Final answer:
To determine the speed of an object attached to a horizontally lying spring on a frictionless surface at a specific stretch, we apply the law of conservation of energy and calculate the distribution of potential and kinetic energy at that point.
Explanation:
The subject of this question is Physics, and it involves a concept known as simple harmonic motion(SHM). To find the speed of the object at the instant when the spring is stretched by 0.041 m relative to its unstrained length, we can use the law of conservation of energy. In SHM, the total mechanical energy (sum of potential and kinetic energy) is conserved if there is no energy loss due to friction or other non-conservative forces.
At the compressed position (0.080 m from equilibrium), all the energy is potential, given by Ep = 1/2 k x2, where k is the spring constant and x is the displacement from equilibrium. At the position where the spring is stretched by 0.041 m, the energy will be partly potential and partly kinetic. The potential energy at this point is Ep = 1/2 k (0.041 m)2 and the kinetic energy is Ek = 1/2 m v2, where m is the mass and v is the velocity of the object.
Conservation of energy gives us:
1/2 k (0.080 m)2 = 1/2 k (0.041 m)2 + 1/2 m v2.
We can solve this equation for v, the speed at 0.041 m stretch, knowing the values of k, m, and x.
What is the kinetic energy of a bike with a mass of 16 kg traveling at 4 m/s?
Answer: 128J
Explanation:
[tex]Formula: E_k=\frac{1}{2}mv^2[/tex]
[tex]E_k=\frac{1}{2}(16kg)(4m/s)^2 \\E_k=\frac{1}{2}(16kg)(16m^2/s^2)\\ E_k=\frac{1}{2}(256kg*m^2/s^2)\\ E_k=128kg*m^2/s^2\\or\\E_k=128J[/tex]
Answer:
That is correct
Explanation:
128 J
A rockets initially at rest , it steady gains speed at a rate of 10m/s^2 for 4.6s during take off .
What was the rocket’s top speed ?
Final answer:
The rocket's top speed is 46 m/s.
Explanation:
The rocket's top speed can be found using the equation v = vo + at, where v is the final velocity, vo is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity (vo) is 0 m/s since the rocket starts at rest, the acceleration (a) is 10 m/s², and the time (t) is 4.6 s.
Plugging these values into the equation, we get:
v = 0 m/s + (10 m/s²)(4.6 s) = 46 m/s.
Therefore, the rocket's top speed is 46 m/s.
An object is thrown with a horizontal velocity of 20 m/s from a cliff that is 125 m above level ground. If air resistance is negligible, the time that it takes the object to fall to the ground from the cliff is most nearly
Given that,
Horizontal velocity of the object, v = 20 m/s
Height of the cliff, h = 125 m
We need to find the time that it takes the object to fall to the ground from the cliff is most nearly. It can be calculated using second equation of motion. Let us consider that the initial speed of the object is 0. So,
[tex]h=ut+\dfrac{1}{2}at^2[/tex]
Here, a = g and u = 0
[tex]h=\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 125}{10}} \\\\t=5\ s[/tex]
So, the object will take 5 seconds to fall to the ground from the cliff.
The time taken for the object to fall to the ground is 5 s
From the question given above, the following data were obtained:
Horizontal velocity (u) = 20 m/sHeight (h) = 125 mAcceleration due to gravity (g) = 10 m/s²Time (t) =?The time taken for the object to fall to the ground can be obtained as follow:
[tex]t = \sqrt{ \frac{2h}{g}} \\ \\ t = \sqrt{ \frac{2 \times 125}{10}} \\ \\ t = 5 \: s \\ \\ [/tex]
Therefore, the time taken for the object to get to the ground is 5 s
Learn more about projectile motion: https://brainly.com/question/15206806
Trace the path of a ray emitted from the tip of the object toward the focal point of the mirror and then the reflected ray that results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray. Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded.
When a ray is emitted from the tip of an object towards the focal point of a mirror, it follows the law of reflection. The location and orientation of the reflected ray depend on the shape of the mirror. Additional rays can be traced from the base of the object to locate the extended image.
Explanation:When a ray is emitted from the tip of an object towards the focal point of a mirror, it follows the law of reflection. For a concave mirror, the reflected ray passes through the focal point, while for a convex mirror, the reflected ray extends backward through the focal point. The location and orientation of the reflected ray depend on the shape of the mirror. To locate the extended image, additional rays can be traced from the base of the object along the optical axis. All four principal rays run parallel to the optical axis, reflect from the mirror, and then run back along the optical axis. The image of the base of the object is located directly above the image of the tip, as the mirror is symmetrical from top to bottom.
Learn more about Ray tracing in mirrors here:https://brainly.com/question/15506761
#SPJ12
Ray tracing involves drawing rays from an object through points of interest like the focal point of a mirror and then determining where the reflected rays intersect to locate the image. It is a crucial method in optics to determine the properties of images formed by mirrors and lenses.
Explanation:Understanding Ray Tracing in MirrorsTo trace the path of a ray emitted from the tip of an object toward the focal point of a mirror, as well as the reflected ray, one must understand the basic principles of ray tracing with mirrors. With a concave mirror, an incident ray that travels parallel to the optical axis will be reflected through the focal point on the same side of the mirror. Conversely, with a convex mirror, a ray that travels parallel to the optical axis will reflect as if it originates from the focal point behind the mirror, forming a virtual focus. To construct the reflected ray, you draw the ray until it reaches the mirror’s surface and then redirect it according to the mirror type's ray tracing rules.
Furthermore, the intersection of the reflected rays (real or virtual) determines the location of the image. If the rays meet in real space, the image is real; if they only appear to intersect upon extension in virtual space, the image is virtual.
To achieve a complete picture of how an image is formed, one must also trace rays from another point on the object. For instance, tracing rays from the base of the object can help determine the orientation of the image. In the case of rays that are collinear with the optical axis, the image will maintain the object's vertical orientation. By tracing at least two different rays following the simple ray tracing rules, one can locate the image formed by mirrors and lenses.
Learn more about Ray Tracing here:https://brainly.com/question/35884592
#SPJ3
1. a) If a particle's position is given by LaTeX: x\:=\:4-12t\:+\:3t^2x = 4 − 12 t + 3 t 2(where t is in seconds and x is in meters), what is its velocity at LaTeX: t=1st = 1 s? b) Is it moving in the positive or negative direction of LaTeX: xx just then? c) What is its speed just then? d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculations.) e) Is there ever an instant when the velocity is zero? If so, give the time LaTeX: tt ; if not, answer no. f) Is there a time after LaTeX: t=3st = 3 s when the particle is moving in the negative direction of LaTeX: xx? If so, give the time LaTeX: tt; if not, answer no. (Hint: Speed= LaTeX: \mid v\mid∣ v ∣)
Answer:
a) v=-6m/s
b) negative direction
c) 6m/s
d) decreasing
e) for t=2s
f) Yes
Explanation:
The particle position is given by:
[tex]x=4-12t+3t^2[/tex]
a) the velocity of the particle is given by the derivative of x in time:
[tex]v=\frac{dx}{dt}=-12+6t[/tex]
and for t=1s you have:
[tex]v=\frac{dx}{dt}=-12+6(1)^2=-6\frac{m}{s}[/tex]
b) for t=1s you can notice that the particle is moving in the negative x direction.
c) The speed can be computed by using the formula:
[tex]|v|=\sqrt{(-12+6t)^2}=\sqrt{(-12+6)^2}=6\frac{m}{s}[/tex]
d) Due to the negative value of the velocity in a) you can conclude that the speed is decreasing.
e) There is a time in which the velocity is zero. You can conclude that because if t=2 in the formula for v in a), v=0
[tex]v=0=-12+6(t)\\\\t=\frac{12}{6}=2[/tex]
f) after t=3s the particle will move in the negative direction, this because it is clear that 4+3t^2 does not exceed -12t.
Answer:
a) v (1) = -6 m/s
b) negative x-direction
c) s ( 1 ) = 6 m/s
d) The speed decreases at t increases from 0 to 2 seconds.
e) At t = 2 s, the velocity is 0
f) No
Explanation:
Given:-
- The position function of the particle:
x (t) = 4 - 12t + 3t^2
Find:-
what is its velocity at t = 1 s? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time t; if not, answer no. (f) Is there a time after t = 3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer no.
Solution:-
- The velocity function of the particle v(t) can be determined from the following definition:
v (t) = d x(t) / dt
v (t) = -12 + 6t
- Evaluate the velocity at time t = 1 s:
v (1) = -12 + 6(1)
v (1) = -6 m/s
- The negative sign of the velocity at time t = 1s shows that the particle is moving in the negative x-direction.
- The speed ( s ( t )is the absolute value of velocity at time t = 1s:
s ( t ) = abs ( v ( t ) )
s ( 1 ) = abs ( v ( 1 ) )
s ( 1 ) = abs ( -6 )
s ( 1 ) = 6 m/s
- The speed of the particle at time t = 0,
s ( t ) = abs ( -12 + 6t )
s ( 0 ) = abs (-12 + 6 (0) )
s ( 0 ) = abs ( -12 )
s ( 0 ) = 12 m/s
- The speed of the particle at time t = 2,
s ( t ) = abs ( -12 + 6t )
s ( 2 ) = abs (-12 + 6 (2) )
s ( 2 ) = abs ( 0 )
s ( 2 ) = 0 m/s
- Hence, the speed of the particle decreases from s ( 0 ) = 12 m/s to s ( 2 ) = 0 m/s in the time interval t = 0 to t = 2 s.
- As the speed decreases as time increases over the interval t = 0 , t = 2 s the velocity v(t) also approaches 0, at time t = 2 s. s ( 2 ) = 0 m/s.
- We will develop an inequality when v (t) is positive:
v (t) = -12 + 6t > 0
6t > 12
t > 2
- So for all values of t > 2 the velocity of the particle is always positive.
Kopnięta poziomo piłka o masie 1,5 kg 1,5 kg w momencie kopnięcia uzyskała przyspieszenie 5 m s 2 5ms2 . Oblicz wartość siły kopnięcia. Po nasiąknięciu wodą masa piłki wzrosła do 2 kg 2 kg . O ile większej siły musi użyć zawodnik, aby nadać jej takie samo przyspieszenie jak poprzednio?
1) 7.5 N
2) 10 N
Explanation:
1)
We can solve this first part of the problem by using Newton's second law of motion, which states that the net force on an object is equal to the product between its mass and its acceleration:
[tex]F=ma[/tex]
where
F is the net force
m is the mass of the object
a is its acceleration
In this problem we have:
m = 1.5 kg is the mass of the ball
[tex]a=5 m/s^2[/tex] is the acceleration
So, the kick force on it was:
[tex]F=(1.5)(5)=7.5 N[/tex]
2)
In this case, the mass of the ball has increased to
m' = 2 kg
We can also solve this part by using again Newton's second law of motion:
[tex]F'=m'a[/tex]
where
F' is the new kick force
m' = 2 kg is the new mass of the ball
a is the acceleration
The acceleration is the same as before,
[tex]a=5 m/s^2[/tex]
Therefore, the new kick force is:
[tex]F'=(2)(5)=10 N[/tex]
Suppose that 3.00 g of hydrogen is separated into electrons and protons. Suppose also that the protons are placed at the Earth's North Pole and the electrons are placed at the South Pole. What is the resulting compressional force on the Earth?
Answer:
Explanation:
Atomic mass of hydrogen is 1 . so 1 g of hydrogen will have 6.02 x 10²³ atoms
3 g of hydrogen will have 3 x 6.02 x 10²³ atoms . This will give 3 x 6.02 x 10²³ protons and same number of electrons
So number of protons at north pole = 3 x 6.02 x 10²³
charge on these protons = 1.6 x 10⁻¹⁹ x 3 x 6.02 x 10²³ C
= 28.9 x 10⁴ C
similarly total charge on electrons at south pole = 28.9 x 10⁴ C
distance between them = diameter of the earth = 2 x 6356 x 10³ m
= 12712 x 10³ m
Attractive force between these charges
= k q₁q₂ / r² , q₁ ,q₂ are charges and r is distance between charges.
= 9 x 10⁹ x (28.9 x 10⁴)² / (12712 x 10³ )²
= 4.6517 x 10⁻⁵ x 10¹¹
= 4.6517 x 10⁶ N
By what factor will the electrostatic force between two charged objects change when the amount of charge on one object doubles?
Answer:
The electrostatic force between two charge doubles when when the amount of charge on one object doubles.
Explanation:
Force between two charge particle [tex]F=\frac{1}{4\pi\epsilon } \frac{q_1\cdot\ q_2}{r^{2} }[/tex]
Where, [tex]\frac{1}{4\pi\epsilon } = constant[/tex]
[tex]q_1,q_2\ magnitude\ of\ charges.\\r\ is\ the\ distance\ between\ the\ particles.[/tex]
Now suppose as per the question
[tex]q_1= 2 q_1\ with\ q_2\ and\ r\ with\ same\ value.\\Hence\ force\ between\ the\ particle\ F_2 =2 F[/tex]
Hence, electrostatic force between two charge doubles when when the amount of charge on one object doubles.
1) what is a calorie?
An open container holds ice of mass 0.550 kg at a temperature of -15.3 âC . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 900 J/minute .The specific heat of ice to is 2100 J/kgâK and the heat of fusion for ice is 334Ã103J/kg.a) How much time tmelts passes before the ice starts to melt?tmelts=___minutesb)From the time when the heating begins, how much time trise does it take before the temperature begins to rise above 0âC?trise=____minutes
The ice in the open container begins to melt after 19.82 minutes of heating, and the temperature of the system starts to rise above 0°C after a total heating time of 223.93 minutes.
Explanation:To answer both parts of this question, we need to calculate the time for two processes: the heating of the ice to 0°C (melting point), and the melting of the ice into water at 0°C.
Firstly, we find the heat (Q) required to raise the temperature of the ice to 0°C using Q=mcΔT, which is 0.55kg * 2100 J/kg°C * (0 - (-15.3°C)) = 17842.5 J. As the heater adds 900J per minute, the time for this is 17842.5J ÷ 900J/min = 19.82 minutes.
Next, we find the heat required to melt ice into water at 0°C using Q=mLf, where Lf = 334000 J/kg. This is 0.55kg * 334000 J/kg = 183700J. The time for this can be found by 183700J ÷ 900J/min = 204.11 minutes.
tmelts (the time before ice begins to melt) is 19.82 minutes, and trise (the total time before the temperature begins to rise above 0°C) is 19.82 min + 204.11 min = 223.93 minutes.
Learn more about Heat Transfer Calculations here:https://brainly.com/question/31080599
#SPJ12
The time before the ice starts to melt (tmelts) is 19.63 minutes, and the total time before the temperature begins to rise above 0 °C (trise) is 223.74 minutes.
Explanation:Calculating Time Until Ice Starts to MeltTo calculate the time tmelts before the ice starts to melt, you firstly need to determine how much heat is necessary to raise the ice from -15.3 °C to 0 °C using the formula Q = m * c * ΔT, where Q is the heat, m is the mass of the ice, c is the specific heat of ice, and ΔT is the change in temperature. In this case, Q would be:
Q = (0.550 kg) * (2,100 J/kg·K) * (15.3 K) = 17,661.5 J.
Since heat is supplied at 900 J/minute, the time tmelts can be calculated as:
tmelts = Q / (heat rate) = 17,661.5 J / (900 J/min) = 19.63 minutes.
Calculating Time Until Temperature Rises Above 0 °CTo calculate the time trise until the temperature begins to rise above 0 °C, we must add the time it takes to melt the ice completely at 0 °C. The heat of fusion (∆Hfus) formula is used here: Q = m * Lf where Lf is the heat of fusion for ice. Considering that the heat of fusion for ice is 334,000 J/kg:
Q = (0.550 kg) * (334,000 J/kg) = 183,700 J.
For melting the ice at 0 °C, it will take:
trise = Q / (heat rate) = 183,700 J / (900 J/min) = 204.11 minutes.
Therefore, the total time before the temperature begins to rise above 0 °C is the sum of the time to warm the ice to 0 °C and the time to melt it completely, which will be 19.63 minutes + 204.11 minutes = 223.74 minutes.
Learn more about Heat Transfer in Ice here:https://brainly.com/question/18940221
#SPJ3
A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/10. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her
Answer:
(a) 1058.4 J
(b) -10584 J
Explanation:
Parameters given:
Mass of astronaut, m = 72 kg
Distance moved by astronaut, d = 15 m
(a) WORK DONE BY FORCE FROM THE HELICOPTER
Work done is given as the product of Force applied to a body and the distance moved by the body:
W = F * d
The force from the helicopter is given as:
F = m * a
where a = acceleration of the astronaut due to the helicopter
Therefore, the work done is given as:
W = m * a * d
W = 72 * g/10 * 15
W = [tex]\frac{72 * 9.8 * 15}{10}[/tex]
W = 1058.4 J
(b) WORK DONE BY FORCE OF GRAVITY
W = F * d
The force of gravity is given as:
F = -m * g
where g = acceleration due to gravity
The negative sign is due to the fact that the astronaut moves in an opposite direction (upwards) to the force of gravity (Gravity acts downwards)
Therefore, the work done is given as:
W = -m * g * d
W = -72 * 9.8 * 15
W = -10584 J
A double-concave (thickest at the edges, thinnest in the middle) thin lens is made of glass with an index of refraction of 1.6. The radius of curvature of the left and right faces of the lens are 65 cm and 75 cm, respectively. The object lies to the left of the lens. What is the focal length of the lens
Answer:
- 58 cm
Explanation:
refractive index, n = 1.6
radius of curvature of left face, R1 = - 65 cm
Radius of curvature of the right face, R2 = 75 cm
Use the lens maker's formula
[tex]\frac{1}{f}=\left ( n-1 \right )\times \left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]
[tex]\frac{1}{f}=\left ( 1.6-1 \right )\times \left ( -\frac{1}{65}-\frac{1}{75} \right )[/tex]
[tex]\frac{1}{f}=\left ( 0.6 \right )\times \left ( \frac{-75-65}{75\times 65}\right )[/tex]
f = - 58 cm
Thus, the focal length of the lens is - 58 cm.
A kettle is rated at 1 kW, 220 V. Calculate the working resistance of the kettle.
Resistance = ...........
When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point. Calculate how much energy has been supplied?
Explanation:
Power of electric kettle, P = 1 kW
Voltage, V = 220 V
(a) Electric power is given by the formula as follows :
[tex]P=\dfrac{V^2}{R}[/tex]
R is resistance
[tex]R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{10^3}\\\\R=48.4\ \Omega[/tex]
(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.
Energy supplied is given by :
[tex]E=P\times t[/tex]
P is power, [tex]P=\dfrac{V^2}{R}[/tex]
[tex]E=\dfrac{V^2}{R}t\\\\E=\dfrac{(220)^2}{48.4}\times 180\\\\E=180000\ J\\\\E=180\ kJ[/tex]
The working resistance of the kettle is 48.35 ohms. The amount of energy supplied is 180,000 J.
Explanation:To calculate the working resistance of the kettle, we can use Ohm's law, which states that resistance (R) is equal to the voltage (V) divided by the current (I).
In this case, we know that the kettle is rated at 1 kW, which is equal to 1000 W. Since power (P) is equal to voltage multiplied by current, we can rearrange the equation to solve for current: I = P/V. Thus, I = 1000 W / 220 V = 4.55 A. Now, we can use Ohm's law to determine the resistance: R = V/I = 220 V / 4.55 A = 48.35 ohms.To calculate the amount of energy supplied, we need to use the formula E = P * t, where E is energy, P is power, and t is time.
Since the kettle is rated at 1 kW or 1000 W, we can calculate the energy supplied over 3 minutes as follows: E = 1000 W * 3 min * (60 s / 1 min) = 180,000 J.Therefore, The working resistance of the kettle is 48.35 ohms. The amount of energy supplied is 180,000 J.
Star A appears brighter than star B, as seen from Earth. Therefore, star A must be closer to Earth than star B. Star A appears brighter than star B, as seen from Earth. Therefore, star A must be closer to Earth than star B. a.True b.False
Answer:
The statement is not true always.
Explanation:
Given:
Star A appears brighter than Star B.
The apparent brightness of a star depends on two factors:
i) the distance of the star from the earth
ii) the limunosity of the star
If two stars are at same distance from the earth and one has more luminosity than other then ir appears to be more luminous. Also if two stars are at a different distance from the earth, then the star having more luminosity appears to be more luminous.
Thus, the statement is not true.
Which element is a metal? A.lithium B.Nitrogen C.sulfur D.chlorine
Answer:
the answer is b
Explanation:
hope it helps ...
Lithium (Li) is a metal located in Group 1 of the periodic table and is known as an alkali metal. It stands out for its low density and capacity to conduct electricity, along with exhibiting the typical properties of metals.
The element lithium (Li) is a metal, and notably, it is the first metal in the periodic table located in Group 1. It has the lowest atomic number (3) among the metals. The defining characteristics of metals include their ability to conduct electricity, possess malleability, and often exhibit a shiny luster. Lithium, being part of the alkali metals category, shares these properties. Furthermore, lithium is the least dense metal and has a unique electronic structure where the third electron occupies an outer shell, differentiating it from nonmetals.
Lithium is used in various applications including batteries and the aerospace industry due to its low density and high reactivity. The chemically similar elements to lithium include other alkali metals such as sodium and potassium, which also share similar chemical properties.
Which factors are most likely to affect whether a galaxy will evolve by merging with a second galaxy? Select the two correct answers.(1 point)
its proportion of gas compared to that of the other galaxy
its distance from the other galaxy
its size compared to that of the other galaxy
its age compared to that of the other galaxy
Answer:
ii. its distance from the other galaxy.
iii. its size compared to that of the other galaxy
Explanation:
A galaxy is a vast constituent of interstellar medium of gas, dust and billions of stars bounded by the gravitational pull. There are tens of billions of galaxies existing in the whole universe. Various numbers of galaxies can merge to form an inter-cluster, which is a more massive and bigger galaxy.
The constituent of an individual galaxy are held together by mutual gravitational pull, but may interact with other galaxies depending on; its distance from other galaxies and its size compared to that of the other galaxy. This process would result into the formation of a giant elliptical galaxy and a number of stars.
Answer:
1.Gravity acts on existing material, shaping it into a new form.
2.a much larger galaxy, a nearby galaxy
3.low gas content, little star formation
Explanation:
i just did the test and its 100%. dont believe me? just do the test and see for your self
In a summer storm, the wind is blowing with a velocity of 24 m/s north. Suddenly in 3 seconds the wind's velocity is 6 m/s. What is the wind's deceleration?
Answer:
The deceleration is [tex]6ms^{-2}[/tex]
Explanation:
Acceleration is change in velocity with respect to time.
[tex]a = \frac{\Delta V}{\Delta t}\\a = \frac{24-6}{3} \\a = 6[/tex]
A student's life was saved in an automobile accident because an airbag expanded in front of his head. If the car had not been equipped with an airbag, the windshield would have stopped the motion of his head in a much shorter time. Compared to the windshield, the airbag:
Answer:
Compared to windshield the airbag exerts much lesser force
Explanation:
Impulse is defined as change in momentum of the object when it is acted upon by a force during interval of time
Impulse = Impulsive force *time
I = F*Δt
If the object should be bought to rest from certain velocity there should be change in momentum. If the duration in which the momentum is increased then there would be less force applied and hence less damage.
Airbags are used to reduce the force experience by the people when they are met with accident by extending the time required to stop the momentum.
During the collision, the passenger is carried towards the windshield and if they are stopped by collision with wind shield the force will be larger and more damage.But if they are hit with airbag then the force will be less due to increased time.
The change is momentum will be the same with or without momentum but its the time that decides the impact of force.By making it longer the force become less.
Thus compared to the windshield the airbag exerts much lesser force.
You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it at position y0. Imagine that you slowly pull down with a spring scale so the block is now at position y bottom, below the equilibrium position y0 where it was hanging at rest. The scale reading when you let go of the block is 3.0 N.
a. Where was the block when you let go? Assume y0 is the equilibrium position of the block and that "down" is a positive direction.
b. Determine the work you did stretching the spring.
Express your answer to two significant figures and include the appropriate units.
c. What was the energy of the spring-Earth system when you let go (assume that zero potential energy corresponds to the equilibrium position of the block)?
Express your answer to two significant figures and include the appropriate units.
d. How far will the block rise after you release it?
Express your answer to two significant figures and include the appropriate units.
Answer:
a) Δy = 0.144 m
b) W = 0.145 J
c) Us = 0.32 J
d) ymax = 0.144 m
Explanation:
a) First let's find the spring constant using Hooke's Law
F = k*Δy ⇒ k = F/Δy
where
F = m*g = 0.1 kg*9.81 m/s² = 0.981 N
and Δy = 0.07 m. Hence
k = 0.981 N/0.07 m = 14.014 N/m ≈ 14 N/m
In order to find the position of the block when we let it go, we need to find the force that caused this expansion in the spring, we know that the reading of the scale was 3 N and this reading includes the force we want to find and the weight of the block, therefore:
f = 3 N - F = 3 N - 0.981 N = 2.019 N
Now that we have found the force we can use Hooke's Law in order to find the position of the block
f = k*Δy ⇒ Δy = f/k
⇒ Δy = 2.019 N/14 N/m
⇒ Δy = 0.144 m
b) First, notice that there are two kind of potential energy: the potential energy in the spring and the potential energy due to the gravitational field:
W = ΔU
W = ΔUs + ΔUg
W = (Usf - Usi) + (Ugf - Ugi)
Notice that
Us = 0.5*k*y²
where
yf = 0.07 m + 0.144 m = 0.214 m and
yi = 0.07 m
and we will take the zero level to be the equilibrium position where the block was hanging at rest. Hence
W = 0.5*k*(yf² - yi²) + m*g*(0 - Δy)
⇒ W = 0.5*14 N/m*((0.214 m)² - (0.07 m)²) + (0.1 kg)*(9.81 m/s²)*(0 - 0.144 m)
⇒ W = 0.145 J
c) When we let the block go the spring was stretched by
y = 0.07 m + 0.144 m = 0.214 m
Therefore:
Us = 0.5*k*y²
⇒ Us = 0.5*14 N/m*(0.214 m)²
⇒ Us = 0.32 J
d) Because the position that we pulled the block to it is considered as the amplitude for the vibrational motion that will happen after we release the block, then the maximum height the particle will reach above the equilibrium position is
ymax = Δy = 0.144 m
Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series with the galvanometer is 1680 Ω for the 20.0-V scale and 2930 Ω for the 30.0-V scale. Determine the coil resistance.
Answer:
Resistance of the circuit is 820 Ω
Explanation:
Given:
Two galvanometer resistance are given along with its voltages.
Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.
⇒ [tex]V=20\ \Omega,\ V_1=30\ \Omega[/tex]
⇒ [tex]G=1680\ \Omega,\ G_1=2930\ \Omega[/tex]
Concept to be used:
Conversion of galvanometer into voltmeter.
Let [tex]G[/tex] be the resistance of the galvanometer and [tex]I_g[/tex] the maximum deflection in the galvanometer.
To measure maximum voltage resistance [tex]R[/tex] is connected in series .
So,
⇒ [tex]V=I_g(R+G)[/tex]
We have to find the value of [tex]R[/tex] we know that in series circuit current are same.
For [tex]G=1680[/tex] For [tex]G_1=2930[/tex]
⇒ [tex]I_g=\frac{V}{R+G}[/tex] equation (i) ⇒ [tex]I_g=\frac{V_1}{R+G_1}[/tex] equation (ii)
Equating both the above equations:
⇒ [tex]\frac{V}{R+G} = \frac{V_1}{R+G_1}[/tex]
⇒ [tex]V(R+ G_1) = V_1 (R+G)[/tex]
⇒ [tex]VR+VG_1 = V_1R+V_1G[/tex]
⇒ [tex]VR-V_1R = V_1G-VG_1[/tex]
⇒ [tex]R(V-V_1) = V_1G-VG_1[/tex]
⇒ [tex]R =\frac{V_1G-VG_1}{(V-V_1)}[/tex]
⇒ Plugging the values.
⇒ [tex]R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}[/tex]
⇒ [tex]R =\frac{(50400 - 58600)}{(-10)}[/tex]
⇒ [tex]R=\frac{-8200}{-10}[/tex]
⇒ [tex]R=820\ \Omega[/tex]
The coil resistance of the circuit is 820 Ω .
Let's say that the number density of galaxies in the universe is, on average, 3 × 10–68 galaxies/m3. If astronomers could observe all galaxies out to a distance of 1010 light-years, how many galaxies would they find? (Note that there are 1016 meters in 1 light-year.)
Answer:
They would find [tex]3.14\times 10^{10}[/tex]galaxies.
Explanation:
Given that,
The number of density of galaxies in the universe is 3×10⁻⁶⁸ galaxies /m³.
Assuming that, the astronomers are observing at the center of sphere.
So, they can observe the sphere of space whose radius 10¹⁰ light years.
1 light year = 10¹⁶ meters
10¹⁰ light years =10¹⁰ .10¹⁶ meters
=10²⁶meters
The volume of the space is
=[tex]\frac43 \pi r^3[/tex]
[tex]=\frac43 \pi (10^{26})^3[/tex] m³
[tex]=\frac43 \pi 10^{78}[/tex] m³.
The number of galaxies
= Volume of the space × density
[tex]=(3\times 10^{-68}\ galaxies /m^3)\times(\frac43 \pi . 10^{78}\ m^3)[/tex]
[tex]=10^{10}\pi[/tex] galaxies
= [tex]3.14\times 10^{10}[/tex]galaxies
They would find [tex]3.14\times 10^{10}[/tex]galaxies.
If astronomers could observe all galaxies out to a distance of 10^10 light-years, they would find approximately 10^100 galaxies.
Explanation:To find the number of galaxies that astronomers would find if they could observe all galaxies out to a distance of 1010 light-years, we can use the average number density of galaxies in the universe. The number density is given as 3 × 10-68 galaxies/m3. We can convert the distance to meters by multiplying by 1016 (since there are 1016 meters in 1 light-year).
Next, we can calculate the volume of space that astronomers would be observing. The volume can be found by multiplying the distance cubed (in meters) by 4/3π. The number of galaxies can then be calculated by multiplying the volume by the number density.
Substituting the given values into the equation, we have:
Volume = (4/3) × π × (1016)3 m3
Number of galaxies = Number density × Volume
After calculating the volume and multiplying it by the number density, we find that astronomers would find approximately 10100 galaxies if they could observe all galaxies out to a distance of 1010 light-years.
Learn more about Galaxy Observations here:https://brainly.com/question/34148541
#SPJ11
Use the following information for determining sound intensity. The number of decibels β of a sound with an intensity of I watts per square meter is given by β = 10 log(I/I0), where I0 is an intensity of 10−12 watt per square meter, corresponding roughly to the faintest sound that can be heard by the human ear. Find the number of decibels β of the sound.
Answer:
120 decibels
Explanation:
we know sound in decibels is given by.
[tex]\beta =10log(\frac{I_{} }{I_{0} } )[/tex]
Where I is intensity of the sound and [tex]I_{0}=10^-12W/m^{2 }[/tex] is reference intensity.
substituting All of this in our decibel formula with I =1W/[tex]m^2[/tex]
gives us 120 decibels.
The sound intensity level of a sound is measured in decibels (dB) using a logarithmic scale that uses a reference intensity of 10^-12 W/m², which is the threshold of human hearing. The decibel level of a sound is determined by the ratio of its intensity to this reference intensity, with each increment of 10 dB corresponding to a tenfold increase in intensity.
Explanation:The sound intensity level ß, measured in decibels (dB), provides a logarithmic measure of sound intensity I in watts per meter squared (W/m²), with the formula
ß = 10 log(I/I0)
where I0 = 10⁻¹² W/m² is the reference intensity, corresponding to the threshold of human hearing. The decibel level of a sound is thus derived from the ratio of its intensity to this reference intensity. A sound with the same intensity as the reference intensity (I = I0) has a decibel level of 0 dB, because log 10(1) = 0.
Interpreting decibels involves understanding that because the formula uses a logarithm, each increment of 10 dB corresponds to a tenfold increase in intensity. Therefore, a sound with an intensity of 10^-11 W/m² would be 10 dB, a sound with an intensity of 10⁻¹⁰ W/m² would be 20 dB, and so on. The decibel scale helps to compress the vast range of sound intensities that human ears can perceive into a more manageable range of numbers.
Learn more about Sound Intensity Level here:https://brainly.com/question/30101270
#SPJ11
All organisms require a set of instructions that specify its traits.The instructions that are responsible for all the inherited traits of an organisms
Answer:
That's DNA
Explanation:
DNA(Deoxyribonucleic acid) is the building blocks of life, the Legos for genetically complex life if you will. All living things have DNA, it is the instruction that that organism must follow to specify what exact traits it'll have, For example, I have Blue eyes because Blue eyes is apart of my DNA. The DNA told my body that I was gonna have blue eyes before I was born.
You are coasting on your 12-kg bicycle at 13 m/s and a 5.0-g bug splatters on your helmet. The bug was initially moving at 1.5 m/s in the same direction as you. If your mass is 70 kg, answer the following questions:a.What is the initial momentum of you plus your bicycle? b.What is the intial momentum of the bug? c.What is your change in velocity due to the collision the bug? d.What would the change in velocity have been if the bug were traveling in the opposite direction?
Answer:
a) Pi,c = 1066 kgm/s
b) Pi,b = 0.0075 kgm/s
c) ΔV = - 0.0007 m/s
d) ΔV = - 0.0008 m/s
Explanation:
Given:-
- The mass of the bicycle, mc = 12 kg
- The mass of passenger, mp = 70 kg
- The mass of the bug, mb = 5.0 g
- The initial speed of the bicycle, vpi = 13 m/s
- The initial speed of the bug, vbi = 1.5 m/s
Find:-
a.What is the initial momentum of you plus your bicycle?
b.What is the initial momentum of the bug?
c.What is your change in velocity due to the collision the bug?
d.What would the change in velocity have been if the bug were traveling in the opposite direction?
Solution:-
- First we will set our one dimensional coordinate system, taking right to be positive in the direction of bicycle.
- The initial linear momentum (Pi,c) of the passenger and the bicycle would be:
Pi,c = vpi* ( mc + mp)
Pi,c = 13* ( 12+ 70 )
Pi,c = 1066 kgm/s
- The initial linear momentum (Pi,b) of the bug would be:
Pi,b = vbi*mb
Pi,b = 0.005*1.5
Pi,b = 0.0075 kgm/s
- We will consider the bicycle, the passenger and the bug as a system in isolation on which no external unbalanced forces are acting. This validates the use of linear conservation of momentum.
- The bicycle, passenger and bug all travel in the (+x) direction after the bug splatters on the helmet.
Pi = Pf
Pi,c + Pi,b = V*(mb + mc + mp)
Where, V : The velocity of the (bicycle, passenger and bug) after collision.
1066 + 0.0075 = V*( 0.005 + 12 + 70 )
V = 1066.0075 / 82.005
V = 12.9993 m/s
- The change in velocity is Δv = 13 - 12.9993 = - 0.00070 m/s
- If the bug travels in the opposite direction then the sign of the initial momentum of the bug changes from (+) to (-).
- We will apply the linear conservation of momentum similarly.
Pi = Pf
Pi,c + Pi,b = V*(mb + mc + mp)
1066 - 0.0075 = V*( 0.005 + 12 + 70 )
V = 1065.9925 / 82.005
V = 12.99911 m/s
- The change in velocity is Δv = 13 - 12.99911 = -0.00088 m/s
Answer:
a. The initial momentum of you and your bicycle is 1066 kgm/s.
b. The initial momentum of the bug is 0.0075 kgm/s.
c. The change in velocity due to the collision with the bug is -0.0008 m/s.
d. If the bug were travelling in the opposite direction, the change in velocity due to the collision would have been -0.0009 m/s.
Explanation:
The initial momentum of you and your bicycle can be easily calculated using the definition of momentum:
[tex]p=mv\\\\p=(m_{you}+m_{bicycle})v\\\\p=(70kg+12kg)(13m/s)\\\\p=1066kgm/s[/tex]
So the initial momentum of you plus your bicycle is 1066 kgm/s (a).
The initial momentum of the bug can be obtained in the same way:
[tex]p=mv\\\\p=(0.005kg)(1.5m/s)\\\\p=0.0075kgm/s[/tex]
Then the initial momentum of the bug is 0.0075 kgm/s (b).
Now, since the mass of the bug is much less than your mass, we can think of this as a perfectly inelastic collision. This means that, after the collision, the velocity of you, the bicycle and the bug is the same. From the conservation of linear momentum, we have:
[tex]p_0=p_f\\\\(m_{you}+m_{bicycle})v_{you}+m_{bug}v_{bug}=(m_{you}+m_{bicycle}+m_{bug})v_f\\\\v_f=\frac{(m_{you}+m_{bicycle})v_{you}+m_{bug}v_{bug}}{m_{you}+m_{bicycle}+m_{bug}}\\\\v_f=\frac{(70kg+12kg)(13m/s)+(0.005kg)(1.5m/s)}{70kg+12kg+0.005kg}\\ \\v_f=12.9992m/s[/tex]
As your initial velocity was 13m/s, the change in velocity is of -0.0008 m/s (c).
If the bug were travelling in the opposite direction, its initial velocity would have been negative. So:
[tex]v_f=\frac{(70kg+12kg)(13m/s)-(0.005kg)(1.5m/s)}{70kg+12kg+0.005kg}\\ \\v_f= 12.9991m/s[/tex]
So, in this case the change in velocity is of -0.0009 m/s (d).
Note that the bug is so small that the change in velocity is negligible in most cases. That's why we don't notice when we hit a bug when riding bicycle.
When light travels from a material with a lower index of refraction to a material with a higher index of refraction, the refracted beam will: a) shift away from the normal b) shift towards the normal
Answer:
Shift towards the normal
Explanation:
Refraction is defined as the change in direction of light rays when passing through from a medium to another.
The ray can either pass through a less dense medium to a denser medium or from a denser medium to a less dense medium. The light ray bends towards or away from the normal ray(ray perpendicular to the plane) depending whether the travels from less dense to denser or otherwise.
Note that if a ray travels from less dense medium which have a low refractive index like air to a more dense medium like water which have a higher refractive index than air, the refracted ray tends to bend towards the normal, otherwise they bend away from the normal.
Answer:
Shift towards the normal
Explanation:
what is the wavelength of microwaves
Answer:
1 mm PLS mark brainliest
Explanation:
Microwaves are electromagnetic waves with wavelengths longer than those of terahertz (THz) wavelengths, but relatively short for radio waves. Microwaves have wavelengths approximately in the range of 30 cm (frequency = 1 GHz) to 1 mm (300 GHz).
Why is a shadow formed
Answer:
Shadows are made by blocking light. Light rays travel from a source in straight lines. If an opaque (solid) object gets in the way, it stops light rays from traveling through it. The size and shape of a shadow depend on the position and size of the light source compared to the object.
Explanation:
Answer:
shadows are made by blocking light
En un planeta distante,un objeto en caida libre cambia su velocidad de 10m/s a 90m/s en 2 segundos ¿Cual es la aceleración gravitacional de dicho planeta?
Answer:
a = 40 m / s²
Explanation:
For this exercise we can use the vertical launch equations to which these are valid in any system with inertial
v = v₀ + a t
a = (v-v₀) / t
We calculate
a = (90 - 10) / 2
a = 80/2
a = 40 m / s²
This is the value of the planet's gravity acceleration
when a cup of hot chocolate cools from 90c to 80c which of the following is happening to the molecules of the liquid
Answer:
I think that the liquids molecules are slowing down. Hope this helps!
Explanation:
Please vote me Brainliest
Which of the following correctly describes which types of waves travel through which types of mediums?
A) Transverse waves travel through solids, liquids, and gases. Longitudinal waves only travel through
solids
B) Longitudinal waves travel through solids, liquids, and gases, Transverse waves only travel through
solids
C) Transverse waves travel through solids, liquids, and gases, Longitudinal waves only travel through
liquids
D) Longitudinal waves travel through solids, liquids, and gases. Transverse waves only travel through
liquids
Answer: B (Longitudinal waves travel through solids, liquids, and gases, Transverse waves only travel through solids)
Explanation: A longitudinal wave alternately compresses the medium and stretches it out. Solids, liquids and gasses all push back when they are compressed— so they are all able to store energy this way and thus transmit the wave. But, Transverse waves can only go through solids because they have enough shear strength, but liquids and gases don't. ( Transverse waves are the transfer of energy in a motion that is perpendicular to the direction the wave is traveling. Only solids are able to switch it's motion to travel through the wave.)
Answer:
b
Explanation: