= 11cm=0.110m
It is hang from a mass of
3.05-kg having a length of L1= 12.40cm= 0.124m
Force required to stretch the spring= Fkx
But weight of mass mg= kx then K= Mg/x
K= 3.05-kg× 9.8)/(0.124m-.110m)
K=2135N
But potential Energy U= 0.5Kx
X=√ 2U/k
√(2*10)/2135
X=0.0967m
The required new length= L2= L0 ±x
=
.110m ± 0.0967m
X= 0.2067m or 0.2067m hence the total lenghth
A coil of area 0.320 m2 is rotating at 100 rev/s with the axis of rotation perpendicular to a 0.430 T magnetic field. If the coil has 700 turns, what is the maximum emf generated in it
Answer:
The maximum emf generated in the coil is 60527.49 V
Explanation:
Given;
area of coil, A = 0.320 m²
angular frequency, f = 100 rev/s
magnetic field, B = 0.43 T
number of turns, N = 700 turns
The maximum emf generated in the coil is calculated as,
E = NBAω
where;
ω is the angular speed = 2πf
E = NBA(2πf)
Substitute in the given values and solve for E
E = 700 x 0.43 x 0.32 x 2π x 100
E = 60527.49 V
Therefore, the maximum emf generated in the coil is 60527.49 V
If you were in a smooth-riding train with no windows, could you sense the difference between uniform motion and rest or between accelerated motion and rest?
1. Both acclerated and uniform motion can be sensed.
2. Only uniform motion can be sensed.
3. Only accelerated motion can be sensed.
4. No motion can be sensed.
Answer:
3. Only accelerated motion can be sensed
Explanation:
Without windows on such a train, you'd have no frame of reference for your speed. By that I mean, without being able to see how fast you are moving past other things, it's almost as if you aren't moving at all... almost.
At rest you obviously aren't moving and in uniform motion, with a constant speed, it would feel as though you aren't moving. But during periods of acceleration you'll feel the force on your body (F=ma) and would be able to tell if you were moving in a particular direction.
You've probably felt this before. Maybe not on a windowless train but perhaps in a car or on a roller coaster. Speeding up makes you go back into your seat a bit and slowing down makes you lean forward a bit. Both speeding up and slowing down are examples of acceleration (just in different directions) and how fast you accelerate will affect how much force you experience.
So the answer would be option 3.
Side note: If the train wasn't smooth riding then there would be some amount of friction going on and you could probably tell if you were in motion by the products of that friction (like sound and vibrations) even at a constant speed.
You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you
Answer:
The speed of light will be c=3x10^8m/s
Explanation:
This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c
You are at the carnival with you your little brother and you decide to ride the bumper cars for fun. You each get in a different car and before you even get to drive your car, the little brat crashes into you at a speed of 3 m/s.
A. Knowing that the bumper cars each weigh 80 kg, while you and your brother weigh 60 and 30 kg,respectively, write down the equations you need to use to figure out how fast you and your brother are moving after the collision.
B. After the collision, your little brother reverses direction and moves at 0.36 m/s. How fast are you moving after the collision?
C. Assuming the collision lasted 0.05 seconds, what is the average force exerted on you during the collision?
D. Who undergoes the larger acceleration, you or your brother? Explain.
Answer:
a) The equation is [tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
b) Your velocity after collision is 2.64 m/s
c) The force you felt is 7392 N
d) you and your brother undergo an equal amount of acceleration
Explanation:
Your mass [tex]m_{y}[/tex] = 60 kg
your brother's mass [tex]m_{b}[/tex] = 30 kg
mass of the car [tex]m_{c}[/tex] = 80 kg
your initial speed [tex]u_{y}[/tex] = 0 m/s (since you've not started moving yet)
your brother's initial velocity [tex]u_{b}[/tex] = 3 m/s
your final speed [tex]v_{y}[/tex] after collision = ?
your brother's final speed [tex]v_{b}[/tex] after collision = ?
a) equations you need to use to figure out how fast you and your brother are moving after the collision is
[tex](m_{y}+m_{c} )u_{y} + (m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
but [tex]u_{y}[/tex] = 0 m/s
the equation reduces to
[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
b) if your little brother reverses with velocity of 0.36 m/s it means
[tex]v_{b}[/tex] = -0.36 m/s (the reverse means it travels in the opposite direction)
then, imputing values into the equation, we'll have
[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
(30 + 80)3 = (60 + 80)[tex]v_{y}[/tex] + (30 + 80)(-0.36)
330 = 140[tex]v_{y}[/tex] - 39.6
369.6 = 140[tex]v_{y}[/tex]
[tex]v_{y}[/tex] = 369.6/140 = 2.64 m/s
This means you will also reverse with a velocity of 2.64 m/s
c) your initial momentum = 0 since you started from rest
your final momentum = (total mass) x (final velocity)
==> (60 + 80) x 2.64 = 369.6 kg-m/s
If the collision lasted for 0.05 s,
then force exerted on you = (change in momentum) ÷ (time collision lasted)
force on you = ( 369.6 - 0) ÷ 0.05 = 7392 N
d) you changed velocity from 0 m/s to 2.64 m/s in 0.05 s
your acceleration is (2.64 - 0)/0.05 = 52.8 m/s^2
your brother changed velocity from 3 m/s to 0.36 m/s in 0.05 s
his deceleration is (3 - 0.36)/0.05 = 52.8 m/s
you and your brother undergo an equal amount of acceleration. This is because you gained the momentum your brother lost
An object has an acceleration of 12.0 m/s/s. If the net force was doubled and the mass were tripled, then the new acceleration would be _____ m/s/s.
✴ Case - I
⟶ Force = F
⟶ Mass = m
⟶ Acceleration = 12m/s²
✴ Case - II
⟶ Force = 2F
⟶ Mass = 3m
To Find :➳ Acceleration in second case.
Concept :⇒ This question is completely based on the concept of newton's second law of motion.
⇒ As per this law, Force is defined as the product of mass and acceleration.
Mathematically, F = ma
Calculation :[tex]\implies\sf\:\dfrac{F_1}{F_2}=\dfrac{m_1\times a_1}{m_2\times a_2}\\ \\ \implies\sf\:\dfrac{F}{2F}=\dfrac{m\times 12}{3m\times a_2}\\ \\ \implies\sf\:\dfrac{1}{2}=\dfrac{4}{a_2}\\ \\ \implies\sf\:a_2=4\times 2\\ \\ \implies\underline{\boxed{\bf{a_2=8\:ms^{-2}}}}[/tex]
New acceleration would be 12 m/s²
Given that;
Acceleration of object = 12 m/s²
New net force = 2f
New mass = 3m
Find:
New acceleration
Computation:
[tex]\frac{F1}{F2} = \frac{m1a1}{m2a2} \\\\\frac{f}{2f} = \frac{m(12)}{(3m)a2} \\\\\frac{1}{2} = \frac{4}{a2} \\\\a2 = 8 m/s^2[/tex]
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The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad/s)t].
(a) What is the speed of the wave?
(b) What are the amplitudes of the electric and magnetic fields of this wave?
(c) What are the frequency, wavelength, and period of the wave? Is this light visible to humans?
Answer:
a) v = 2,9992 10⁸ m / s , b) Eo = 375 V / m , B = 1.25 10⁻⁶ T,
c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz , T = 1.05 10⁻¹⁵ s , UV
Explanation:
In this problem they give us the equation of the traveling wave
E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]
a) what the wave velocity
all waves must meet
v = λ f
In this case, because of an electromagnetic wave, the speed must be the speed of light.
k = 2π / λ
λ = 2π / k
λ = 2π / 1.99 10⁷
λ = 3,157 10⁻⁷ m
w = 2π f
f = w / 2 π
f = 5.97 10¹⁵ / 2π
f = 9.50 10¹⁴ Hz
the wave speed is
v = 3,157 10⁻⁷ 9.50 10¹⁴
v = 2,9992 10⁸ m / s
b) The electric field is
Eo = 375 V / m
to find the magnetic field we use
E / B = c
B = E / c
B = 375 / 2,9992 10⁸
B = 1.25 10⁻⁶ T
c) The period is
T = 1 / f
T = 1 / 9.50 10¹⁴
T = 1.05 10⁻¹⁵ s
the wavelength value is
λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm
this wavelength corresponds to the ultraviolet
lock of mass m2 is attached to a spring of force constant k and m1 . m2. If the system is released from rest, and the spring is initially not stretched or com- pressed, find an expres- sion for the maximum displacement d of m2
Answer:
The maximum displacement of the mass m₂ [tex]= \frac{2(m_1-m_2)g}{k}[/tex]
Explanation:
Kinetic Energy (K) = 1/2mv²
Potential Energy (P) = mgh
Law of Conservation of energy states that total energy of the system remains constant.
i.e; Total energy before collision = Total energy after collision
This implies that: the gravitational potential energy lost by m₁ must be equal to sum of gravitational energy gained by m₂ and the elastic potential energy stored in the spring.
[tex]m_1gd = m_2gd+\frac{1}{2}kd^2\\\\m_1g = m_2g+\frac{1}{2}kd\\\\d = \frac{2(m_1-m_2)g}{k}[/tex]
d = maximum displacement of the mass m₂
A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s at a distance of 120 m from its starting point. When the truck has travelled a distance of 60 m from its starting point, its speed is v1 m/s.
Which of the following statements concerning v1 is true?
a. v1< 12.5m/s
b. v1= 12.5m/s
c. v1 >12.5m/s
Answer:
the correct answer is c v₁> 12.5 m / s
Explanation:
This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.
v² = v₀² + 2 a₁ x
as part of rest v₀ = 0
a₁ = v² / 2x
a₁ = 25² / (2 120)
a₁ = 2.6 m / s²
now we can find the velocity for the distance x₂ = 60 m
v₁² = 0 + 2 a1 x₂
v₁ = Ra (2 2,6 60)
v₁ = 17.7 m / s
these the speed at 60 m
we see that the correct answer is c v₁> 12.5 m / s
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to (a) (1/3), (b) (1/10)
Answer:
35.3°
18.4°
Explanation:
a.
The first polariser polarises the unpolarised light reducing its intensity from I0 to I0/2. We have to reduce the intensity from I0/2 to I0/3.
Using to Law of Malus, I=I0cos²θ
cos²θ=I/I0=(I0/3)/I0/2 ,
cosθ=√2/3−−√=0.6667−−−−−√=0.8165
θ=cos−1(0.8165)=35.3∘
B.
Cos²θ=I/Io =Io/10/Io9
Cosθ= √9/10= 0.9487
= cos−10.9487
=18.4°
(a) The angle of polaroid such that intensity reduces by 1/3 is 35.26°
(b) The angle of polaroid such that intensity reduces by 1/10 is 63.43°
Angle of polarisation:According to the Malus Law: The intensity of light when passing through a polarizer is given by:
I = I₀cos²θ
where θ is the angle of the polarizer axis with the direction of polarization of the light
I₀ is the initial intensity
When an unpolarised light passes through a polarizer, θ varies from 0 to 2π, so the intensity after passing the first polarizer is :
I = I₀<cos²θ> { average of cos²θ, for 0<θ<2π}
I = I₀/2
Now, this emerging light passes through a second polarizer such that:
(a) the intensity is I' = I₀/3
From Malus Law:
I' = Icos²θ
I₀/3 = (I₀/2)cos²θ
cos²θ = 2/3
θ = 35.26°
(b) the intensity is I' = I₀/10
From Malus Law:
I' = Icos²θ
I₀/10 = (I₀/2)cos²θ
cos²θ = 1/5
θ = 63.43°
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Tarik winds a small paper tube uniformly with 163 turns of thin wire to form a solenoid. The tube's diameter is 6.13 mm and its length is 2.49 cm . What is the inductance, in microhenrys, of Tarik's solenoid?
Answer:
The inductance is [tex]L = 40\mu H[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 163 \ turns[/tex]
The diameter is [tex]D = 6.13 \ mm = 6.13 *10^{-3} \ m[/tex]
The length is [tex]l = 2.49 \ cm = 0.0249 \ m[/tex]
The radius is evaluated as [tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{6.13 *10^{-3}}{2}[/tex]
[tex]r = 3.065 *10^{-3} \ m[/tex]
The inductance of the Tarik's solenoid is mathematically represented as
[tex]L = \frac{\mu_o * N^2 * A }{l }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
A is the area which is mathematically evaluated as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [ 3.065*10^{-3}]^2[/tex]
[tex]A = 2.952*10^{-5} \ m^2[/tex]
substituting values into formula for L
[tex]L = \frac{ 4\pi *10^{-7} * [163]^2 * 2.952*10^{-5} }{0.0249 }[/tex]
[tex]L = 40\mu H[/tex]
An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.25 m/s2. Determine the orbital period of the satellite.
Answer:
118 minutes( 2 hours approximately )
Explanation:
Here, we are interested in calculating the orbital period of the satellite
Please check attachment for complete solution
Answer:
T = 7101 s = 118.35 mins = 1.9725 hrs
Explanation:
To solve the question, we apply the formula for gravitational acceleration
a = GM/r², where
a = acceleration due to gravity
G = gravitational constant
M = mass of the earth
r = distance between the satellite and center of the earth
Now, if we make r, subject of formula, we have
r = √(GM/a)
Recall also, that
a = v²/r, making v subject of formula
v = √ar
If we substitute the equation of r into it, we have
v =√a * √r
v =√a * √[√(GM/a)]
v = (GM/a)^¼
Again, remember that period,
T = 2πr/v, we already have v and r, allow have to do is substitute them in
T = 2π * √(GM/a) * [1 / (GM/a)^¼]
T = 2π * (GM/a³)^¼
T = 2 * 3.142 * [(6.67*10^-11 * 5.97*10^24) / (6.25³)]^¼
T = 6.284 * [(3.982*10^14) / 244.140]^¼
T = 6.284 * (1.63*10^12)^¼
T = 6.284 * 1130
T = 7101 s
T = 118.35 mins
T = 1.9725 hrs
In the figure, suppose the length L of the uniform bar is 3.2 m and its weight is 220 N. Also, let the block's weight W = 270 N and the angle θ = 45˚. The wire can withstand a maximum tension of 450 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x, what are the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A?
Answer:
a) x = 2.46 m
b) 318.2 N
c) 177.8 N
Explanation:
Need to resolve the tension of the string at end say B.
The vertical upward force at B due to tension is 450 sin 45°.
Using Principle of Moments, with the pivot at A,
Anti clockwise moments = Clockwise moments
450 sin 45° X 3.2 = 220 X (3.2/2) + (270 X x)
x = 2.46 m
(b) The horizontal force is only due to the wire's tension, so it is
450 cos 45° = 318.2 N
(c) total downward forces = 270 + 220 = 496 N
Total upward forces = 450 sin 45° (at B) + upForce (at A)
Equating, upForce = 496 - 318.2
= 177.8 N
An object with a mass of 5.5 kg is allowed to slide from rest down an inclined plane. The plane makes an angle of 30o with the horizontal and is 72 m long. The coefficient of friction between the plane and the object is 0.35. The speed of the object at the bottom of the plane is:_________.
a. 24 m/s.
b. 11 m/s.
c. 15 m/s.
d. 5.3 m/s.
e. 17 m/s.
Answer:
The speed will be "16.67 m/s".
Explanation:
The given values are:
Distance
= 72 m
Angle
= 30°
Acceleration
= [tex]g(sin \theta-ucos \theta)[/tex]
= [tex](9.8\times sin30^{\circ}) - (0.53\times cos30^{\circ})[/tex]
= [tex]1.929 \ m/s^2[/tex]
Let the speed be "v".
⇒ [tex]v^2=u^2+2as[/tex]
⇒ [tex]v^2=0(2\times 1.929\times 72)[/tex]
⇒ [tex]v^2=277.226[/tex]
⇒ [tex]v=\sqrt{277.776}[/tex]
⇒ [tex]v=16.67 \ m/s[/tex]
Learning Goal:
To understand the use of Hooke's law for a spring.
Hooke's law states that the restoring force F on a spring when it has been stretched or compressed is proportional to the displacement x of the spring from its equilibrium position. The equilibrium position is the position at which the spring is neither stretched nor compressed.
Recall that F∝x means that F is equal to a constant times x . For a spring, the proportionality constant is called the spring constant and denoted by k. The spring constant is a property of the spring and must be measured experimentally. The larger the value of k, the stiffer the spring.
In equation form, Hooke's law can be written
F =−kx .
The minus sign indicates that the force is in the opposite direction to that of the spring's displacement from its equilibrium length and is "trying" to restore the spring to its equilibrium position. The magnitude of the force is given by F=kx, where x is the magnitude of the displacement.
In Haiti, public transportation is often by taptaps, small pickup trucks with seats along the sides of the pickup bed and railings to which passengers can hang on. Typically they carry two dozen or more passengers plus an assortment of chickens, goats, luggage, etc. Putting this much into the back of a pickup truck puts quite a large load on the truck springs.
Part A
A truck has springs for each wheel, but for simplicity assume that the individual springs can be treated as one spring with a spring constant that includes the effect of all the springs. Also for simplicity, assume that all four springs compress equally when weight is added to the truck and that the equilibrium length of the springs is the length they have when they support the load of an empty truck.
A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.4×10−2 m . What is the effective spring constant of the spring system in the taptap?
2.9x10^4
Part B
After driving a portion of the route, the taptap is fully loaded with a total of 23 people including the driver, with an average mass of 70 kg per person. In addition, there are three 15-kg goats, five 3-kgchickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed?
Part C
Whenever you work a physics problem you should get into the habit of thinking about whether the answer is physically realistic. Think about how far off the ground a typical small truck is. Is the answer to Part B physically realistic?
Answer:
A) k = 2,858 10⁴ N / m , B) x_total = - 5,812 10⁻¹ m
C) it is very possible that the obtained value is not realistic for small vehicles
Explanation:
Part A
In this case we can use Hooke's law
F = - k x
force is the weight of the driver
F = W
mg = - k x
k = - mg / x
the springs are compressed x = - 2,4 10⁻² m
k = - 70 9.8 / (-2.4 10⁻²)
k = 2,858 10⁴ N / m
Part B
Since we have the spring constant we must find the complete weight, for this we look for the total masses
each mass is the number of element by the mass of an element
M = 23 70 + 3 15 + 5 3 + 1 25
M = 1695 kg
F = -k x
F = W = M g
Mg = - k x_total
x_total = -M g / k
x_total = -1695 9.8 / 2,858 10⁴
x_total = - 5,812 10⁻¹ m
The negative sign indicates that the springs are compressing
Part C
The truck has lowered 0.58 m = 58 cm
This drop is very large probably in a real vehicle with this drop it would be touching the ground or very close, therefore it is very possible that the obtained value is not realistic for small vehicles
A truck accidentally rolls down a driveway for 8.0\,\text m8.0m8, point, 0, start text, m, end text while a person pushes against the truck with a force of 850\,\text N850N850, start text, N, end text to bring it to a stop. What is the change in kinetic energy for the truck?
Answer:
Explanation:
According to work energy theorem
change in kinetic energy of truck = work done against it
work done against it = force x displacement
= - 850 x 8 = 6800 J
change in kinetic energy of truck = - 6800 J .
energy will be reduced by 6800 J
Answer:-6800
Explanation:
To compensate for acidosis, the kidneys will
Answer:
Acidosis is defined as the formation of excessive acid in the body due to kidney disease or kidney failure.
In order to compensate acidosis, the kidneys will reabsorb more HCO3 from the tubular fluid through tubular cells and collecting duct cell will secret more H+ and ammoniagenesis, which form more NH3 buffer.
10 pts! :) If Kyla picks up a grocery bag, using 10 N of force to lift it 1.5 m off the floor, how much work did Kyla do on the bag?
Explanation:
work = force x Distance
w = 10 x 1.5 = 15Nm
The amount of work done by Kyla in lifting the bag is 15 J.
What is meant by work done ?Work done on an object is defined as the cross product of the force applied on the object and the vertical displacement of the object.
Here,
Force applied by Kyla to pick up the bag, F = 10 N
Vertical displacement of the bag, s = 1.5 m
The work done by Kyla in lifting the bag,
W = F x s
W = 10 x 1.5
W = 15 J
Hence,
The amount of work done by Kyla in lifting the bag is 15 J.
To learn more about work done, click:
https://brainly.com/question/29989410
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You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By monitoring the capacitor voltage on an oscilloscope, you see that the voltage decays to half its initial value in 3.40 msms . You then use the resistor and capacitor to make a low-pass filter. What is the crossover frequency fcfc
Answer:
The frequency is [tex]f = 0.221 \ Hz[/tex]
Explanation:
From the question we are told that
The time taken for it to decay to half its original size is [tex]t = 3.40 \ ms = 3.40 *10^{-3} \ s[/tex]
Let the voltage of the capacitor when it is fully charged be [tex]V_o[/tex]
Then the voltage of the capacitor at time t is said to be [tex]V = \frac{V_o}{2}[/tex]
Now this voltage can be mathematical represented as
[tex]V = V_o * e ^{-\frac{t}{RC} }[/tex]
Where RC is the time constant
substituting values
[tex]\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }[/tex]
[tex]- \frac{0.5}{RC} = ln (0.5)[/tex]
[tex]-\frac{0.5}{RC} = -0.6931[/tex]
[tex]RC = 0.721[/tex]
Generally the cross-over frequency for a low pass filter is mathematically represented as
[tex]f = \frac{1}{2 \pi * RC }[/tex]
substituting values
[tex]f = \frac{1}{2* 3.142 * 0.72 }[/tex]
[tex]f = 0.221 \ Hz[/tex]
Search Results Web results A car of mass 650 kg is moving at a speed of 0.7
Answer:
W = 1413.75 J
Explanation:
It is given that,
Mass of car, m = 650 kg
Initial speed of the car, u = 0.7 m/s
Let a man pushes the car, increasing the speed to 2.2 m/s, v = 2.2 m/s
Let us assume to find the work done by the man. According to the work energy theorem, work done is equal to the change in kinetic energy.
[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times 650\times ((2.2)^2-(0.7)^2)\\\\W=1413.75\ J[/tex]
So, the work done by the car is 1413.75 J.
5) A coil of wire consists of 20 turns, each of which has an area of 0.0015 m2. A magnetic field is perpendicular to the surface with a magnitude of B = 4.91 T/s t – 5.42 T/s2 t2. What is the magnitude of the induced emf in the coil?
Answer:
1.5x10^-1 V
Explanation:
See attached file
Answer:
The magnitude of the induced emf in the coil is 15.3 mV
Explanation:
Given;
number of turns, N = 20 turns
Area of each coil, A = 0.0015 m²
initial magnitude of magnetic field at t₁, B₁ = 4.91 T/s
final magnitude of magnetic field at t₂, B₂ = 5.42 T/s
The magnitude of the induced emf in the coil is given by;
[tex]E = -N\frac{\delta \phi}{\delta t} \\\\E =-N (\frac{\delta B}{\delta t} )A\\\\E = -NA(\frac{B_1-B_2)}{\delta t} \\\\E = NA(\frac{B_2-B_1)}{\delta t} \\\\E = 20(0.0015)(5.42-4.91)\\\\E = 0.0153 \ V\\\\E = 15.3 \ mV[/tex]
Therefore, the magnitude of the induced emf in the coil is 15.3 mV
A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 20.5 meters?
Answer:
The final speed of the crate after being pulled these 20.5 meters is 13.82 m/s
Explanation:
I'll assume that the correct question is
A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 22.5 meters?
mass of box = 51 kg
for the first 12 m, it is pulled with a constant force of 240 N
The acceleration of the box for this first 12 m will be
from F = ma
a = F/m
where F is the pulling force
m is the mass of the box
a is the acceleration of the box
a = 240/51 = 4.71 m/s^2
Since the body started from rest, the initial velocity u = 0
applying Newton's equation of motion to find the final velocity at the end of the first 12 m, we have
[tex]v^{2}= u^{2}+2as[/tex]
where v is the final velocity
u is the initial velocity which is zero
a is the acceleration of 4.71 m/s^2
s is the distance covered which is 12 m
substituting value, we have
[tex]v^{2}[/tex] = 0 + 2(4.71 x 12)
[tex]v^{2}[/tex] = 113.04
[tex]v = \sqrt{113.04}[/tex] = 10.63 m/s
For the final 10.5 m, coefficient of friction is 0.21
from f = μF
where f is the frictional force,
μ is the coefficient of friction = 0.21
and F is the pulling force of the box 240 N
f = 0.21 x 240 = 50.4 N
Net force on the box = 240 - 50.4 = 189.6 N
acceleration = F/m = 189.6/51 = 3.72 m/s^2
Applying newton's equation of motion
[tex]v^{2}= u^{2}+2as[/tex]
u is initial velocity, which in this case = 10.63 m/s
a = 3.72 m/s^2
s = 10.5 m
v = ?
substituting values, we have
[tex]v^{2}[/tex] = [tex]10.63^{2}[/tex] + 2(3.72 x 10.5)
[tex]v^{2}[/tex] = 112.9 + 78.12
v = [tex]\sqrt{191.02}[/tex] = 13.82 m/s
A tornado passes in front of a building, causing the pressure to drop there by 25% in 1 second. Part A If a door on the side of the building is 8.1 feet tall and 3 feet wide, what is the net force on the closed door.
Answer:
F_net = 264, 26 pound
Explanation:
For this exercise we use Newton's second law
F_net = F_int - F_outside
where the force can be found from the definition of pressure
P = F / A
F = P A
we substitute
F_net = P_inside A - P_outside A
F_net = A (P_inside - P_outside)
indicate that the pressure on the outside is 25% less than the pressure on the inside
P_outside = 0.25 P_inside
The area is
A = L W
we substitute
F_net = L W P_inside (1-0.25)
let's calculate
suppose the pressure inside is atmospheric pressure
P_inside= P_atmospheric = 1,013 10⁵ Pa = 14.7 PSI
F_net = 8.1 3 14.5 0.75
F_net = 264, 26 pound