To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant angular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.8 m . Part A If the thrower takes 1.2 s to complete one revolution, starting from rest, what will be the speed of the discus at release
Answer:
9.42 m/s
Explanation:
a) Using Newton's law of motion formula:
[tex]\theta=\frac{(\omega+\omega_o)}{2}t\\\\where \ \theta=angular\ displacement=1\ rev =2\pi, w_o=initial\ velocity\ of\ discus\\=0\ rad/s, \omega=angular\ speed\ of\ discus\ at\ release,t=time\ = 1.2\ s.\\\\Hence:\\\\2\pi=\frac{(0+\omega)}{2}(1.2)\\\\\omega=\frac{2*2\pi}{1.2} \\\\\omega=10.47\ rad/s\\[/tex]
The speed of the discus at release (v) is:
v = ωr; where r = radius of discus
diameter = 1.8 m, r = diameter / 2= 1.6 / 2 = 0.9 m
v = ωr = 10.47 * 0.9
v = 9.42 m/s
What is the acceleration when a force of 2 N is applied to a ball that has a mass of 0.60 kg?
Answer:
3.333
Explanation:
Acceleration is force divided by mass. So divide the force, 2, by the mass, 0.60, and you will get 3.333. I hope this helps :)
The acceleration of an object is the force divided by time. The acceleration of the ball of 0.60 kg when a force of 2N is applied, is 3.33 m/s².
What is acceleration ?Acceleration of an object is the measure of rate of change in its velocity. Like velocity acceleration is a vector quantity having both magnitude and direction.
Acceleration is defined as the ratio of change in velocity to the change in time. However, according to Newton's second law of motion fore applied on an object is the product of its mass and acceleration.
Hence, F = m a .
Given that, force applied on the ball = 2 N
mass of the ball = 0.60 Kg.
Then acceleration a = force/mass
a = 2 N/ 0.60 Kg = 3.33 m/s²
Therefore, the acceleration of the ball is 3.33 m/s².
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Which hormone do ovaries release?
A. estrogen
B. glucagon
C. insulin
D. testosterone
Answer:
A. estrogen
Explanation:
This is released in the female reproductive organ.
A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.76 rad/sec. The moment of inertia of the student plus the stool is 5 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.33 m from the rotation axis.
Required:
a. Find the new angular speed of the student.
b. Find the kinetic energy of the student before and after the objects are pulled in.
Answer:
a) the new angular speed of the student is 0.9642 rad/s
b)
the kinetic energy of the student before the objects are pulled in is 1.9119 J
the kinetic energy of the student after the objects are pulled in is 2.4252 J
Explanation:
Given that;
mass of each object m = 1 kg
distance of objects from axis of rotation r = 0.9 m
Moment of inertia of each object initially [tex]I_{oi}[/tex]
[tex]I_{oi}[/tex] = mr² = 1kg ×(0.9m)² = 1 kg × 0.81 m² = 0.81 kg.m²
moment of inertia of each object finally [tex]I_{of}[/tex]
[tex]I_{of}[/tex] = mr² = 1kg × (0.33 m)² = 0.1089 kg.m²
Now
moment of inertia of student plus stool [tex]I_{}[/tex] = 5 kg.m²
initial angular speed ω₀ = 0.76 rad/sec
final angular speed ω = ?
Now using conservation of angular momentum;
([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀ = ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω
so we substitute
(5 + 2 (0.81) )0.76 = (5 + 2 (0.1089) )ω
5.0312 = 5.2178 ω
ω = 5.0312 / 5.2178
ω = 0.9642 rad/s
Therefore, the new angular speed of the student is 0.9642 rad/s
b)
K.E of student before = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀²
= (0.5) (5 + 2 (0.81) )(0.76)²
= 0.5 × 6.62 × 0.5776
= 1.9119 J
Therefore, the kinetic energy of the student before the objects are pulled in is 1.9119 J
KE of student finally = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω²
= (0.5) (5 + 2 (0.1089) ) (0.9642)²
= 0.5 × 5.2178 × 0.9296
= 2.4252 J
Therefore, the kinetic energy of the student after the objects are pulled in is 2.4252 J
Two masses are suspended by cord that passes over a pulley with negligible mass. The cord also has negligible mass. One of the masses, m1, has a mass of 7.0 kg and the other mass, m2, has a mass of 3.0 kg. The pulley turns on a shaft through the center of the pulley and supports the pulley and all the masses. The vertical force of the shaft on the pulley that supports the whole system is
Answer: F = 98N
Explanation: The shaft have to sustain the pulley, the cord and the two masses. The pulley and the cord have negligible masses, so, they have negligible weight.
The two masses have two vertical forces acting on them: force of traction because of the cord and force due to gravitational force, also known as weight.
So, the vertical force the shaft has to support is the sum of the weight of each mass:
[tex]F_{net}=F_{g}_{1}+F_{g}_{2}[/tex]
[tex]F_{net}=m_{1}.g+m_{2}.g[/tex]
[tex]F_{net}=g(m_{1}+m_{2})[/tex]
[tex]F_{net}=9.8(7+3)[/tex]
[tex]F_{net}=[/tex] 98
The vertical force that supports the whole system is 98 N.
When you jump, you push down on the earth and it pushes back up against you. The earth pushing up against you is what causes you to go into the air. Why doesn’t your push cause the earth to go down if your push on the earth is equal and opposite of the earth's push on you?
That's a great question !
The answer is: It does !
A push on an object causes the object to accelerate in the direction of the force.
The less mass the object has, the more the force accelerates it.
Now, when you jump, the forces on you and on the Earth are equal forces.
The up force on you causes you to accelerate up by some amount.
The down force on the Earth causes the Earth to accelerate down by some amount.
The Earth's mass is something like 5,972,000,000,000,000,000,000,000 kg, while your mass is something like 50 kg.
The Earth has something like 119,400,000,000,000,000,000,000 times as much mass as you have.
So your acceleration is something like 119,400,000,000,000,000,000,000 times as great as the Earth's acceleration.
==> The Earth's downward acceleration, caused by your jump, is there. It's just too small to notice.
BUT . . . That's the reason why seismometers (instruments to detect and measure the vibrations from distant earthquakes) have to be located as far as possible from cities and busy roads.
In places that are too close to cities and roads, the Earth's surface is always vibrating, wiggling, jiggling, heaving and weaving, in reaction to the forces of people walking around, cars and trucks driving around, even rain falling down. And kids jumping up and down !
In such places, these people-motions are louder and stronger than the vibrations coming from distant earthquakes. Seismometers wouldn't work there.
Which three statements are true of all matter?
A.
It is filled with air.
B.
It takes up space.
C.
It contains aluminum.
D.
It has mass.
E.
It is made up of atoms
Answer:
B, D and E, not all matter can be filled with air
First to answer gets brainliest
Answer:
Sodium (K)
Explanation:
You push a box across the floor at a constant speed of 1 m/s, applying a horizontal constant force of magnitude 20 N. Your friend pushes the same box across the same floor at a constant speed of 2 m/s, applying a horizontal force. What is the magnitude of the force that your friend applies to the box
Answer:
the force your friend applied on the box is 40 N.
Explanation:
Given;
speed of the box, v₁ = 1 m/s
force applied to the box, F₁ = 20 N
the speed of the box when your friend pushes it, v₂ = 2 m/s
then your friends applied force, F₂ = ?
Assuming the time, t, through which both forces were applied and mass of the box, m, to be constant;
[tex]F_1 = \frac{mv_1}{t} \\\\\frac{m}{t} = \frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
[tex]F_2 = \frac{F_1v_2}{v_1} \\\\F_2 = \frac{20\times 2}{1} \\\\F_2 = 40 \ N[/tex]
Therefore, the force your friend applied on the box is 40 N.
Explain how you could use iron filings and a piece of paper to help reveal the effect of a magnetic field.
Answer:
you could put the iron filings on the peace of paper and hover a magnet over top of the paper and the iron filings would stand up, or even stick to the magnet
Explanation:
A characteristic of a nebula is that it-
Answer:
Center of solar system
Explanation:
Answer: b
Explanation:
Heeeeeeeeelp please
OK please your picture not perfect please try again
A point charge of -11 [Coulombs] is placed inside a spherical conducting shell with net charge of 5 [Coulombs]. Calculate the net charge on the outer surface of the conducting shell. Enter your answer without units (example 100 for 100 [Coulombs] or -100 for -100 [Coulombs] ).
Answer:
20 C
Explanation:
To do this, is pretty easy, we just need to do a little reasoning of what is happening.
When any charge called q is placed inside this metallic shell which is spheric, all the opposite and even equal charges are induced on the inner and outer surface of the shell. Hence, we can say that if in the inner shell we have +q, in the outer will be -q.
Now, here we have the shell with 5 C, and when the charge of -11 C is placed inside the shell we can have the following changes on the inner surface and the outer surface:
Inner surface: +11 C
Outer surface: 9 + 11 = 20 C
Net charge on the outer surface: 20 CHope this helps
The cart is given an initial push up the ramp. After this push, as the car moves up the ramp, the direction of the acceleration of the cart is ________ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _______.
Answer:
Explanation:
The only force acting on the cart is a component of its weight parallel to ramp downwards . No other force acts parallel to the ramp .
Even when the cart is moving up after the initial push , its weight is acting downwards so acceleration is acting downwards .
When the cart is stationary at the top position , its weight is acting downwards so acceleration is downwards at that moment also . When the cart is going downwards , still its weight is acting down so acceleration is acting downwards .
After this push, as the car moves up the ramp, the direction of the acceleration of the cart is _down _______ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is _down ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _downwards ______.
The equilibrant is the equal to the resultant magnitude but opposite in direction.
True
False
Answer and I will give you brainiliest
Answer:
The answer is False......
Answer:
true
it is equal but opposite
Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, as if they had been migrating. For years curiosity mounted about why the stones moved. One explanation was that strong winds during the occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about 0.80. What horizontal force is needed on a 25 kg stone (a typical mass) to maintain the stone's motion once a gust has started it moving
Answer:
F = 196 N
Explanation:
For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically
Y axis
N- W = 0
N = mg
X axis
F -fr = ma
In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.
F- fr = 0
F = fr
the friction force has the equation
fr = μ N
fr = μ mg
we substitute
F = μ mg
let's calculate
F = 0.80 9.8 25
F = 196 N
How would you compare the acceleration between the unbalanced net force of 100 N and of 50 N
Answer:
The acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.
Explanation:
Given;
first net force, F₁ = 100 N
second net force, F₂ = 50 N
If we consider equal mass for the two net forces, and apply Newton's second law of motion, the acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.
Let a₁ be the acceleration produced by the first net force
then, a₂ be the acceleration produced by the second net force
Thus, a₁ = 2a₂
You are working in a shoe test laboratory measuring the coefficients of friction for running shoes on a variety of surfaces. The shoes are pushed against the surface with a downward force of 400 N, and a sample of the surface material is then pulled out from under the shoe by a machine. The machine pulls with a force of 300 N before the material begins to slide. When the material is sliding, the machine only has to pull with a force of 200 N to keep the material moving. What is the coefficient of static friction between the shoe and the material
Answer:
0.75
Explanation:
Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.
Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).
So, μ = F/N
= 300 N/400 N
= 3/4
= 0.75
So, the coefficient of static friction μ = 0.75
Electric power is to be generated by installing a hydraulic turbine-generator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 2300 kg/s steadily. Determine the power generation potential.
Answer:
the power generation potential is 2.705 x 10⁶ J/s.
Explanation:
Given;
height below the free surface of a large water reservoir, h = 120 m
mass flow rate of the water, m' = 2300 kg/s
The power generation potential is calculated as;
[tex]Power = \frac{Energy}{time} = \frac{F\times h}{t} = \frac{(mg) \times h}{t} = \frac{m}{t}\times gh = m' \times gh\\\\Power = m' \times gh\\\\Power = 2300 \ kg/s \ \times \ 9.8 \ m/s^2 \ \times \ 120 \ m\\\\Power = 2.705 \times 10^6 \ J/s[/tex]
Therefore, the power generation potential is 2.705 x 10⁶ J/s.
A Car is moving at a speed of 20 m/s. How Much Distance it will cover in 1 min? Express the answer in km.
Answer:
d=20m/sx60s=1200m=1200/1000Km=1.2km
Explanation:
6. A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at
30.00°C). The final temperature of the system is 40.22°C. Find the specific heat of iron.
90
Answer:
[tex]c=0.45\ J/g^{\circ} C[/tex]
Explanation:
Given that,
A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at 30.00°C). The final temperature of the system is 40.22°C.
We need to find the specific heat of iron.
It can be calculated as:
Cooler water gains = hot metal loses
mc∆T = - mc∆T
Put all the values,
[tex]200g(4.184\ J/g^{\circ} C)(T_f-T_i) = -25g(c)(T_f-T_i) \\\\200g(4.184 )( 40.22-30.00) = -25\times (c)\times (40.22-800.00)\\\\8552.096 = 18994.5c\\\\c=\dfrac{8552.096 }{18994.5}\\\\c=0.45\ J/g^{\circ} C[/tex]
So, the specific heat of iron is [tex]0.45\ J/g^{\circ} C[/tex]
1.0 kg clay ball traveling straight down at -10 m/s hits the floor and and sticks on it
Answer:
What am I suppose to solve
Explanation:
A circuit has 12 Amps and 220 Volts. What is the Resistance of the circuit?
Answer:
:To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)
To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)
To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)
To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)
True or false it is impossible to determine weather you are moving unless you can touch another object
Answer: false
Explanation:
Answer:
false
Explanation:
According to Newton's First Law of motion, an object remains in the same state of motion unless a resultant force acts on it.
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta leading away from the heart. Since blood within the heart is essentially stationary, this pressure difference can be inferred from a measurement of the speed of blood flow in the aorta. Take the speed of sound in stationary blood to be c.
a. Sound sent by a transmitter placed directly inline with the aorta will be reflected back to a receiver and show a frequency shift with each heartbeat. If the maximum speed of blood in the aorta is v, what frequency will the receiver detect? Note that you cannot simply use the textbook Doppler Shift formula because the detector is the same device as the source, receiving sound after reflection.
b. Show that in the limit of low blood velocity (v <
f= 2fo v/c
Answer:
a) f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex] , b) Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo[tex]\frac{c+v}{c}[/tex]
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’ [tex]\frac{c}{ c-v}[/tex]
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀ [tex]\frac{c}{c-v}[/tex]
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex]
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
[tex]( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}[/tex]
f ’’ = fo [tex]( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )[/tex]
f ’’ = fo [tex]( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )[/tex]
leave the linear term
f ’’ = f₀ + f₀ 2[tex]\frac{v}{c}[/tex]
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
Why do people eat bo oty
Answer: I don't know my dude
Explanation:
Determine the gravitational potential energy, in kJ, of 1 m3 of liquid water at an elevation of 30 m above the surface of Earth. The acceleration of gravity is constant at 9.7 m/s2 and the density of the water is uniform at 1000 kg/m3. Determine the change in gravitational potential energy if the elevation decreases by 10 m.
Answer:
Explanation:
Gravitational potential energy = mgh where m is mass , g is acceleration due to gravity and h is height from the ground .
In the first case mass m = volume x density
= 1 x 1000 = 1000 kg
height h = 30 m
potential energy = 1000 x 30 x 9.8 = 294000 J = 294 kJ .
When height decreases by 10 m , potential decreases as follows .
Decrease in potential energy
= mass x gravitational energy x decrease in height
= 1000 x 9.8 x 10
= 98000 J
= 98 kJ .
keli learned that an air mass is a very large body of air with similar temperature humidity and pressure and the air mass are constantly in motion she knows that you're messing depending on the temperature and moisture content tent of region where they form she looked up more information about what makes them move what are the major causes for moving & Masten North America choose two that apply.
Answer choices
A. changing humidity
B. low temperature
C. jet storm
D. prevailing westerlies
Air masses from the tropics and the equator are warm as they form over lower latitudes. The major causes for moving air masses North America exists jet storm.
What is meant by air mass?An air mass is a volume of air that in meteorology is identified by its temperature and humidity. Many hundreds or thousands of square miles are covered by air masses, which adjust to the properties of the land underneath them. Latitude and their continental or maritime source regions are used to categories them.
Warmer air masses are referred to as tropical, whilst colder air masses are referred to as polar or arctic. Superior and maritime air masses are moist, whereas continental and superior air masses are dry. Air masses with various densities are divided by weather fronts. Once an air mass has left its original location, nearby plants and bodies of water can quickly change the way it behaves. Classification systems address both the properties and modification of an air mass.
Air masses from the tropics and the equator are warm as they form over lower latitudes. They move poleward along the southern edge of the subtropical ridge and are drier and hotter than those that originate over seas. Trade air masses are another name for tropical maritime air masses. The Caribbean Sea, southern Gulf of Mexico, and tropical Atlantic Oceans, east of Florida via the Bahamas, are the origins of maritime tropical air masses that have an impact on the United States.
Monsoon air masses are moist and unstable. Rarely do dry superior air masses touch the ground. A trade wind inversion, which is a warmer and drier layer over the more moderately moist air mass below, is typically created over maritime tropical air masses when they are located above them.
Therefore, the correct answer is option C. jet storm.
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Which statement applies only to magnetic force instead of both electric and magnetic forces? O A. It acts between a north pole and a south pole. O B. It can push objects apart. O C. It can pull objects together. D. It acts between objects that do not touch.
Answer:
the answer would be A. electricity don't specify the direction of any cardinal points the flow of charges moves.
Answer:
A
Explanation:
I did the test on ap3x
An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA. The current density is uniformly distributed through the beam. The kinetic energy of each proton is 20 MeV. The beam strikes a metal target and is absorbed by the target. (a) What is the number density of the protons in the beam
Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = [tex]\frac{1}{2}[/tex][tex]m_{p}[/tex]v²
v = √( 2K / [tex]m_{p}[/tex] )
lets relate the cross-sectional area A of the beam to its diameter D;
A = [tex]\frac{1}{4}[/tex]πD²
now, we substitute for v and A
n = I / [tex]\frac{1}{4}[/tex]πeD² ×√( 2K / [tex]m_{p}[/tex] )
n = 4I/π eD² × √([tex]m_{p}[/tex] / 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³