A 3.0-kilogram mass is traveling in a circle of
0.20-meter radius with a speed of 2.0 meters per
second. What is its centripetal acceleration?
(1) 10. m/s
(3) 60. m/s2
(2) 20. m/s2
(4) 6.0 m/s2
Answer:
[tex]a=20\ m/s^2[/tex]
Explanation:
Given that,
The mass of an object, m = 3 kg
The radius of a circle, r = 0.2 m
The speed of the object, v = 2 m/s
We need to find the centripetal acceleration. Its formula is given by :
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{0.2}\\\\a=20\ m/s^2[/tex]
So, the centripetal acceleration is [tex]20\ m/s^2[/tex].
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man is walking due east at the rate of of 4kmph and the rain is falling 30° east of vertical with a velocity of 6kmph the velocity of rain relative to the man will be?
Answer:
No answer
Explanation:
no explanation
Integrate your expressions for dEx and dEy from θ=0 to θ=π. The results will be the x-component and y-component of the electric field at P
.
Express your answers separated by a comma in terms of some, all, or none of the variables Q
and a and the constants k and π.
Answer:
hello your question is incomplete below is the missing part
Ex = 0
Ey = [tex]\frac{-2kQ}{\pi a^2}[/tex]
Explanation:
Attached below is a detailed solution showing the integration of the expression dEx and dEy from ∅ = 0 to ∅ =π
Ex = 0
Ey = [tex]\frac{-2kQ}{\pi a^2}[/tex]
231 91 Pa has__neutrons
Answer:
140
Explanation:
A squirrel drops an acorn onto the head of an unsuspecting dog. The acorn falls 20 m before it lands on the dog. Ignore air resistance. How many seconds did the acorn fall?
Answer:
2 seconds
Explanation:
From the question,
Applying the equation of motion for a body falling under gravity
s = ut+1/2gt²................................. Equation 1
Where, s = hieght of fall, u = initial velocity, t = time, g = acceleration due to gravity.
Given: s = 20 m, u = 0 m/s, t = ?, g = 10 m/s²
Substitute these values into equation 1
10 = 0(t²)+1/2(10×t)
10 = 5t
Solve for t
5t/5 = 10/5
t = 2 seconds
a. Using the ideas of electric field and force, explain what would happen to an electron if released from rest at r=2.0m?
b. Would the electron released from rest move to a region of higher electrical potential or lower electrical potential?
c. Would the electron released from rest move such that the system would have higher potential energy or lower potential energy?
You are standing on the bottom of a lake with your torso above water. Which statement is correct?
a. You feel a buoyant force only when you momentarily jump up from the bottom of the lake.
b. There is a buoyant force that is proportional to the weight of your body below the water level.
c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.
d. There is no buoyant force on you since you are supported by the lake bottom.
Answer:
c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.
Explanation:
Buoyancy can be defined as a force which is created by the water displaced by an object.
Simply stated, buoyancy is directly proportional to the amount of water that is being displaced by an object.
Hence, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up.
The buoyancy of an object is given by the formula;
[tex] Fb = pgV [/tex]
[tex] But, \; V = Ah [/tex]
[tex] Hence, \; Fb = pgAh [/tex]
Where;
Fb = buoyant force of a liquid acting on an object.
g = acceleration due to gravity.
p = density of the liquid.
v = volume of the liquid displaced.
h = height of liquid (water) displaced by an object.
A = surface area of the floating object.
The unit of measurement for buoyancy is Newton (N).
In this scenario, you are standing on the bottom of a lake with your torso above water. Thus, there is a buoyant force that is proportional to the volume of your body that is below the level of the water.
if a body of mass m is placed on earth ,what is the amount of potential energy possessed by it (g:-9.8m/s
Answer:
mgh
Explanation:
Assume the height of the body is 1.8m.
The gravity?of the body is G=mg
the height of the gravity center is about 0.9m
E=mgh
=m*9.8m/s*0.9m
= 8.82mJ
PLEASE HELP ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!
What 2 forces would be responsible for exerting forces to the left on a cyclist that is already in motion moving to the right?
(Select 2 of the choices below)
A. gravity
B. normal force
C. air resistance
D. friction
Answer:
Friction and Air resistence
Explanation:
i already passed this grade years ago...
The forces responsible for exerting forces to the left on a cyclist are -
Air resistanceFrictionWhat is air resistance?Air resistance describes the forces that are in opposition to the relative motion of an object as it passes through the air. We can write air resistance as -
[tex]$F_{D}=\frac{1}{2} \rho v^{2} C_{D} A[/tex]
where -
F{D} = drag
{ρ} = density of fluid
{v} = speed of the object relative to the fluid
C{D} = drag coefficient
{A} = cross sectional area
Given is to find what 2 forces would be responsible for exerting forces to the left on a cyclist that is already in motion moving to the right.
The forces responsible for exerting forces to the left on a cyclist are -
Air resistanceFrictionTherefore, the forces responsible for exerting forces to the left on a cyclist are -
Air resistanceFrictionTo solve more questions on forces, visit the link -
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A airplane accelerated from 59 m/s to 95 m/s in a distance of 123 meters what was its acceleration in m/s^2, assumed constant?
Answer:
22.54 m/s^2
Explanation:
vf = final velocity = 95 m/s
vi = initial velocity = 59 m/s
d = displacement = 123 m
a = acceleration, unknown
Use this kinematics equation to find a:
vf^2 = vi^2 + 2a*d
95^2 = 59^2 + 2*a*123
22.54 m/s^2 = a
Hope this helps!! :)
Steve is planning his annual Spring Break road trip. He pulls out his map and draws out his route to visit the five locations that he has planned for this year. They go in a counterclockwise loop and he ends up at home, where he started, just in time to start classes again. Whenever he is on the road he travels a constant 60 miles/hour. When Steve adds up the total distance traveled, as measured by his odometer, and divides it by the time that his trip took, he has measured what quantity?
a. His average velocity.
b. His average speed.
c. His instantaneous velocity.
d. His instantaneous speed.
Steve’s average velocity for the whole trip is:______
a. greater than 60 miles/hour.
b. equal to 60 miles/hour.
c. less than 60 mile/hour, but greater than zero.
d. exactly zero.
Answer:
Part 1
Steve is measuring his average speed
Part 2
Average velocity is equal to 60 miles per hour
Explanation:
Part 1
Average velocity is equal to total distance travelled divided by total time taken. It also takes into consideration the change of direction through out the journey.
Hence, Steve is measuring his average speed
Option A is correct
Part 2
Average velocity is equal to 60 miles per hour only because velocity is a vector quantity
Option B is correct
reason why the center of gravity must not be at 50cm
Answer:
hope this helps
hope this is what u want
The masses of astronauts are monitored during long stays in orbit, such as when visiting a space station. The astronaut is strapped into a chair that is attached to the space station by springs and the period of oscillation of the chair in a friction-less track is measured.
(a) The period of oscillation of the 10.0 kg chair when empty is 0.750 s. What is the effective force constant of the springs?
(b) What is the mass of an astronaut who has an oscillation period of 2.00 s when in the chair?
(c) The movement of the space station should be negligible. Find the maximum displacement of the 100,000 kg sace station if the astronaut's motion has an amplitude of 0.100 m.
Answer:
a) k = 701.8 N / m, b) m_{ast} = 61.1 kg, c) v ’= -1.3 10⁻⁴ m / s
Explanation:
a) For this exercise let's use the relationship of the angular velocity
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
k = w² m
the angular velocity is related to the period
w = 2π / T
we substitute
k = 4 π² [tex]\frac{m}{T^2}[/tex]
let's calculate
k = 4 π² 10 /0.75²
k = 701.8 N / m
b) now repeat the measurement with an astronaut on the chair
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
where the mass Month the mass of the chair plus the mass of the astronaut
M = m + [tex]m_{ast}[/tex]
M = k / w²
w = 2π / T
let's calculate
w = 2π / 2
w = π rad / s
M = 701.8 /π²
M = 71,111 kg
now we use that
M = m + m_{ast}
m_{ast} = M - m
m_{ast} = 71.111 - 10.0
m_{ast} = 61.1 kg
c) if the astronaut's movement is simple harmonic
x = A cos wt
therefore the speed is
v = [tex]\frac{dx}{dt}[/tex]
v = -Aw sin wt
maximum speed is
v = - Aw
v = 0.100 π
v = 0.31416 m / s
we can suppose that the movement of the space station and the astronaut is equivalent to division of the same
initial instant. Before the move
p₀ = 0
final instant. When the astronaut is moving
p_f = M_station v’+ m_{ast} v
the moment is preserved
p₀ = pf
0 = M__{station} v ’+ m_{ast} v
v ’= - [tex]\frac{m_{ast} }{M_{station} } \ v[/tex]
we substitute
v ’= [tex]\frac{61.1 }{ 100000 } \ 0.31416[/tex]
v ’= -1.3 10⁻⁴ m / s
the negative sign indicates that the station is moving in the opposite direction from the astronaut
A 69.0 kg ice skater moving to the right with a velocity of 2.61 m/s throws a 0.22 kg snowball to the right with a velocity of 25.2 m/s relative to the ground. (a) What is the velocity of the ice skater after throwing the snowball
Answer:
0.08m/s
Explanation:
Given data
M1= 69kg
v1= 2.61m/s
M2= 0.22kg
v2= 25.2m/s
Before snowball is thrown:
Total mass of skater + snowball = 69+ 0.22 = 69.22kg
Total Momentum of skater + snowball = mv = 69.22 x 2.61 = 180.7 kgm/s
After snowball is thrown:
Let's call the velocity of the skater V.
Total momentum = momentum of skater + momentum of snowball
=69.22V + (5.544)
= 69.22V + 5.544
So:
180.7 = 69.22V+5.544
180.7- 5.544= 69.22V
175.156= 69.22V
V= 175.156/69.22
V = 2.53m/s
The total momentum after catching the snowball is mV or:
(69.0 + 0.22) x V
So:
5.544= 69.22V
V= 5.544/69.22
V=0.08m/s
The velocity of the ice skater after throwing the snowball is 0.08m/s
4. Draw conclusions: What is the minimum energy required to break the egg?
.
Answer:
0.25 J
Explanation:
The strength of the egg shell, the size of the egg, and the force used to break it are just some of the variables that affect how much energy is needed to crack an egg. When an object hits the egg with an impact energy of 12–26 mJ, cracks occur.
What is energy?The capacity of a system or object to do work is called energy, which is a fundamental term in physics. Kinetic energy, potential energy, heat energy, electromagnetic energy, and nuclear energy are just a few of the different forms of energy.
While potential energy is the energy possessed by an object as a result of its position or position, kinetic energy is the energy of motion. While electromagnetic energy is energy carried by electromagnetic waves like light, thermal energy is energy related to the temperature of a substance. The energy stored in the nucleus of an atom is called nuclear energy.
Therefore, when an object hits the egg with an impact energy of 12–26 mJ, cracks occur.
Learn more about Energy, here:
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what is the relation of pressure of a liquid with its depth and density?
Answer:
★ Pressure and depth have a directly proportional relationship. This is due to the greater column of water that pushes down on an object submerged. Conversely, as objects are lifted, and the depth decreases, the pressure is reduced.
Explanation:
Hope you have a great day :)
A 10-kg rock falls from a height of 8-m above the ground. What is the kinetic energy of the rock just before it hits the ground?
Answer: 800
Explanation:
1/2 x m x v^2 = m x g x h
KE = 10 x 10 x 8
KE= 800
The energy of the body by the virtue of its motion is known as the kinetic energy of the body. The kinetic energy of the rock just before it hits the ground will be 784.8 J.
What is kinetic energy?The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of the velocity.
According to the law of conservation of energy, energy can not be created nor be destroyed can be transferred from one form to another form.
Kinetic energy= potential energy
Kinetic energy= mgh
Kinetic energy= 10×9.81×8
Kinetic energy=784.8 J
Hence the kinetic energy of the rock just before it hits the ground will be 784.8 J.
To learn more about kinetic energy refer to the link;
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Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces off a cement floor, and in Case 2, the ball bounces off a piece of stretchy rubber. Two cases of a ball dropping and bouncing off of a surface. In case one, the ball approaches a cement floor, comes in contact, momentarily comes to rest, and then bounces off with a velocity smaller than the approach velocity. In case 2, the ball approaches a piece of stretchy rubber. When the ball comes in contact with the rubber, it deforms the rubber and comes momentarily to rest. The ball bounces off again with a velocity smaller than the approach velocity, and the rubber regains its original shape. In both cases, the balls are dropped from the same initial height and reach the same final height. In which case is the magnitude of the ball's change in momentum the greatest
Answer:
the impulse must be the same in these two cases F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
Explanation:
For this exercise we use the relationship between momentum and momentum
I = Δp
F t = m v_f - m v₀
To know the speed we use the conservation of energy
starting point. Highest point
Em₀ = U = m g h
fincla point. Just before the crash
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
we substitute in the impulse relation
F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases
A cylinder containing the air comprises the systemm. Cycle is completed as follows : (i) 82000 N-m of work is done by the piston on the air during compression stroke and 45 kJ of heat are rejected to the surroundings. (ii) During expansion stroke 100000 N-m of work is done by the air on the piston. Calculate the quantity of heat added to the system؟?
Answer & Explanation:
1 N-m = 1 Joule
So 82 kJ of energy put into the system during (i).
45 kJ of heat leaves the system, so 82 kJ - 45 kJ = 37 kJ is remaining.
(ii) requires 100 kJ of energy but only 37 kJ is available, so 100 kJ - 37 kJ = 63 kJ of heat energy must be added to the system.
The heat given would be equal to the heat emitted from the system and by providing some external source of energy the volume or temperature of the system may increase.
The amount of heat added to the system is 63kJ.
The energy can be estimated as:
Given,
Work done by piston = 82000 NmHeat rejected in surrounding = 45 kJWork done during expansion stroke = 100000 NmQuantity of added heat = ?During compression stroke:
Work done by the piston [tex]\rm (W_{1-2})[/tex] = - 82000Nm or - 82kJHeat rejected to the system [tex]\rm (Q_{1-2})[/tex] = - 45kJWe know that,
[tex]\rm Q_{1-2} = (U_{2} - U_{1}) + W[/tex]
Therefore,
[tex]\begin{aligned}-45 &= \rm (U_{2} - \rm U_{1}) + (-82)\\\\\rm (U_{2}-U_{1}) &= 37\rm (U_{2} - U_{1}) = 37\; kJ \end{aligned}[/tex] (equation 1 )
During Expansion system:
Work done by the piston [tex]\rm (W_{2-1})[/tex] = 100000 Nm or 100 kJ
Now putting values in the equation:
[tex]\begin{aligned}\rm Q_{2-1} &= \rm U_{1} - U_{2} + W\\\\&=\rm (U_{1} - U_{2}) + W\end{aligned}[/tex]
Substituting value from equation 1:
[tex]\begin{aligned}\rm Q_{2-1} &= - 37+100\rm \;kJ\\&= 63\rm \; kJ\end{aligned}[/tex]
Therefore, 63kJ of energy is added to the system.
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When talking about the products of photosynthesis, why is it a HUGE issue when a vast amount of the rainforests is dying along with vast amounts of forests from fires/deforestation? Remember, we know that this hurts the ecosystems of animals living there, but what else does it effect?
Question 5 options:
the creation of too much smoke
the creation of carbon dioxide
the creation of carbon monoxide
the creation of oxygen
Answer:
the creation of oxygen
Explanation: i just took the k12 test :)
Given that Carbon-14 has a half-life of 5700 years, determine how long it would take for
this reduction to occur.
Answer:It will take about 3000 years
Explanation:
How could being mindful in conversation be helpful?
You can see if the person your talking to and being mindfull they will see respect towards you or they see you as a mindful person.It shows a person that you care.
Explanation:
A liquid fueled rocket is red on a test stand. The rocket nozzle has an exit diameter of 30 cm and the combustion gases leave the nozzle at a velocity of 3800 m/s and a pressure of 100 kPa, which is the same as the ambient pressure. The temperature of the gases in the combustion area is 2400 C. Find (a) the temperature of the gases at the nozzle exit plane, (b) the pressure in the combustion area, and (c) the thrust developed. Assume that the gases have a speci c heat ratio of 1.3, and a molar mass of 9. Assume that the ow in the nozzle is isentropic.
Answer:
1. Temperature= 869.35 K
2. Pressure of combustion = 12994.043 kpa
3. Thrust = 127x10⁶N
Explanation:
this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.
1.
The temperature = (273+2400k) - (3800)²/2(4003)
= 2673 - 14440000/8006
= 2673 - 1803.65
= 869.35 K
Approximately 869.4K
2. We first get mach number
= 3800/√1.3(923.8)(869.35)
= 3800/1021.78
= 3.719
Pressure = 100kpa[1+2.07464415]^1.3/0.3
= 12995.043kpa
C. Thrust
Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)
= 12678.621
= 126.781 kN
Thrust is approximately 127kN = 127x10⁶N
what are the importance of regulare
health examination.
Answer:
The purpose of regular health examination is to evaluate health status, screen for risk factors and disease, and provide preventive counseling interventions. The major benefits of regular health examination is early detection of treatable disease.
A makeshift sign hangs by a wire that is extended over an ideal pulley and is wrapped around a large potted plant on the
roof as shown in the figure below. When first set up by the shopkeeper on a sunny and dry day, the sign and the pot are in
equilibrium. The mass of the sign is 27.5 kg, and the mass of the potted plant is 67.5 kg.
Plant
sale
today!
(a) Assuming the objects are in equilibrium, determine the magnitude of the static friction force experienced by the
potted plant.
N
(b) What is the maximum value of the static friction force if the coefficient of static friction between the pot and the
roof is 0.707?
N
Answer:I know the answer for B cus I’m doing the same problem. For B, you would only take the coefficient of friction given and then multiply it by the Normal Force, which in this case is the same as the Gravitational Force.
Explanation:
Mr. Voytko wants to know how high in meters he can lift an 0.3 kg apple with 7.35 joules?
Answer:
the height above the ground through Mr. Voytko lifted the apple is 2.5 m.
Explanation:
Given;
energy of Mr. Voytko, E = 7.35 J
mass of the apple, m = 0.3 kg
Apply the principle of conservation of energy.
Energy of Mr. Voytko = Potential energy of the apple due to its height above the ground.
E = mgh
where;
h is the height above the ground through Mr. Voytko lifted the apple.
g is acceleration due to gravity = 9.8 m/s²
h = E / (mg)
h = 7.35 / (0.3 x 9.8)
h = 2.5 m
Therefore, the height above the ground through Mr. Voytko lifted the apple is 2.5 m.
We assume the foam plate has a positive charge when rubbed with paper towels.
Lift the pan away from the charged plate using the styrofoam cup. Briefly touch the rim of the pan to neutralize it. Place the neutralized pan on the plate and observe the tape rise. When the pan is on the plate, the rim of the plate has a _____________. This means that the pan base is ________________ charged because the net charge on the pan is __________. You know that this must be the case because as you lift the pan with the cup away from the plate, the tape on the rim goes down.
Answer:
POSITIVE CHARGE, NEGATIVE CHARGE, ZERO
Explanation:
To solve this completion exercise, we must remember that charges of the same sign repel each other and in a metallic object (frying pan) the charge is mobile.
Let's analyze the situation when we touch the pan, the charges are neutralized, therefore when we bring the pan to the plate that has a positive charge, it attracts the mobile negative charges in the pan, until it is neutralized, therefore on the opposite side of the pan. pan (edge with a glued tape) is left with a positive charge; therefore the edge and the tape, which is very light, have positive charges and repel each other.
We must assume that the frying pan is insulated so that the net charge is zero, since the induction process.
Consequently the words to complete the sentence are
When the pan is on the plate, the edge of the plate has a _POSITIVE CHARGE_____.
This means that the base of the container is loaded NEGATIVE CHARGE_____ because the net charge of the container is ___ZERO_
What is any push or pull on an object called?
Answer:
A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.
Explanation:
A 450.0 kg roller coaster is traveling in a circle with radius 15.0m. Its speed at point A is 28.0m/s and its speed at point B is 14.0 m/s. At point A the cart is already moving with circular motion. a) Draw free bodydiagramsfor the cartatpointsAand B(two separate free body diagrams). b) Calculate the acceleration of the cartat pointsAandB(magnitude and direction). c) Calculate the magnitude of the normal force exerted by the trackson the cartat point A. d) Calculate the magnitude of the normal force exerted by the tracks on the cart at point B.
Answer:
b) a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N
Explanation:
a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle
b) Let's start at point A
Let's use that the acceleration is centripetal
a = v² / r
let's calculate
a = 28² / 15.0
a = 52.26 m / s²
as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards
Point B
a ’= 142/15
a ’= 13.06 m / s²
in this case the acceleration is vertical downwards
c) The values of the normal force
point A
let's use Newton's second law
∑ F = m a
N- W = m a
N = mg + ma
N = m (g + a)
N = 450.0 (9.8 + 52.25)
N = 2.79 10⁴ N
d) Point B
-N -W = m (-a)
N = ma -m g
N = m (a-g)
N = 450.0 (14.0 - 9.8)
N = 1.89 10³ N
