Hi there!
Part A:
The only quantities explicitly given to us are:
3. mass (m)
4. Maximum velocity (vmax)
5. Amplitude (A)
8. Position x from equilibrium
Part B:
To solve, we must begin by calculating the force constant, 'k'.
We can use the following relationship:
[tex]v = \sqrt{\frac{k}{m}(A^2-x^2)[/tex]
We are given the max velocity which occurs at a displacement of 0 m, because the mass is the fastest at the equilibrium point. We can rearrange the equation for k/m:
[tex]\frac{v^2}{(A^2-x^2)} = \frac{k}{m}[/tex]
[tex]\frac{3.2^2}{(0.06^2-0)} = \frac{k}{m} = 2844.44[/tex]
Now, we can find the velocity at 4.7cm (0.047m) using the equation:
[tex]v = \sqrt{(2844.44)(0.06^2-0.047^2)} = 1.989 m/s[/tex]
Plug this value into the equation for kinetic energy:
[tex]KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(0.05)(1.989^2) = \boxed{0.0989 J}[/tex]
Part C:
The potential energy of a spring is given as:
[tex]U = \frac{1}{2}kx^2[/tex]
Find 'k' using the derived quantity above:
[tex]\frac{k}{m} = 2844.44\\\\k = 2844.44m = 142.22 N/m[/tex]
Now, calculate potential energy:
[tex]U = \frac{1}{2}(142.22)(0.047^2) = \boxed{0.157 J}[/tex]
A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that is wound on the wheel and attached to it (see figure). The wheel is released from rest and the block descends 1.5 m in 2.00 s without any slipping of the rope. The tension in the rope during the descent of the block is 20 N. What is the moment of inertia of the wheel?
The moment of inertia of the wheel is 4.27 kg.m²
The kinematics equation explains the variables associated and related of motion.
From the information given, applying the kinematic equation of motion to determine the acceleration of the block, we have:
[tex]\mathbf{y = ut + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = (0)t + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = \dfrac{1}{2}at^2}[/tex]
Making acceleration (a) the subject, we have:
[tex]\mathbf{a = \dfrac{2y}{t^2}}[/tex]
where;
y = 1.5 mt = 2.0 s[tex]\mathbf{a = \dfrac{2\times 1.5 }{2.0^2}}[/tex]
a = 0.75 m/s²
The angular acceleration of the wheel can be estimated by the formula:
[tex]\mathbf{\alpha = \dfrac{a}{r}}[/tex]
[tex]\mathbf{\alpha = \dfrac{0.75 \ m/s^2}{0.40 \ m}}[/tex]
[tex]\mathbf{\alpha = 1.875 \ rad/s^2}[/tex]
Finally, the torque acting on the wheel is:
[tex]\mathbf{\tau = I \alpha}[/tex]
[tex]\mathbf{Tr = I \alpha}[/tex]
where;
T = tensionr = radiusI = moment of inertia∝ = angular acceleration∴
[tex]\mathbf{I =\dfrac{T\times r}{\alpha} }[/tex]
[tex]\mathbf{I =\dfrac{20 \ N\times 0.40 \ m}{1.875 \ rad/s^2} }[/tex]
I = 4.27 kg.m²
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you are standing ata known distance from the statue of liberty describe how you could determine its height using only a meter stick
Use the meter stick to measure your height to the level of your eyes, then use trigonometry ratio formula to calculate the height from your eyes' level and above of the statue of liberty with angle of elevation, then, add the two heights.
If you are standing at a known distance from the statue of liberty, a meter stick can be used to measure your known distance away from the statue of liberty.
To determine its height using only a meter stick, the angle at which you look at the peak of statue of liberty must be measured or known. The height of the statue of liberty can be calculated if you know the angle of elevation at which you look at the peak of the statue, and the availability of the meter stick.
Use the meter stick to measure your height to the level of your eyes, then use trigonometry ratio formula to calculate the height from your eyes' level and above of the statue of liberty. That is,
Tan Ф = opposite / adjacent
Tan Ф = H/d
H = d x TanФ
Where
H = calculated height from the eyes level and above
Ф = angle of elevation
d = known distance away from the statue
Let h = Your measured height of your body to the eyes level
Then,
The height of the statue of liberty = H + h
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if the Periodic time of an oscillating object Triples then its frequency will?
Answer:
it would decrease
Explanation:
f=1/T
A solar collector for a hot water system absorbs solar radiation at the rate of 660 W/m^2
. Its collecting area is 3.8m^2 . Cold water at 15degrees enters the collector. What volume of
water per minute at an output temperature of 60 degrees can this collector deliver?
This question involves the concepts of the law of conservation of energy and specific heat capacity.
This collector can deliver "7.96 x 10⁻⁴ m³/min" of water at an output temperature of 60°C.
According to the law of conservation of energy:
Solar Energy = Energy Required to raise the temperature of the water
Solar Power = Energy Required to raise the temperature of the water
[tex]IA=\frac{mC\Delta T}{t}\\\\\frac{m}{t}=\frac{IA}{C\Delta}[/tex]
where,
[tex]\frac{m}{t}[/tex] = mass flow rate = ?
I = solar radiation = 660 W/m²
A = Area = 3.8 m²
ΔT = change in temperature = 60°C - 15°C = 45°C
C = specific heat capacity = 4200 J/kg.°C
Therefore,
[tex]\frac{m}{t}=\frac{(660\ W/m^2)(3.8\ m^2)}{(4200\ J/kg.^oC)(45^oC)}\\\\\frac{m}{t}=(0.0133\ kg/s)(\frac{60\ s}{1\ min})\\\\\frac{m}{t}=0.796\ kg/min[/tex]
Now, the volume flow rate will be:
[tex]\frac{V}{t}=\frac{(\frac{m}{t})}{(density\ of\ water)}=\frac{(0.796\ kg/min)}{(1000\ kg/m^3)}\\\\\frac{V}{t}=7.96\ x\ 10^{-4}\ m^3/min[/tex]
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The attached picture explains the law of conservation of energy.
What happens to the iron in the coilgun if the electricity in the coil was turned on
please help me
please help me
please help me
Answer:
do it got a picture
on the edge
Explanation:
What happens when the object is placed at F? Explain
your answer.
Answer:
Sample Response: No image will be formed because the rays will not converge to or diverge from a common point.
Explanation:
After passing point 2 the hill becomes frictionless and the ring's rotational velocity remains constant. What is the linear velocity of the ring at point 3 in m/s
The energy in the system is given by the initial potential energy at the point 1.
The linear velocity at point 3, is approximately 33.59 m/s.
Reasons:
The parameters are;
Height at point 1, h₁ = 83 m
Radius of the ring = 8 cm
Mass of the ring, M = 8 kg
Height at point 2, h₂ = 32 m
At point 2, we have;
Change in potential energy = Kinetic energy
Which gives;
(83 - 32) × 9.81 × 8 = 0.5 × 8 × v² + 0.5 × 8 × 0.08² × (v/0.08)²
Which gives;
v ≈ 22.37 m/s
At point 3, the rotational kinetic energy remains constant while the
translational kinetic energy increases as follows;
K.E. at point 3 = Initial kinetic energy + Change in potential energy
Which gives;
K.E. at point 3 = 0.5 × 8 × v₃³ ≈ 0.5×8×22.37² + 32×9.81×8
[tex]v_3^2 = \dfrac{0.5 \times 8 \times 22.37^2 + 32 \times 9.81 \times 8}{0.5 \times 8} = 1128.15[/tex]
v₃ ≈ √(1128.15) ≈ 33.59
The linear velocity at point 3, v₃ ≈ 33.59 m/s
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The probable question parameters as obtained from a similar question online are;
Height at point 1, h₁ = 83 m
Radius of the ring = 8 cm
Mass of the ring, M = 8 kg
Height at point 2, h₂ = 32 m
A mass undergoes SHM with amplitude of 4 cm. The energy is 8.0 J at this time. The mass is cut in half, and the system is again set in motion with amplitude 4.0 cm. What is the energy of the system now?
Hi there!
[tex]\large\boxed{E_{total} = 8.0 \text{ J}}[/tex]
For a mass undergoing SHM, the total energy of the system is given as:
[tex]ME = \frac{1}{2}kA^2[/tex]
Where:
k = Spring constant (N/m)
A = amplitude (m)
There is no quantity of mass in the equation, so the total mechanical energy of the system is NOT impacted by the object's mass.
Thus, the energy of the system will still be 8.0 J.
A 5kg mass is pushed with a force of 10N for a distance of 2.5 meters. The work done is
W = 25 J
Explanation:
Work done on an object is defined as
[tex]W = Fd = (10\:\text{N})(2.5\:\text{m}) = 25\:\text{J}[/tex]
Four small 0.600-kg spheres, each of which you can regard as a point mass, are arranged in a square 0.400 m on a side and connected by light rods. Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane at point O.
Answer:
.192 kg x m^2
Explanation:
I= mass of a times radius of a squared + mass of b times radius of b squared +...
I= .6 kg x .4m^2 + .6 kg x .4m^2
= .192 kg x m^2
Hope this helps :)
A 0.035-kg bullet is fired vertically at 214 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball
Answer:
Explanation:
conservation of momentum during the collision
0.035(214) + 0.15(0) = 0.185v
v = 40.486 m/s
The kinetic energy after impact will convert to gravity potential energy
(ignoring air resistance)
mgh = ½mv²
h = v²/2g
h = 40.486² / (2(9.8))
h = 83.6303...
h = 84 m
An object moving at a constant velocity of 5.4 m/s travels for 12 s. How far will it move during that time?
Free-fall Acceleration is -10 m/s^2
Answer:
we know that
s=vt
given
v=5.4 m/s
t=12 s
s=5.4 m/s*12 s=64.8m
Explanation:
Hope this helps:)
Saturn's mass is 5.68 x 1024 kg and its radius is 6.03 x 107 m. A. Calculate the gravitational field strength at Saturn's surface. (2 marks) B. Calculate the force of gravity at Saturn's surface on an object with a mass of 50 kg.
Hi there!
A.
We can calculate the gravitational field strength using the following equation:
[tex]g = \frac{Gm_p}{r^2}[/tex]
G = Gravitational Constant
mp = mass of planet (kg)
r = radius (m)
Plug in the given values:
[tex]g = \frac{(6.67*10^{-11})*(5.68*10^{24})}{(6.03*10^7)^2} = \boxed{0.104 N/kg}[/tex]
B.
The force can be calculated using:
[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]
Plug in the values:
[tex]F_g = \frac{(6.67*10^{-11})(5.68*10^{24})(50)}{(6.04*10^7)^2} = \boxed{5.209N}[/tex]
Answer:
[tex]\boxed {\boxed {\sf g=0.104 \ N/kg \ and \ F_g= 5.2 \ N }}[/tex]
Explanation:
A. Gravitational Field Strength
The gravitational field strength can be calculated using the following formula:
[tex]g= \frac{Gm}{r^2}[/tex]
G, or the universal gravitational constant, is 6.67 × 10⁻¹¹ N*m²/kg². The mass of Saturn is 5.68 × 10²⁴ kilograms. The radius of Saturn is 6.03×10⁷ meters.
Substitute these values into the formula.
[tex]g= \frac{ (6.67 \times 10^{-11} \ N*m^2/kg^2) (5.68 \times 10^{24} \ kg)}{(6.03 \times 10^{7} \ m )^2}[/tex]
Multiply the numerator and square the denominator.
[tex]g= \frac{ 3.78856 \times 10^{14} \ N *m^2/kg }{3.63609 \times 10^{15} \ m^2}[/tex]
Divide.
[tex]g= 0.1041932405 \ N/kg[/tex]
The original measurements of mass and radius have 3 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 1 in the ten-thousandth place tells us to leave the 4 in the thousandth place.
[tex]\boxed {g \approx 0.104 \ N/kg}[/tex]
B. Force of Gravity
The force of gravity is calculated using the following formula:
[tex]F_g= mg[/tex]
The mass of the object is 50 kilograms. We just calculated the gravitational field strength, which is 0.104 Newtons per kilogram. Substitute these values into the formula.
[tex]F_g= (50 \ kg)(0.104 \ N/kg)[/tex]
Multiply. The units of kilograms cancel.
[tex]\boxed {F_g=5.20 \ N}[/tex]
HELP PLS!!
A 3 kg mass is raised a distance of 14 m above the earth by a vertical force of 93 N.
a
The gain in potential energy of the mass, to 3 significant figures, is:
Hi there!
We know that:
U (Gravitational Potential Energy) = mgh
Where:
g = acceleration due to gravity (m/s²)
m = mass (kg)
h = height/displacement (m)
Plug in the values:
U = 3 × 9.8 × 14 = 412 J
A flywheel having constant angular acceleration requires 4.70 s to rotate through 164 rad . Its angular velocity at the end of this time is 101 rad/s . Find the angular velocity at the beginning of the 4.70 s interval. Find the angular acceleration of the flywheel.
Answer:
A) -31.2 rad/s
B) 28.1 rad/s^2
Explanation:
Air is pumped into the tyre to inflate it.
This increases the temperature and the pressure of the air in the tyre.
Use ideas about molecules to explain why the air pressure in the tyre increases. *
Objects 1 and 2 attract each other with a gravitational force of 34 units. If the distance separating objects 1 and 2 is changed to one-third the original value, then the new gravitational force will be ____ units.
Answer:
F12 = G M1 M2 / R12^2
F12' = G M1 M2 / R12'^2
F12' / F12 = R12'^2 / R12^2 = (1/3)^2
F12' = 1/9 F12
The new force is 1/9 the of the old force
what is kinetic friction ? what causes it ? what does it generate ?
Answer:
What is kinetic friction?
Kinetic friction is defined as a force that acts between moving surfaces. A body moving on the surface experiences a force in the opposite direction of its movement.
What causes it?
When the mass is not moving, the object experiences static friction. The friction increases as the applied force increases until the block moves. After the block moves, it experiences kinetic friction, which is less than the maximum static friction.
What does it generate?
When surfaces in contact move relative to each other, the friction between the two surfaces converts kinetic energy into thermal energy (that is, it converts work to heat). This property can have dramatic consequences, as illustrated by the use of friction created by rubbing pieces of wood together to start a fire.
An object will begin moving from rest when acted upon by which forces?
A. Forces that are slightly less than the force of friction
B. Forces that result in a net force of zero
C. Forces that are equal and act in opposite directions
D. Forces that are greater in one direction than in any other direction
Answer:
D
Explanation:
Process of elimnination + it's the only one that makes sense
An object will begin moving from rest when acted upon by forces that are greater in one direction than in any other direction. Hence, Option (D) is correct.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
When forces that are greater in one direction than in any other direction, resultant will be unbalanced forces. Unbalanced forces are those acting on a body when the net force acting on the body is greater than zero. The body alters its state of motion when unbalanced forces act on it.
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Objects 1 and 2 attract each other with a gravitational force of 179 units. If the distance separating objects 1 and 2 is changed to four times the original value (i.e., quadrupled), then the new gravitational force will be ______ units.
Explanation:
Fgravity = G*(mass1*mass2)/D²
G is the gravitational constant throughout the universe.
D is the distance between the 2 objects.
the distance is now quadrupled.
Fgravitynew = G*(mass1*mass2)/(4D)² =
= G*(mass1*mass2)/(16D²) =
= (G*(mass1*mass2)/D²) / 16 = Fgravity/16
the new gravitational force will be 179/16 = 11.1875 units
Two objects are a distance of 1.7 x 103 meters apart. One object has a mass of 3 x 107 kg and the other has a mass of 6 x 108. Determine the gravitational force between the objects.
Answer: You need to use Newton's law for the equation --->
Explanation: G × M × m / separation. Thats how youll get your answer !!
Peregrine falcons, which can dive at 200 mph (90 m/s), grab prey birds from the air. The impact usually kills the prey. Suppose a 480 g falcon diving at 75 m/s strikes a 240 g pigeon, grabbing it in her talons. We can assume that the slow-flying pigeon is stationary. The collision between the birds lasts 15 ms.
Answer:
What is the average force of collision?
Explanation:
The velocity of the combined mass after impact is found by conservation of momentum
0.480(75) + 0.240(0) = (0.480 + 0.240)v
v = 50 m/s
An impulse results in a change of momentum
FΔt = mΔv
F = mΔv/Δt
for the pigeon
F = 0.240(50 - 0)/0.015
F = 800 N
for the falcon
F = 0.480(50 - 75)/0.015
F = -800 N
The final speed(v) of the Peregrine falcons and pigeon is 50 m/s
The objective of this question is to determine the final speed(v) of the Peregrine falcons and pigeon
From the information given:
the mass of the falcon m_f = 480 gthe speed of the falcon v_f = 75 m/sthe mass of the pigeon m_p = 240 gthe collision time = 15 ms = 0.015 sAccording to the conservation of momentum, we can say that the totality of momentum before and after the collision is the same. As such;
[tex]\mathbf{m_fv_f = (m_f+m_p) v}[/tex]
[tex]\mathbf{480 g \times 75 m/s = (480 + 240) \ g \times v}[/tex]
[tex]\mathbf{v = \dfrac{480 g \times 75 m/s}{(480 + 240) \ g}}[/tex]
[tex]\mathbf{v = \dfrac{36000 \ m/s}{ 720}}[/tex]
v = 50 m/s
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The element which does not show variable valency a) AI b)Fe c) Cu d) Hg
Answer:
None of these elements.
A 25-kg box of books is dropped on the floor from a height of 1.1 m and comes to rest. What impulse did the floor exert on the box
The impulse the floor exert on the box is 116 kgm/s.
The given parameters;
mass of the books, m = 25 kgheight of the books, h = 1.1 mThe final velocity of the box when it dropped to the floor is calculated as follows;
[tex]\frac{1}{2} mv^2 = mgh\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 1.1} \\\\v = 4.64 \ m/s[/tex]
The impulse the floor exert on the box is calculated as follows;
the impulse the floor exert on the box is equal to change in momentum of the book
[tex]J = \Delta P\\\\J = \Delta Mv\\\\J = M(v_f - v_0)\\\\J = 25(4.64 - 0)\\\\J = 116 \ kgm/s[/tex]
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The 6 strings on a guitar all have about the same length and are stretched with about the same tension. The highest string vibrates with a frequency that is 4 times that of the lowest string. 1)If the strings are made of the same material, how would you expect the diameters of the lowest and highest strings to compare
Answer:
not sure i need points
Explanation:
Please help me as quick as possible!!!!!!! Please, please please!!!!!!
Answer:
23
Explanation:
3x-4
Please Help
A projectile fired over level ground has an initial total velocity of 41.3 m/s. It is in the air for 5.1 s. What is the x-component of the projectile's initial velocity?
Answer:
Explanation:
In the vertical analysis assuming launch from ground level.
0 = 0 + (41.3sinθ)(5.1) + ½(-9.8)5.1²
(41.3sinθ)(5.1) = ½(9.8)5.1²
(41.3sinθ) = ½(9.8)5.1
sinθ = ½(9.8)5.1/41.3
sinθ = 0.60508...
θ = 37.235°
vx = 41.3cos37.235
vx = 32.881452...
vx = 32.9 m/s
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According to "Patterns of Change," select the ways that rocks are changed naturally over time. Choose three answers. O A. water O B. location O C. animals D. wind O E. people O F temperature
Answer: A , D & F
Explanation:
You spit a wad of paper horizontally
from a height of 1.8 m. The wad
leaves your mouth with a velocity of
6 m/s. How long does it take the
wad to hit the ground?