Answer:
Generally, the lowest overtone for a pipe open at one end and closed would be at y / 4 where y represents lambda, the wavelength.
Since F (frequency) = c / y Speed/wavelength
F2 / F1 = y1 / y2 because c is the same in both cases
F2 = y1/y2 * F1
F2 = 3 F1 = 750 /sec
Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe
and y1 = 3 y2
The second harmonic will be three times the first harmonic. The answer is 750 Hz
VIBRATION OF WAVES IN PIPESClosed pipes have odd multiples of frequencies or harmonics. That is,
If [tex]F_{0}[/tex] = fundamental frequency = first harmonic
[tex]F_{1}[/tex] = 3[tex]F_{0}[/tex] = second harmonic
[tex]F_{2}[/tex] = 5[tex]F_{0}[/tex] = third harmonic
[tex]F_{3}[/tex] = 7[tex]F_{0}[/tex] = fourth harmonic
Let assume that the first harmonic is 250 Hz, If you blow it much harder, second, third or fourth harmonic can be produced.
By using the formula above,
second harmonic will be 3 x 250 = 750Hz
Therefore, the frequency of the next harmonic heard if you blow much harder will be 750 Hz
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Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80×106 N , one at an angle 14.0 ∘ west of north, and the other at an angle 14.0 ∘ east of north, as they pull the tanker a distance 0.800 km toward the north. Part A What is the total work done by the two tugboats on the supertanker?
Answer:
W = 1,049 10⁹ J
Explanation:
Work is defined by the relation
W = F. d = F d cos θ
where tea is the angle between the forces and the displacement.
The total work is the sum of the work of each tug.
Tug 1
W₁ = F d cos θ₁
the angle measured from the positive side of the x-axis is
θ₁ = 14 + 90 = 104º
tugboat 2
W₂ = F d cos θ₂
θ₂ = 14
we substitute
W = F d cos θ₁ + F d cos θ₂
W = F d (cos θ₁ + cos θ₂)
let's calculate
W = 1.80 10⁶ 800 (cos 104 + cos 14)
W = 1,049 10⁹ J
A uniform-density 7 kg disk of radius 0.20 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 42 N through a distance of 0.9 m. Now what is the angular speed
Answer:
The angular speed is 23.24 rad/s.
Explanation:
Given;
mass of the disk, m = 7 kg
radius of the disk, r = 0.2 m
applied force, F = 42 N
distance moved by disk, d = 0.9 m
The torque experienced by the disk is calculated as follows;
τ = F x d = I x α
where;
I is the moment of inertia of the disk = ¹/₂mr²
α is the angular acceleration
F x r = ¹/₂mr² x α
The angular acceleration is calculated as;
[tex]\alpha = \frac{2Fr}{mr^2} \\\\\ \alpha = \frac{2F}{mr}\\\\\alpha = \frac{2 \times 42 }{7 \times 0.2} \\\\\alpha = 60 \ rad/s^2[/tex]
The angular speed is determined by applying the following kinematic equation;
[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta[/tex]
initial angular speed, ωi = 0
angular distance, θ = d/r = 0.9/0.2 = 4.5 rad
[tex]\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta} \\\\\omega _f = \sqrt{2 \times 60 \times 4.5} \\\\\omega _f = 23.24 \ rad/s[/tex]
Therefore, the angular speed is 23.24 rad/s.
5. How much heat is needed to warm .052 kg of gold from 30°C to 120°C? Note: Gold has a specific heat of 136
J/kg °C
Answer:
Q = 636.48 J
Explanation:
Given that,
The mass of gold, m = 0.052 kg
The temperature increase from 30°C to 120°C.
The specific heat of gold is 136 J/kg °C.
We need to find the heat needed to warm the gold. The formula for heat needed is given by :
[tex]Q=mc\Delta T\\\\Q=0.052\times 136\times (120-30)\\\\Q=636.48\ J[/tex]
So, 636.48 J of heat is needed to warm gold.
Name one similarity and one difference between a set and a bump in volleyball??
One similarity is the use of physical body whereas one difference is that one is exercise and the other is a sport.
One similarity and one differenceOne similarity between a set and a bump in volleyball is the movement and use of legs and hands.
Whereas one difference between a set and a bump in volleyball is that completing several reps of a particular exercise in a row is called a set while on the other hand, the basic pass in volleyball is known as bump.
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If matter cannot be created or destroyed, then how do you end up with
rust? Below is the equation for rust.
4Fe + 302 → 2Fe203
oxygen from the air
water in the atmosphere
oxygen from in the metal
there shouldn't be any oxygen
Your question is a "non sequitur", which means "it doesn't follow".
Your "then" doesn't contradict your "If", so no mystery is implied.
Maybe you're trying to say that matter is somehow not conserved in the equation . . . 4Fe + 302 → 2Fe203 . But it is. There are 4 Irons and 6 Oxygens on each side, so conservation is not violated here.
I looked up "rust" on Floogle, and got slapped with pages and pages of chemistry that I don't completely understand. But what it's saying is that rusting is a very complex chemical process, AND it doesn't happen unless there's some water involved.
So the bottom line is that there's a lot more going on than simply
4Fe + 302 → 2Fe203 ,
there's water going in and out of the process at every stage, and when it's all over, you have rusty iron, and mass has been conserved.
When measuring espresso for a drink, which instrument would give the
greatest precision?
How many mL is an espresso?
One shot of espresso is generally about 30–50 ml (1–1.75 oz), and contains about 63 mg of caffeine (3). Important point: The “golden ratio” for espresso is this: a single shot is 30 to 44 mL (1 to 1.5 ounces) of water and 7 grams of coffee
A group of particles of total mass 48 kg has a total kinetic energy of 320 J. The kinetic energy relative to the center of mass is 80 J. What is the speed of the center of mass?
A force of 10 N is applied at right angles to the handle of a spanner, 0.5 m from the centre of a nut. The
moment on the nut is:
20 Nm
50 Nm
5 Nm
Explanation:
the movement of the nut is 20Nm
What is diffraction of light
Answer:
According to "http://ww2010.atmos.uiuc.edu" Diffraction is the slight bending of light as it passes around the edge of an object.
Some examples of Light Defraction would be..
-CD reflecting rainbow colours
-Sun appears red during sunset
-From the shadow of an object
Discuss how the pressure cooker is designed to achieve temperatures above 100°C.
With rising heat, the steam pressure inside the pot builds up beyond atmospheric pressure, allowing the temperatures to rise well above boiling point. This design enables to save time, energy, and resources. The temperature inside a pressure cooker can well go beyond 110° C, which reduces the time needed to cook food.
A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of4.00 kg-m/s, directed along the positive y-axis. The final momentum of the first particle is 3.00 kg-m/s, directed 45.0 above the positive x-axis.
a. the magnitude and direction (angle expressed counter-clockwise with respect to the positive x-axis) of the final momentum for the second particle
b. assuming that these particles have the same mass, % loss of their total kinetic energy after they collided
Answer:
a) p₂ = 1.88 kg*m/s
θ = 273.4 º
b) Kf = 37% of Ko
Explanation:
a)
Assuming no external forces acting during the collision, total momentum must be conserved.Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:[tex]p_{o1x} = 2.00 kg*m/s (1)[/tex]
[tex]p_{o1y} = 0 (2)[/tex]
We can do the same for the particle moving along the positive y-axis:[tex]p_{o2x} = 0 (3)[/tex]
[tex]p_{o2y} = 4.00 kg*m/s (4)[/tex]
Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:[tex]p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)[/tex]
[tex]p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)[/tex]
Now, the total initial momentum, along these directions, must be equal to the total final momentum.We can write the equation for the x- axis as follows:[tex]p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)[/tex]
We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:[tex]p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)[/tex]
Now, we can repeat exactly the same process for the y- axis, as follows:[tex]p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)[/tex]
We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:[tex]p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)[/tex]
Since we have the x- and y- components of the final momentum of the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:[tex]p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)[/tex]
We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:[tex]tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)[/tex]
The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.b)
Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:[tex]\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)[/tex]
So, the final kinetic energy has lost a 37% of the initial one.A force of 12 N changes the momentum of a toy car from 3kgm/s t0 10kgm/s. Calculate the time the force took to produce this change in momentum.
Answer:
Time = 0.58 seconds
Explanation:
Given the following data;
Initial momentum = 3 kgm/s
Final momentum = 10 kgm/s
Force = 12 N
To find the time required for the change in momentum;
First of all, we would determine the change in momentum.
[tex] Change \; in \; momentum = final \; momentum - initial \; momentum [/tex]
[tex] Change \; in \; momentum = 10 - 3 [/tex]
Change in momentum = 7 kgm/s
Now, we can find the time required;
Note: the impulse of an object is equal to the change in momentum experienced by the object.
Mathematically, impulse (change in momentum) is given by the formula;
[tex] Impulse = force * time [/tex]
Making "time" the subject of formula, we have;
[tex] Time = \frac {impulse}{force} [/tex]
Substituting into the formula, we have;
[tex] Time = \frac {7}{12} [/tex]
Time = 0.58 seconds
PLEASE HELP
Section 1 - Question 6
Wave Movement Through Media
What could be happening to the wave as it travels from left to right?
A
It's moving through a medium whose density stays the same
B
It's moving from a low density medium to a high density medium.
С
It's moving from a high density medium to a low density medium.
D
It's moving from a low density medium, to a high density medium, and then back to a low density medium
Answer: B
Explanation:
If you swing an object on the end of a string around a circle, the string pulls on the object to keep it moving in a circle. What is the name of this force?
A. inertial
B. centripetal
C. resistance
D. gravitational
Answer:
B
Explanation:
The centripetal force keeps an object moving in a circular path. Therefore option (B) is correct.
What is centripetal force?A centripetal force can be described as a force that makes a body follow a curved path and its direction is orthogonal to the motion of the body. Gravity offers the centripetal force causing astronomical orbits.
The centripetal force is directed perpendicular to the direction of the displacement of an object. It always acts towards the center of the circle on an object moving in a circular path. For example, When spinning a ball on a string, the tension on the rope pulls the object toward the center.
The Centripetal Force can be described as the product of mass and velocity squared, divided by the radius.
F = mv²/r
Where F is the Centripetal force, m is the mass, r is the radius of the circle and v is the velocity of the object.
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What type of weather would a continental Polar air mass bring
Answer:
Continental polar ( cp):
Explanation:
Cold and dry, originating from high latitudes, typically as air flowing out of the polar highs. This air mass often brings the rattleing cold, dry and clear weather on a perfect winter day and also dry and warm weather on a pleasant day in summer.
A 38.0 kg box initially at rest is pushed 4.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find the following. (a) the work done by the applied force J (b) the increase in internal energy in the box-floor system due to friction J (c) the work done by the normal force J (d) the work done by the gravitational force J (e) the change in kinetic energy of the box J (f) the final speed of the box m/s
Answer:
a) Wapp = 520 N
b) ΔUf = 447 N
c) Wn = 0
d) Wg = 0
e) ΔK = 73 J
f) v = 1.96 m/s
Explanation:
a)
Applying the definition of work, as the dot product between the applied force and the displacement, since both are parallel each other, the work done on the box by the applied force can be written as follows:[tex]W_{app} = F_{app} * \Delta X = 130 N * 4.0 m = 520 N (1)[/tex]
b)
The change in the internal energy due to the friction, is numerically equal to the work done by the force of friction.This work is just the product of the kinetic force of friction, times the displacement, times the cosine of the angle between them.As the friction force opposes to the displacement, we can find the work done by this force as follows:[tex]W_{ffr} = F_{fr} * \Delta X * cos (180) (2)[/tex]
The kinetic force of friction, can be expressed as the product of the kinetic coefficient of friction times the normal force.If the surface is level, this normal force is equal to the weight of the object, so we can write (2), as follows:[tex]W_{ffr} = F_{fr} * \Delta X * cos (180) = -\mu_{k} * m* g* \Delta X = \\ -0.300*38.0kg9.8 m/s2*4.0m = -447 J (3)[/tex]
So, the increase in the internal energy in the box-floor system due to the friction, is -Wffr = 447 Jc)
Since the normal force (by definition) is normal to the surface, and the displacement is parallel to the surface, no work is done by the normal force.d)
Since the surface is level, the displacement is parallel to it, and the gravitational force is always downward, we conclude that no work is done by the gravitational force either.e)
The work-energy theorem says that the net work done on the object, must be equal to the change in kinetic energy.We have two forces causing net work, the applied force, and the friction force.So the change in kinetic energy must be equal to the sum of the work done by both forces, that we found in a) and b).So, we can write the following expression:[tex]\Delta K = W_{app} + W_{ffr} = 520 J - 447 J = 73 J (4)[/tex]
f)
Since the object starts at rest, the change in kinetic energy that we got in e) is just the value of the final kinetic energy.So, replacing in (4), we finally get:[tex]\Delta K = 73 J = \frac{1}{2}*m*v^{2} (5)[/tex]
Solving for v:[tex]v_{f} = \sqrt{\frac{2*\Delta K}{m} } = \sqrt{\frac{2*73J}{38.0kg}} = 1.96 m/s (6)[/tex]a) The work done by the applied force [tex]W_{AP}=520\ J[/tex]
b) The change in the internal energy [tex]\Delta U=447\ J[/tex]
c) Work done by normal force [tex]W_n=0[/tex]
d) Work done by gravitation [tex]W_g=0[/tex]
e) The change in KE will be [tex]\Delta KE=73\ J[/tex]
f) The final speed v = 1.96 m/s
What will be the work done?The work done on any object can be defined as the force applied on the object and its displacement due the effect of the force.
If the object achieve movement due to the work then the energy in the object will be kinetic energy.
If the object attains some height against the gravity then the energy in the object will be potential energy.
Now it is given in the question that
The horizontal force [tex]F_h=130\N[/tex]
mass of the object m= 38 kg
Coefficient of friction [tex]\mu=0.3[/tex]
Displacement of the object [tex]\delta x=4\ m[/tex]
(a) The work done will be
[tex]W=F_h\tines \Delta x[/tex]
[tex]W=130\times 4=520\ J[/tex]
(b) The increase in the internal energy
The increase in the internal energy of the box is due to the energy generated by the force of friction so
[tex]W_f=F_f\times \Delta x\times Cos(180)[/tex]
here [tex]F_f[/tex] is the frictional force and is given as
[tex]\mu=\dfrac{F_f}{R}[/tex]
Here R is the normal reaction and its value will be weight of the box in opposite direction.
[tex]\mu=\dfrac{F_f}{-mg}[/tex]
[tex]F_f=-mg\times \mu[/tex]
[tex]W_f=F_f\times \Delta x\times Cos180=-mg\times\mu \times cos180[/tex]
[tex]W_f=-38\times 9.81\times 0.3\times4=-447\J\ J[/tex]
(c) The work done by the normal force will be zero because the displacement is horizontal against the normal work so the work done will be zero.
(d) The work done by the gravitational force will also be zero. Because the displacement is horizontal and the gravitational force acts downward.
(e) The change in the KE of the box.
The change in the KE of the box will be the net energy of the box so from the work energy theorem the net energy will be
[tex]\Delta KE =W_{AP}-W_f=520-447=73\ J[/tex]
(f) The speed of the box
The KE of the box will be
[tex]KE=\dfrac{1}{2} mv^2[/tex]
[tex]v=\sqrt{ \dfrac{2\times KE}{m}[/tex]
[tex]v=\sqrt{\dfrac{2\times73}{38} }=1.96\ \dfrac{m}{s}[/tex]
Thus
a) The work done by the applied force [tex]W_{AP}=520\ J[/tex]
b) The change in the internal energy [tex]\Delta U=447\ J[/tex]
c) Work done by normal force [tex]W_n=0[/tex]
d) Work done by gravitation [tex]W_g=0[/tex]
e) The change in KE will be [tex]\Delta KE=73\ J[/tex]
f) The final speed v = 1.96 m/s
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If you push with a power of 20 Watts
on a 150 Newton object, how long would
it take to push it over the 4.3 m?
Answer:
32.25 s
Explanation:
From the question,
P = W/t.............. Equation 1
Where P = Power, W = work done, t = time.
But
W = F×d................. Equation 2
Where F = force and d = distance
Substitute equation 2 into equation 1
P = F×d/t............... Equation 3
make t the subject of euqation 3
t = (F×d)/P............. Equation 4
Givn: F = 150 N, d = 4.3 m, P = 20 watts.
Substitute these values into equation 4
t = (150×4.3)/20
t = 32.25 s
An object is accelerated by a net force in which direction?
A. at an angle to the force
B. in the direction of the force
C. in the direction opposite to the force
D. Any of these is possible.
Answer:
B. in the direction of the force
Explanation:
Sana nakatulong
Two 20kg spheres are placed with their
Centres 50cm apart. What is the magnitude of
gravitational force each exerts on the other?
Answer:
F = 1.07 x 10⁻⁷ N
Explanation:
The gravitational force of attraction between two objects can be found by the use of Newton's Gravitational Law:
[tex]F = \frac{Gm_{1}m_{2}}{r^2}\\\\[/tex]
where,
F = Gravitational Force of attraction = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
m₁ = m₂ = mass of spheres = 20 kg
r = distance between the objects = 50 cm = 0.5 m
Therefore,
[tex]F = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(20\ kg)(20\ kg)}{(0.5\ m)^2}\\\\[/tex]
F = 1.07 x 10⁻⁷ N
(will give brainliest to whoever is correct and shows reasoning) What is the acceleration of an object that has a velocity of 60m/s and is moving in a circle of radius 50m?
Answer:
5.0/s
Explanation:
Answer:
b and a it is this that abewsr
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One of the greatest dangers in a tornado is from flying objects. A 15 pound piece of lumber can turn into a flying missile that could severely damage walls and homes. A piece of steel with a mass of 200 pounds and travelling at the same velocity would cause even more damage. Select any evidence from the list below that you could use to explain why a 200 pound piece of steel would cause more damage than a 15 pound piece of wood travelling at the same velocity.
As the kinetic energy of an object increases, the force it can exert on another object decreases.
As the kinetic energy of an object increases, the force it can exert on another object increases.
Objects with more mass have less kinetic energy.
Objects with more mass have more kinetic energy.
As the velocity of an object increases, its kinetic energy decreases.
As the velocity of an object increases, its kinetic energy increases.
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When magma cools quickly, what kind of texture or
crystals does a rock have?
a. Fine-grained or small crystals
b. a mixture of all types of grains or sizes of crystals
c. Medium-grained or medium crystals
d. Coarsc-grained or large crystals
Answer:
a
Explanation:
when magma cools Crystal's form because the solution is super saturated with respect to some minerals if the magma cools quickly the crystals do not have much time to form hence they are small and also the resulting rock is fine grained
Time of the day when the Sun does not shine (___time)
N____N
When6-2 He He-6 undergoes beta decay, the daughter is?
Answer: The daughter is named Susie.
Explanation: LIL SUSIE!!!
HUH? DIDN'T UNDERSTAND THE QUESTION!
HAVE A GREAT DAY!!!!!
Answer:6/3 Li
Explanation:
I’m not sure what the person under me is talking about but yeah
The speed limmit on an interstate highway is posted at 75mi/h. What is the speed in kilometers per hour? In feet per second?
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Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths from the bubbles' walls. For a soap solution with an index of refraction of 1.21, find the minimum wall thickness that will enhance the reflection of light of wavelength 711 nm in air.
Answer:
the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Explanation:
Given the data in the question;
At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.
At the second interface, no shift occurs,
condition for constructive interference;
t = ( m + 1/2) × λ/2n
where m = 0, 1, 2, 3 . . . . . .
now, the condition for the constructive interference;
t = mλ/2n
where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.
so the minimum thickness of the film which will enhance reflection of light will be;
t[tex]_{min[/tex] = ( m + 1/2) × λ/2n
we substitute
t[tex]_{min[/tex] = ( 0 + 1/2) × 711 /2(1.21)
t[tex]_{min[/tex] = 0.5 × 711/2.42
t[tex]_{min[/tex] = 0.5 × 293.80165
t[tex]_{min[/tex] = 146.9 nm
Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm
A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 38.0 ∘ above the horizontal. The glider has mass 9.00×10−2 kg. The spring has 590 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.70 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring.
Required:
a. What distance was the spring originally compressed?
b. When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?
Answer:
x = 0.056 m
ΔKE = 0.489 J
Explanation:
Given that
Angle, θ = 38°
Length, L = 1.7 m
Mass, m = 0.09 kg
Spring constant, K = 590 N/m
If we use the Work-Energy theorem, then we know that Potential Energy, PE = Kinetic Energy, KE
This is mathematically written as
1/2kx² = mgH
The height, H we can get by using the relation
H = L.Sinθ
H = 1.7 * Sin 38
H = 1.7 * 0.6157
H = 1.047 m
Next, we use the Work-Energy theorem
1/2kx² = mgH
1/2 * 590 * x² = 0.09 * 9.8 * 1.047
295 * x² = 0.9234
x² = 0.9235 / 295
x² = 0.00313
x = √0.00313
x = 0.056 m
If the spring loses contact at x = 0.056, definitely, it will also lose contact at x = 0.8
Then we use the formula
ΔKE = mg(H - H1)
ΔKE = mg(xsinθ - x2.sinθ)
Where, x = 1.7 , x2 = 0.8
ΔKE = 0.09 * 9.8 (1.7 * sin 38 - 0.8 * sin 38)
ΔKE = 0.882(1.047 - 0.493)
ΔKE = 0.882 * 0.554
ΔKE = 0.489 J
Which of these is the BEST answer for why science is important?
Science can take us to other planets, even if it’s only through a telescope.
Science is part of human nature; it helps answer questions about how the world works.
Science helps us learn to think more critically and weigh evidence better.
Science gives us better tablet computers and games.
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An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravitational attraction between the two? Since there is no force opposing him, he will accelerate toward the ship. Find his acceleration.
Answer:
1. 2.5×10¯⁹ N
2. 3.33×10¯¹¹ m/s²
Explanation:
1. Determination of the force of attraction.
Mass of astronaut (M₁) = 75 Kg
Mass of spacecraft (M₂) = 125000 Kg
Distance apart (r) = 500 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force of attraction (F) =?
The force of attraction between the astronaut and his spacecraft can be obtained as follow:
F = GM₁M₂ /r²
F = 6.67×10¯¹¹ × 75 × 125000 / 500²
F = 2.5×10¯⁹ N
Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N
2. Determination of the acceleration of the astronaut.
Mass of astronaut (m) = 75 Kg
Force (F) = 2.5×10¯⁹ N
Acceleration (a) of astronaut =?
The acceleration of the astronaut can be obtained as follow:
F = ma
2.5×10¯⁹ N = 75 × a
Divide both side by 75
a = 2.5×10¯⁹ / 75
a = 3.33×10¯¹¹ m/s²
Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²
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