The expression for the centripetal acceleration allows to find the results for the questions are:
A) The acceleration varies INVERSELY with the radius of curvature.
B) The acceleration varies DIRECTLY with the speed.
C) The acceleration becomes four times greater.
D) The acceleration becomes nine times greater.
E) The acceleration is reduced to half.
F) The acceleration is reduced to a third.
In circular motion there must be an acceleration towards the center of the circle, it is called cenripetal acceleration, in this case all the energy supplied to the system is used to change the direction of the speed even when its magnitude remains constant.
[tex]a_c = \frac{v^2}{r}[/tex]
Where [tex]a_c[/tex] the centripetal acceleration, v is the speed and r the radius of curvature of the circle.
Now we can answer the questions about centripetal acceleration.
A) For the same speed, the acceleration varies INVERSELY with the radius of curvature.
B) For the same radius of curvature the acceleration varies DIRECTLY with the speed.
C) The speed is doubled
v = 2 v₀
[tex]a_c = \frac{(2v_o)^2 }{r}[/tex]
[tex]a_c = 4 \ \frac{v_o^2 }{r}[/tex]
The acceleration becomes four times greater than the original value
D) The speed is tripled
v = 3 v₀
[tex]a_c = 9 \frac{ v_o^2}{r}[/tex]
Acceleration becomes nine times greater than the original
E) the radius of curvature is doubled
r = 2 r₀
[tex]a_c = \frac{v_o^2}{2 r_o }[/tex]
[tex]a_c = \frac{1}{2} a_o[/tex]
Acceleration is reduced to half the original value
F) The radius of curvature is tripled
r = 3 r₀
[tex]a_c = \frac{v_o^2 }{3 r_o} \\ \\a_c = \frac{1}{3} a_o[/tex]
The acceleration is reduced to a third of the initial one.
In conclusion using the expression for the centripetal acceleration we can find the answers for the questions are:
A) The acceleration varies INVERSELY with the radius of curvature.
B) The acceleration varies DIRECTLY with the speed.
C) The acceleration becomes four times greater.
D) The acceleration becomes nine times greater.
E) The acceleration is reduced to half.
F) The acceleration is reduced to a third.
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A string is wrapped around a solid cylinder with mass M and radius R. The free end of the string is held in place, allowing the cylinder to fall. Recall that the moment of inertia of a solid cylinder rotated about its center is given by MR2/2. All answers to this problem should be symbolic, purely in terms of the variables M, R, and g. (a) Find the linear acceleration (in m/s2) of the cylinder and the tension in the string (in Newtons) as the cylinder falls. (b) Now suppose the cylinder is hollow instead of solid. The moment of inertia of a hollow cylinder rotated about its center is given by MR2. What is the acceleration and tension in this case?
Answer:
I will use (a / R) for alpha the angular acceleration
T R = I a / R torque equals angular acceleration for cylinder
M g - T = M a linear acceleration of center of mass
T = M (g - a) = I a / R^2 from first equation
If I = 1/2 M R^2 then M ( g - a) = M a / 2 from above
or g = 3 a / 2 and a = 2 g / 3
Also we have T = M (g - a) = M (g - 2 g / 3) = g / 3
Substitute I = M R^2 for the hollow cylinder
Looks like a = g/2 for hollow cylinder
The seat on a carnival ride is fixed on the end of an 12.60-m-long beam, pivoted at the other end. If the beam sweeps through an angle of 141°, what is the distance through which the rider moves?
The distance through which the rider at the end of the beam moves is;
L = 15.5 m
We are told that the beam on which the carnival ride is fixed is 12.6m in length.
Since the seat is at the end of the beam with the other end pivoted and the beam sweeps through an angle of 141°, then we can say that the radius of this arc formed by the swing is;
Radius; r = 12.6 m
Also, θ = 141°
The distance through which the driver moves will be the length of the arc formed by the beam at angle of 141°.
Formula for length of arc is given as;
L = 2πr(θ/360)
Plugging in the relevant values gives;
L = 2π × 12.6 × 141/360
L = 15.5 m
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How can I create a whirlpool of water?
Answer:
To make a small scale whirlpool of water, take a cup and fill it with water and stir. It will create a vortex that pulls particles down.
Or you could see this when draining a tub of water.
What shape is the graph produced by a force vs acceleration graph?
A. Linear
B. Quadratic
C. Circular
D. Triangular
The answer to the question What shape is the graph produced by a force vs acceleration graph is A. Linear
Since Force, F = ma where m = mass and a = acceleration. For constant mass, F ∝ a. That is, F is directly proportional to acceleration, a.
Since this is a linear relationship, the graph of force vs acceleration will be linear.
The answer to the question What shape is the graph produced by a force vs acceleration graph is A. Linear
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An astronaut standing on a platform on a foreign planet drops a hammer. If the hammer falls 9.0 meters vertically in 2.5 seconds, what is the acceleration due to gravity on that planet?
Answer: 1.646 m/s²
Explanation:
The distance that is traveled by the astronaut given that the motion is free-fall can be calculated through the equation,
d = Vot + 0.5at²
where d is the distance, Vo is the initial velocity, t is the time, and a is the acceleration. Substituting the known,
6 = (0 m/s)(2.7 s) + 0.5(a)(2.7 s)²
Determining the value of a,
a = 1.646 m/s²
in order to keep heat in or out, you need a(n)
Answer:
What is instillation
Explanation:
Which of the following statements about electromagnetic radiation it true? A.electromagnetic waves with long wavelength are more energetic then electromagnetic waves with short wavelength. B.all electromagnetic radiation carries the same amount of energy. C.electromagnetic radiation in a vacuum can change frequently to become more or less energetic. D.electromagnetic waves with high frequency are more energetic then electromagnetic waves with low frequency
Given what we know, we can confirm that option D, Electromagnetic waves with high frequency are more energetic than electromagnetic waves with low frequency is true.
Why are high-frequency waves more energetic?High-frequency waves are synonymous with short wavelengths. This means that the waves are oscillating much quicker and therefore carry more kinetic energy within them. This is transformed and released as electromagnetic radiation, which is the reason why high-frequency waves are more energetic than low-frequency electromagnetic waves.
Therefore, we can confirm that the statement "Electromagnetic waves with high frequency are more energetic than electromagnetic waves with low frequency" is true.
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A meteor falls from the sky to the Earth. The meteor already had an initial velocity downward when it was spotted. If it hit the Earth at 335 m/s after being seen for 30 seconds, then what was the initial velocity of the meteor?
Answer:
335 - 30 g
Explanation:
V = Vo + g*T = 335
Vo + 9.8*30 = 335
Vo =
Incisors (front teeth) act as wedges in the human body. Which change would increase the mechanical advantage of incisors?
1 a decrease in the distance of the teeth from the fulcrum,
2 an increase in the distance of the teeth from the fulcrum,
3 an increase in the thickness at the widest part of the teeth and no change in the length of the teeth,
4 an increase in the length of the teeth and no change in the thickness at the widest part of the teeth
To increase the mechanical advantage of incisors (front teeth) so that they act as wedges in the human body is through 2. an increase in the distance of the teeth from the fulcrum.
The decrease in the distance will decrease the mechanical advantage of incisors. The increase in the thickness at the widest part of the teeth will not necessarily support the incisors to become more effective wedges.
Thus, incisors will acquire some mechanical advantage by increasing the force that it exerts. This advantage can be achieved by increasing the distance of incisors from the fulcrum.
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Answer:
4 aka D
Explanation:
when air gets hot, it rises and forms…
Answer:
Gas
Explanation:
Hot air rises because gases expand as they heat up. When air heats up and expands, its density also decreases.
9) A policeman investigating an accident measures the skid marks left by a car. He determines that 9) the distance between the point that the driver slammed on the brakes and the point where the car came to a stop was 28.0 m. From a reference manual he determines that the coefficient of kinetic friction between the tires and the road under the prevailing conditions was 0.300. How fast was the car going wh~er applied the brakes? (This car was not equipped with anti-lock brakes.)
Answer:
Explanation:
Provided the car did not collide with anything during its slide,
Kinetic energy will convert to work of friction
½mv² = Fd
½mv² = μmgd
v = √(2μgd)
v = √(2(0.300)(9.81)(28.0))
v = 12.8377...
v = 12.8 m/s
The figure shows a jet engine suspended beneath the wing of an airplane. The weight of the engine is 14900 N and acts as shown in the figure. In flight the engine produces a thrust of 61500 N that is parallel to the ground. The rotational axis in the figure is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust.
We have that for the Question, it can be said that With respect to this axis, the magnitude of the torque due to the weight and ,the thrust is
TW=19740N-mTT=130387.39N-mFrom the question we are told
The figure shows a jet engine suspended beneath the wing of an airplane. The weight of the engine is 14900 N and acts as shown in the figure. In flight the engine produces a thrust of 61500 N that is parallel to the ground. The rotational axis in the figure is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust.
a)
Generally the equation for the Torque due to weight is mathematically given as
[tex]TW=Engine weight*2.50*sin32\\\\TW=14900*2.50*sin32[/tex]
TW=19740N-m
b)
Generally the equation for the Torque due to thrust is mathematically given as
[tex]TT=Engine thrust*2.50*cos32\\\\TT=61500*2.50*cos32\\\\[/tex]
TT=130387.39N-m
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The box is pushed to the left with 30 N of force, but does not move. What is the Friction Force?
However here coefficient of friction is not given so unable to calculate.
You can calculate it by below formula.
The formula given by
[tex]\\ \sf\longmapsto F_f=\mu N[/tex]
Or
[tex]\\ \sf\longmapsto F_f=\mu mg[/tex]
u is coefficient of friction
N is normal reaction.
Here..
box doesn't move means the force if limiting friction is greater than applied force.
A trouble-making youth is standing on a bridge, and wants to drop a water balloon on an unsuspecting passerby. A man is jogging on a path below the bridge with a constant speed of 4.2 m/s. The bridge is 11.6 m above the ground. If the balloon is to land right at the jogger's feet, at what horizontal distance x from the bridge should he be when the youth drops the balloon?
Answer:
Explanation:
Time needed for a balloon to drop from vertical rest a distance of 11.6 m
t = √(2h/g) = √(2(11.6)/9.8) = 1.538618
d = vt = 4.2(1.538618) = 6.462197...
d = 6.5 m
Horizontal distance x = 6.5 m from the bridge he should be when the youth drops the balloon with velocity 4.2 m/s.
What is velocity?When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.
Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.
Given in the question time needed for a balloon to drop from vertical rest a distance of 11.6 m
t = √(2h/g) = √(2(11.6)/9.8) = 1.538618
d = vt = 4.2(1.538618) = 6.462197...
d = 6.5 m
Horizontal distance x = 6.5 m from the bridge he should be when the youth drops the balloon with velocity 4.2 m/s.
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In these images taken a few days apart, the light part of the Moon appeared to get smaller
over time. Why did this happen? (1)
a) The Moon moved farther away from the sun so less sunlight reached the Moon's
surface.
b) Earth moved so its shadow was blocking more of the Moon so the student astronomer
was able to see less of it.
c) The Moon rotated so that less of the light-colored rock on the Moon's surface faced
Earth
d) The Moon moved so that the student astronomer was able to see less of the half that
facer the sun
Answer:
Explanation:
d) The Moon moved so that the student astronomer was able to see less of the half that faced the sun.
The lantern rises up because the __________ of the _________inside the lantern is _________ compared to the air around it
liquid
Gas
Density
fire
less dense
the same density
more dense
paper
Answer:
gass fire and the same density
Explanation:
A faulty fireworks rocket launches but never discharges. If the rocket launches
with an initial velocity of 33.0 ft/s at an angle of 85.0°, how far away from the
launch site does the rocket land?
Answer:
a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0∘ above the horizontal. During the flight, the rocket explodes into two pieces of equal mass (see Fig. 8.32). (a) What horizontal distance from the launch point will the center of mass of the two pieces be after both have landed on the ground? (b) If one piece lands a horizontal distance of 26.0 m from the launch point, where does the other piece land?
can anyone heelp me pls pls
Answer:
Explanation:
Answer in the pic.. hope it helps
1.protozoo ore ?
A. multicellular organisms B. one - Celled animals
C. amembers of the group Protocista
D. unicellulor Plants
Answer:
c
Explanation:
amembers of the group protocista
A 500-kg golf cart is moving on level ground in a circular path of radius 10 m with a speed of 12 m/s. What is the centripetal force acting on the cart?
Answer:
7200N
Explanation:
Centripetal force is directly proportional to the product of the mass and the square of the velocity and inversely proportional to the radius given.
describe an experiment to demonstrate surface tension in a liquid
Answer:
mę břöőda I dunno this answer
Explanation:
hope you find it
A baseball is hit straight up in the air with an initial velocity of 38 m/s
How long does it stay in the air?
How high does it go?
The time of motion or time spent in air by the baseball is 7.76 seconds.
The maximum height attained by the baseball is 73.67 m.
The given parameters;
initial velocity of the baseball, u = 38 m/sThe time of motion of the baseball is calculated using the first kinematic equation as follows;
v = u - gt
where;
t is the time to reach maximum heightv is the final velocity of the ball at the maximum height = 00 = u -gt
gt = u
[tex]t = \frac{u}{g} \\\\\t = \frac{38}{9.8} \\\\t = 3.88 \ s[/tex]
The time spent in the air = time of entire motion = 2(3.88 s) = 7.76 seconds
The maximum height attained by the ball is calculated using the second kinematic equation as shown below;
[tex]h = ut -\frac{1}{2} gt^2\\\\h = (38\times 3.88)\ - \ (0.5 \times 9.8 \times 3.88^2)\\\\h = 73.67 \ m[/tex]
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Bob is riding in a roller coaster, and is at the top of a vertical loop (shown below). The radius of the loop is 18 meters. If his apparent weight is just 28% of his actual weight, what is the speed of the roller coaster? Show all your steps.
Ex 1) A Major League baseball pitcher throws a baseball straight up into the air and the ball travels for 4
seconds reaching a height of 52 m before returning to the pitchers hand exactly at the height that he released
it. With what velocity does the baseball strike the pitchers hand on the way back down?
Answer:
52 / 4 = 13 m/s
Explanation:
If you traveled 50m/s for 60 seconds, how far did you travel? Remember speed=distance/time
Question options:
300 m/s
500 m/s
3,000 m/s
300 km/h
Answer:
3,000 m
Explanation:
speed = distance/time
distance = speed * time
= 50 * 60
= 3000 m
The term that is used to quantity the amount of heat in the evaporator is
Answer:
Enthalpy
The quantity of heat contained in a material as measured from a particular base or point.
Starting from rest, a student pulls a (25.0 A) kg box a distance of (4.50 B) m across the floor using a (134 C) N force applied at 35.0 degrees above horizontal. The coefficient of kinetic friction is 0.220 between the floor and the box. Find the work done by the pulling force. Give your answer in joules (J) and with 2 significant figures.
A 25.0 kg box is moved 4.50 m by a force of 134 N applied at 35.0° above horizontal. The work done by the pulling force is 4.9 × 10² J.
Starting from rest, a student pulls a 25.0 kg box a distance of 4.50 m across the floor using a 134 N force applied at 35.0 degrees above horizontal.
We can calculate the work (W) done by the pulling force using the following expression.
[tex]W = F \times d \times cos\theta = 134 N \times 4.50 m \times cos35.0\° = 4.9 \times 10^{2} J[/tex]
where,
F: puling forced: displacementθ: angle between F and dThe friction between the floor and the box. does not influence the work done by the pulling force.
A 25.0 kg box is moved 4.50 m by a force of 134 N applied at 35.0° above horizontal. The work done by the pulling force is 4.9 × 10² J.
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What happened to an enzyme’s structure as it exceeds the typical human body temperature
Enzymes. ... This is because heat energy causes more collisions, with more energy, between the enzyme molecules and other molecules. However, if the temperature gets too high, the enzyme is denatured and stops working. A common error in exams is to write that enzymes are killed at high temperatures.
A race car rounding a corner at a constant speed of 200 miles per hour.
Answer:
Yes since it is changing direction
Explanation:
A race car rounding a corner at a constant speed of 200 miles per hour
Ex 3) A tennis player lobs the tennis ball up in the air during a serve at a rate 3 m/s.
How long will it take the tennis ball to reach its maximum height?
Explanation:
s=(v^2-u^2)/2g
=(0-3^2)/2*10
= 0.45m