Answer:
Spring's displacement, x = -0.04 meters.
Explanation:
Let the spring's displacement be x.
Given the following data;
Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg
Number of shrews, n = 49
Spring constant, k = 24 N/m
We know that acceleration due to gravity, g is equal to 9.8 m/s².
To find the spring's displacement;
At equilibrium position:
Fnet = Felastic + Fg = 0
But, Felastic = -kx
Total mass, Mt = nm
Fg = -Mt = -nmg
-kx -nmg = 0
Rearranging, we have;
kx = -nmg
Making x the subject of formula, we have;
[tex] x = \frac {-nmg}{k} [/tex]
Substituting into the formula, we have;
[tex] x = \frac {-49*0.002*9.8}{24} [/tex]
[tex] x = \frac {-0.9604}{24} [/tex]
x = -0.04 m
Therefore, the spring's displacement is -0.04 meters.
Increasing the telescope diameter beyond the value found in part (a) will increase the light-gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of 10-m diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope (5-m diameter) on Palomar Mountain in California
Answer:
Ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
Explanation:
If increasing the Diameter of a Telescope beyond a given value will increase the ability of the telescope to capture more light and also capture astronomical objects located in a very distant position without improving resolution.
Hence the superiority of Keck telescope atop Mauna Kea over Hale Telescope atop Palomar mountain in California is the ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
Can someone take there time and answer this :)
Answer: I think B.)
Explanation:
A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest. It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.48 kg and 2.77 kg, and the length of the wire is 1.11 m. Find the velocity of the ball just after the collision.
Answer: Velocity of the ball just after the collision is -1.414 m/s.
Explanation:
As energy is conserved in a reaction so here, energy before collision will be equal to the energy after collision.
[tex]E_{before} = mgh = E_{after} = \frac{1}{2}mv_{o}^{2}[/tex]
where,
m = mass
g = gravitational energy = [tex]9.8 m/s^{2}[/tex]
h = height or length
[tex]v_{o}[/tex] = initial velocity
Also here, height is the length of wire. Let the height be denoted by 'L'. Therefore,
[tex]\frac{1}{2}mv_{o}^{2} = mgL\\v_{o}^{2} = 2gL\\v_{o} = \sqrt{2gL}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.11 m}\\= 4.66 m/s[/tex]
Formula used to calculate velocity after the collision is as follows.
[tex]v_{f ball} = v_{o} [\frac{m_{ball} - m_{block}}{m_{ball} + m_{block}}][/tex]
where,
[tex]v_{f ball}[/tex] = final velocity of ball after collision
[tex]m_{ball}[/tex] = masses of ball
[tex]m_{block}[/tex] = masses of block
Substitute the values into above formula as follows.
[tex]v_{f ball} = v_{o} [\frac{m_{ball} - m_{block}}{m_{ball} + m_{block}}]\\= 4.66 m/s [\frac{1.48 kg - 2.77 kg}{1.48 kg + 2.77 kg}]\\= 4.66 m/s \times (-0.303)\\= -1.414 m/s[/tex]
Thus, we can conclude that velocity of the ball just after the collision is -1.414 m/s.
describe brefly how you can charge a body
Answer:
charge or change?
Explanation:
I'll write my answer in the comments when you reply
How does the presence of a nucleus provide a method of basic cell
classification? *
Answer:
The nucleus-containing cells are called eukaryotic cells. Eukaryote means having membrane-bound organelles.
Explanation:
I hope this is what you were looking for?!
Hope this helps!
Have a great day!
-Hailey!
The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.
What is a nucleus?
The nucleus is the controlling center of the body. It performs all the major activities in the cell. The cell which possesses a nucleus is called a eukaryote. The cell which does not have a nucleus is called prokaryote.
A prokaryotic cell is a straightforward, one-celled (unicellular) organism that is devoid of a nucleus or any other organelle that is membrane-bound. The nucleoid, a dark area in the center of the cell, is where prokaryotic DNA is located.
Therefore, The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.
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Application question: In Lancaster county, Pennsylvania, it is common for members of the Amish community to use windmills to pump water from underground to fill a tank for drinking water. The wind causes the turbine blades to spin, rotating a shaft, which is transferred through some gears to operate a pump, which pumps water up from deep below the ground to fill an above ground tank. Identify the energy conversions happening at each step below.
Wind blows______causing the turbine to turn, rotating shaft works pump_____energy.
Motion of water moving up from well______energy.
Water in tank which is positioned 5 feet above the ground level______potential energy.
Answer:
Wind blows__wind energy ____causing the turbine to turn, rotating shaft works pump__mechanical___energy.
Motion of water moving up from well___kinetic energy___energy.
Water in tank which is positioned 5 feet above the ground level______potential energy
Explanation:
Wind has in it wind energy which is then used to rotate the turbine shaft which is a form of mechanical work and hence possess mechanical energy
Moving water posses kinetic energy and when this water is kept at some height it possess potential energy
When 16.35 moles of SI reacts with 11.26 moles of N2, how many moles of SI3N4 are formed
Answer:
5.45 moles
Explanation:
The chemical balanced equation of this reaction is;
3Si + 2N2 → Si3N4
From the balanced equation, we can see that that 3 moles of Si reacts with 2 moles of N2 to produce 1 mole of Si3N4.
Thus imies that the molar ration of Si to N2 is 3:2.
Now, we are told that 16.35 moles of Si reacts with 11.26 moles of N2.
16.35
Thus, using the ratio 3:2, we can say that moles of 16.35 miles of Si will react completely with (16.35 × ⅔) = 10.9 moles of N2.
Remaining N2 = 11.26 - 10.9 = 0.36 will be the excess.
From our balanced chemical equation, we saw that;
3 moles of Si produced 1 mole of Si3N4.
Thus; 16.35 moles of Si will produce;
no. of moles of Si3N4 produced = (1 × 16.35)/(3.0) = 5.45 moles
A circular cylinder has a diameter of 3.0 cm and a mass of 25 g. It floats in water with its long axis perpendicular to the water's surface. It is pushed down into the water by a small distance and released; it then bobs up and down. Part A What is the oscillation frequency
Answer:
f = 5.3 Hz
Explanation:
To solve this problem, let's find the equation that describes the process, using Newton's second law
∑ F = ma
where the acceleration is
a = [tex]\frac{d^2 y}{dt^2 }[/tex]
B- W = m \frac{d^2 y}{dt^2 }
To solve this problem we create a change in the reference system, we place the zero at the equilibrium point
B = W
In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains
B’= m [tex]\frac{d^2 y'}{dt^2 }[/tex]
the thrust is given by the Archimedes relation
B = ρ_liquid g V_liquid
the volume is
V = π r² y'
we substitute
- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }
[tex]\frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi r^2/m ) y' \ =0[/tex]
this differential equation has a solution of type
y = A cos (wt + Ф)
where
w² = ρ_liquid g π r² /m
angular velocity and frequency are related
w = 2π f
we substitute
4π² f² = ρ_liquid g π r² / m
f = [tex]\frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \ \pi r^2 \ g}{m } }[/tex]
calculate
f = [tex]\frac{1}{2 \pi } \sqrt{ \frac{ 1000 \ \pi \ 0.03^2 \ 9.8 }{0.025} }[/tex]
f = 5.3 Hz
In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 4.07 kg. They then hang the object on a pivot located 0.155 m from the object's center of mass and set it swinging at a small amplitude. As two of the team members carefully count 113 cycles of oscillation, the third member measures a duration of 247 s. What is the moment of inertia of the object with respect to its center of mass about an axis parallel to the pivot axis
Answer:
I = 0.65 kgm²
Explanation:
Since the mass is an inertial pendulum, we use the formula for the period, T of an inertial pendulum.
T = 2π√(I/mgh) where I = moment of inertia of object about pivot point, m = mass of object5 = 4.07 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.155 m.
Given that the team measures 113 cycles of oscillation in 247 s, the period, T = time of oscillations/total number of oscillations = 247 s/113 oscillations = 2.186 s/oscillation
So, T = 2.186 s
We now find I by making it subject of the formula in the equation for T.
So,
T = 2π√(I/mgh)
dividing both sides by 2π, we have
T/2π = √(I/mgh)
squaring both sides, we have
(T/2π)² = [√(I/mgh)]²
T²/4π² = I/mgh
multiplying both sides by mgh, we have
T²mgh/4π² = I
I = T²mgh/4π²
substituting the values of the variables into the equation, we have
I = T²mgh/4π²
I = (2.186 s)² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²
I = 4.778 s² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²
I = 29.539 kgm²/4π²
I = 0.748 kgm²
Now I = I' + mh² (parallel axis theorem) where I' = moment of inertia of object about its center of mass, m = mass of object = 4.07 kg and h = distance of center of mass object from pivot point.
So, I' = I - mh²
Substituting the values of the variables into the equation, we have
I' = I - mh²
I' = 0.748 kgm² - 4.07 kg × (0.155 m)²
I' = 0.748 kgm² - 4.07 kg × 0.02403 m²
I' = 0.748 kgm² - 0.098 kgm²
I = 0.65 kgm²
Urgent!!!!! A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes. The data are shown in the table. Estimate the average temperature of the air in the room at 20 min. explain your answer.
The average temperature of the air in the room at 20 min is 23°C.
What is temperature?Temperature is the degree of hotness or coldness of the object.
A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes.
The surrounding is vast, its temperature does not get affected by small amount of water. So, the temperature of air remains constant.
Thus, the average temperature of the air in the room at 20 min is 23°C.
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What happens when Earth rotates on its axis and how long does it take
Answer:
You get Day and Night
It takes 24 hour
Answer:
Explanation:
The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.
In medieval times it was believed that projectiles were pushed through the air until they reached their impetus.
a. True
b. Fals
Answer:
false
Explanation:
whem completing an emergency Roaside stop,it is necessary to put on your parking brake
A. True
B. False
Answer:
trueeeeeeee..........mmmm...........
An object changes velocity from 100m/s
to 76m/s in 6 seconds. If it has a mass of
20kg, what is the net force?
Answer:
1-1=0
Explanation:
ahriqqwertyuioo
A rabbit dashes across a field in a zig-zag pattern trying to outrun a fox. If the rabbit’s average is speed is 9.8 m/s it reaches the hole in 8.5 s what is the total distance it ran?
Answer:
83.3 m
Explanation:
From the question,
Applying on of the formula of velocity
V = d/t.................. Equation 1
Where V = average velocity, d = distance, t = time.
make d the subject of the equation
d = V×t................. Equation 2
Given: V = 9.8 m/s, t = 8.5 s
Substitute these values into equation 2
d = 9.8×8.5
d = 83.3 m
Hence the total distance covered by the rabbit is 83.3 m
When two substances that cannot dissolve each other are mixed, a ________ mixture is formed
Answer: hetero i think i dont know
Explanation:
Answer:
When two substances that cannot dissolve each other are mixed, a mixture is formed.
i hope this helps a little bit.
The law of conservation of angular momentum states that if no external force acts on an object, then its angular momentum does not change. true or false
Answer:
the answer is false.
Explanation:
i took the test and it is false trust me!!!!!!!!!
g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If a charge concentration of 25.0 C is above the plane at a height of 3000 m within the cloud and a charge concentration of -40.0 C is at height 850 m, what is the electric field at the aircraft
Answer:
[tex]523269.9\ \text{N/m}[/tex]
Explanation:
q = Charge
r = Distance
[tex]q_1=25\ \text{C}[/tex]
[tex]r_1=3000\ \text{m}[/tex]
[tex]q_2=40\ \text{C}[/tex]
[tex]r_2=850\ \text{m}[/tex]
The electric field is given by
[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]
The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]
You and a friend are playing with a Coke can that you froze so it's solid to demonstrate some ideas of Rotational Physics. First, though, you want to calculate the Rotational Kinetic Energy of the can as it rolls down a sidewalk without slipping. This means it has both linear kinetic energy and rotational kinetic energy. [The freezing only matters because if there is liquid inside, the calculation for the Moment of inertia becomes more complicated]. A Coke can can be modeled as a solid cylinder rotating about its axis through the center of the cylinder. This can has a mass of 0.33 kg and a radius of 3.20 cm. You'll need to look up the equation for the Moment of Inertia in your textbook. It is rotating with a linear velocity of 6.00 meters / second in the counter-clockwise (or positive) direction. You can use this to determine the angular velocity of the can (since it is rolling without slipping). What is the Total Kinetic Energy of the Coke can
Answer:
K_{total} = 8.91 J
Explanation:
In this exercise you are asked to find the kinetic energy of the can of coca-cola
K_total = K_ {Translation} + K_ {rotation}
the translational kinetic energy is
K_ {translation} = ½ m v²
the kinetic energy of rotation is
K_ {rotation} = ½ I w²
The moment of inertia of a cylinder is
I = ½ m r²
we substitute
K_ {total} = ½ m v² + ½ (½ m r²) w²
angular and linear velocity are related
v = w r
we substitute
K_ {total} = ½ m v² + ¼ m r² v² / r²
K_ {total} = m v² (½ + ¼)
K_ {total} = ¾ m v²
let's calculate
K_ {total} = ¾ 0.33 6.00²
K_{total} = 8.91 J
4. Speedy leaves the ground with an initial vertical velocity of 53 m/s and a horizontal velocity of 42 m/s.
How much time does he spend in the air?
How far (horizontally) does he travel during this time?
5. The Angry Bird is fired at an angle of 35 above the horizontal at a speed of 72 m/s.
Draw the initial velocity vector
Determine the initial horizontal velocity
Determine the initial vertical velocity
How much time does it spend in the air?
What horizontal distance does it go?
physics grade9 teacher guide
Answer:
huh
Explanation:
Question 2 of 20 :
Select the best answer for the question.
2. Which of the following is considered to be a vector?
A. Velocity
B. Time
C. Temperature
D. Mass
Mark for review (Will be highlighted on the review page)
The answer is A. velocity .
Velocity is a vector quantity because it need magnitude as well as direction to be defined.
A geologist notices that a river is eroding its valley at a constant rate. Knowing the height of the valley walls, how could the geologist figure out when the river started carving the valley?
A.
Count growth rings of trees growing on the valley floor.
B.
Divide the height of the valley walls by the rate of erosion.
C.
Fill up the river valley with rocks, and time how long it takes the rocks to wash out.
D.
Sit and observe the river for a few hours until the valley walls double in height.
Answer:
B.
Divide the height of the valley walls by the rate of erosion.
Explanation:
There is a relationship between the rate of erosion and the hieght at which it is eroded according to Newton's law of motion. In the case of the scenario above, the best way to determine the time the river started carving the valley would be the division of the height of the valley walls by the rate of erosion.
Carbon-14 is the typical radioisotope used to date materials; however, it has a limitation to 40,000 years. A scientist who wants to date materials older than 40,000 years would most likely use which radioisotope?
Answer:
the decay of uranium ending in lead, of potassium (40K) that becomes argon, the decay of rubidium
Explanation:
For the radioactive dating process, a material is needed that has a known average life time and that we can know the amount of material at a given moment,
In the case of carbon 14 (14C), living beings stop capturing it from the air and plants when they die, so knowing the amount they currently have, it is possible to calculate the time in which they stopped absorbing, but the life time average is 5730 years, the maximum time that can be used is up to about 10 average visa times
To analyze extra samples have high half-life times
* the decay of uranium ending in lead
* the decay of potassium (40K) that becomes argon T1 / 2 = 1,251 10⁹ years
* the decay of rubidium (87Ru) which becomes strontium T1 / 2 = 4.92 10¹⁰ years
Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellite X has an orbital radius of 3R, and satellite Y has an orbital radius of 4R. The kinetic energy of satellite X is Kx . Satellite X is moved to the same orbit as satellite Y by a force doing work on the satellite. In terms of Kx , the work done on satellite X by the force is
Answer:
The work down on satellite X by the force in terms of Kx is [tex]\dfrac{-K_x}{4}[/tex].
Explanation:
The work done is given as in terms of
[tex]W=\Delta TE[/tex]
Where ΔTE is the change in total energy.
This is given as
[tex]W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}[/tex]
Rearranging it in terms of K_x gives
[tex]W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}[/tex]
An ideal horizontal spring-mass system has a mass of 1.0 kg and a spring with constant 78 N/m. It oscillates with a period of 0.71 seconds. When this same spring-mass system oscillates vertically instead, the period is _______ seconds. Enter 2 significant figures (a total of three digits) and use g = 10.0 m/s2 if necessary.
Answer:
T = 0.71 seconds
Explanation:
Given data:
mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.
We have to calculate time period when this same spring-mass system oscillates vertically.
As we know
[tex]T = 2\pi \sqrt{\frac{m}{K} }[/tex]
This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.
Therefore, T = 0.71 seconds
If the acceleration of the body is towards the center, what is the direction of the unbalanced force ? Using a complete sentence , describe the direction of the net force that causes the body to travel in a circle at a constant speed.
Accelerating objects are changing their velocity. Velocity is often thought of as an object's speed with a direction. Thus, objects which are accelerating are either changing their speed or changing their direction. They are either speeding up, slowing down or changing directions. Changing the velocity in any one of these three ways would be an example of an accelerated motion.
A wheel has eight spokes and a radius of 30 cm. It is mounted on a fixed axle and is spinning at 2.5 rev/s. You want to shoot a 24- cm arrow parallel to this axle and through the wheel without hitting any of the spokes Assume that the arrow and the spokes hitting any of the spokes. Assume that the arrow and the spokes are very thin. (a) What minimum speed must the arrow have to pass through (a) What minimum speed must the arrow have to pass through without contact
Answer:
4.8 m/s
Explanation:
Given: angular velocity of wheel ω = 2.5 rev/sec
radius r = 30 cm
length of arrow = 24 cm
For arrow to pass through spinning ring it has to pass between any two spokes of the wheel.
angle between two spokes = π/4
time taken by a spook to reach the position of adjacent spoke t =θ/ω
= π/4/(2.5×2π) = 1/20 sec
for the arrow to pass through the spokes of the wheel it should take time t <1/20 sec to pass through the wheel
a) therefore, minimum speed = (24/100)/(1/20) = 4.8 m/s
The unit of work done is called derived unit why
A doorknob is a type of wheel and axle. In a doorknob, the door handle acts as the wheel. The handle is connected to a cylinder, called a spindle, which acts as the axle. When the handle turns, the spindle turns in the same direction. The spindle is located inside the handle and pulls the latch back and forth, allowing the door to open.
In the doorknob shown above, when the handle is rotated a distance of 84 millimeters, the spindle is rotated a distance of 14 millimeters. What is the mechanical advantage of this doorknob?
A. 504
B. 6
C. 84
D. 14