Answer:
Explanation:
300000(9.8) = ω²(0.0160)
ω = 13555 rad/s
13555 rad/s (60 s/min/ 2π rad/rev) = 129,445 rev/min
what is the relationship between work and mechanical advantage
As shown in the figure below, a uniform solid sphere rotates about a vertical support on a frictionless bearing. A light cord passes around the equator of the sphere, over a uniform solid disk, and is attached to a hanging mass. The sphere, disk and hanging mass all have equal mass M. The sphere and disk have the same radius R. Find the acceleration of the hanging mass if the string does not slip on the sphere or the disk. Express your answer in terms of M, R and g as needed...
Answer:
As shown in the figure below, a uniform solid sphere rotates about a vertical support on a frictionless bearing. A light cord passes around the equator of the sphere, over a uniform solid disk, and is attached to a hanging mass. The sphere, disk and hanging mass all have equal mass M. The sphere and disk have the same radius R. Find the acceleration of the hanging mass if the string does not slip on the sphere or the disk. Express your answer in terms of M, R and g as needed...
Explanation:
As shown in the figure below, a uniform solid sphere rotates about a vertical support on a frictionless bearing. A light cord passes around the equator of the sphere, over a uniform solid disk, and is attached to a hanging mass. The sphere, disk and hanging mass all have equal mass M. The sphere and disk have the same radius R. Find the acceleration of the hanging mass if the string does not slip on the sphere or the disk. Express your answer in terms of M, R and g as needed...
[tex] \boxed{ \sf \: better \: luck \: at \: your \: class}[/tex]
A uniform solid sphere revolves around a vertical support on a frictionless bearing.
What is Solid sphere?A solid disk is covered with a light string that circles the sphere's equator, over which is suspended a mass. The mass M of the hanging mass, disk, and sphere is the same.
The radius R of the sphere and disk is the same. If the string does not slip on the sphere or disk, determine the acceleration of the hanging mass. If necessary, explain your response in terms of M, R, and g.
A uniform solid sphere revolves around a vertical support on a frictionless bearing, as seen in the illustration below. Around the sphere's equator, a thin cable travels over a solid surface.
Therefore, A uniform solid sphere revolves around a vertical support on a frictionless bearing.
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An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. Find its density.
Answer:
Weight of object = 11.2 N
Apparent weight = 3.83 N when immersed
Weight of water displaced = 11.2 - 3.83 = 7.37 N
d (density) W / V weight / volume the weight density
Wo = Vo do weight of object
Ww = Vo dw where Ww is weight of equivalent volume of water = 7.37
Wo / Ww = do / dw dividing previous equations
do = 11.2 / 7.37 dw = 1.52 dw
The density of the object is 1.52 that of water
The density of water is 1000 kg / m^3 * 9.8 m/s^2 = 9800 N/m^3
So the weight density is 14900 N/m^3
An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. It's density can be calculated as 1523 kg/m³.
To find the density, the given values are,
Weight in air = 11.20 N
Weight in water = 3.83 N
density of water = 1000 kg/m³
What is meant by Density?According to the Archimedes principle, when a body is immersed in a liquid partly or wholly, it experiences an upward force which is called buoyant force. The buoyant force is equal to the loss in weight of the body.
Loss in weight of the object = Weight of object in air - weight of object in water
Loss in weight = 11.20 - 3.83 = 7.37 N
Volume of body x density of water x g = 7.37
Let V be the volume of body
V x 1000 x 9.8 =7.37
V = 7.5× 10⁻⁴ m³
Weight in air = Volume of body x density of body x g
11.20 = 7.5× 10⁻⁴ x d x 9.8
d = 1523 kg/m³.
Thus, the density of body is 1523 kg/m³.
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PLEASE HELP FOR PHYSICS!
All objects exert a gravitational force on all other objects. This force is given by, F = GMm r2 , where the value of G = 6.673 × 10–11 N–m2/kg2 , M is the mass of the heavier object, m is the mass of the lighter object, and r is the distance between the two objects.
What is the force of gravity between two balls of mass 50 kg each if the distance between them is 25 m. Assume that there is no interference from any other gravitational field.
Hi there!
Recall Newton's Law of Universal Gravitation:
[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]
Where:
Fg = Force of gravity (N)
G = Gravitational Constant
m1, m2 = masses of objects (kg)
r = distance between objects (m)
Plug in the given values stated in the problem:
[tex]F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}[/tex]
A 5.0 m length of rope, with a mass of 0.52 kg, is pulled taut with a tension of 46 N. Find the speed of waves on the rope
Answer:
Speed of waves on the rope is 21 m/s
Explanation:
Length of the rope (l) = 5.0 m
Mass of the rope (m) = 0.52 kg
Tension in the rope (T) = 46 N
Formula of speed of waves on the rope:
[tex] \bold{v = \sqrt{\dfrac{T}{\mu}}} [/tex]
[tex] \mu [/tex] = Mass per unit length of the rope (m/l)
By substituting the values in the formula we get:
[tex] \implies \rm v = \sqrt{\dfrac{T}{ \dfrac{m}{l} }} \\ \\ \implies \rm v = \sqrt{\dfrac{Tl}{m}} \\ \\ \implies \rm v = \sqrt{ \dfrac{46 \times 5}{0.52} } \\ \\ \implies \rm v = \sqrt{ \dfrac{230}{0.52} } \\ \\ \implies \rm v = \sqrt{442.3} \\ \\ \implies \rm v = 21 \: m {s}^{ - 1} [/tex]
Speed of waves on the rope (v) = 21 m/s
A 1.95 kg box sits on an incline of 24 ° with the horizontal. If the box accelerates down the incline at 0.245 m/s 2 , what is the coefficient of kinetic friction between the box and the inclined plane?
Answer:
Explanation:
F = ma
mgsinθ - μmgcosθ = ma
gsinθ - μgcosθ = a
μgcosθ = gsinθ - a
μ = (gsinθ - a) / gcosθ
μ = (9.81sin24 - 0.245) / 9.81cos24
μ = 0.4178906...
μ = 0.418
The coefficient of kinetic friction will be equal to 0.418.
What is friction?Friction is the force that prevents solid surfaces, fluid layers, and material elements from sliding against each other. There are various kinds of friction: Dry friction is the force that opposes the relative lateral motion of two in-touch solid surfaces.
Given that a 1.95 kg box sits on an incline of 24 ° with the horizontal. If the box accelerates down the incline at 0.245 m/s 2
The coefficient of kinetic friction will be calculated as,
F = ma
mgsinθ - μmgcosθ = ma
gsinθ - μgcosθ = a
μgcosθ = gsinθ - a
Solve for the value of the coefficient of friction,
μ = (gsinθ - a) / gcosθ
μ = (9.81sin24 - 0.245) / 9.81cos24
μ = 0.4178906...
μ = 0.418
Therefore, the coefficient of kinetic friction will be equal to 0.418.
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A car starting from rest has a constant acceleration of 5 m/s2. How long will it take to reach a speed of 20 m/s
Answer:
Explanation:
v = u + at
20 = 0 + 5t
t = 20/5
t = 4 s
Question No. 1 Marks = = 5 +5 +2 = 12
The driver of a 2.0 × 103 kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car in front of him, which had come to rest because of blocking ahead as shown in above Fig. After the brakes are applied, a constant friction force of 7.5 × 103 N acts on the car. Ignore air resistance.
(a) Determine the least distance should the brakes be applied to avoid a collision with the other vehicle?
(b) If the distance between the vehicles is initially only 40.0 m, at what speed would the collision occur?
(c) Write your conclusive observations on the result obtained from this numerical. i.e. the importance of Physics in daily life.
Question No. 2 Marks = 8
Consider an automobile moving at v mph that skids d feet after its brakes lock. Calculate how far it would skid if it was moving at 2v and the brakes were locked.
The kinematics and Newton's second law we can find the results for the questions about the braking movement of the car are;
Question 1.
a) The stopping distance is: x = 270 m
b) The initial velocity is: v₀ = 17.3 m / s
c) Concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.
Question 2.
The stopping distance is: x = 4d
Given parameters
Mass of the red carriage m1 = 2,0 10³ kg Red car speed vo = 45 m / s Friction force fr = 7.5 10³ N.To find
Question 1.
a) Minimum braking distance.
b) If the distance is x = 40.0 m, what speed should the vehicles have?
c) Conclusive importance of physics in daily life.
Question 2.
The distace to stop.
Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.
v² = v₀² - a2 x
Where v and v₀ are the current and initial velocity, respectively, at acceleration and x the distance traveled.
Newton's second law states that the net force is proportional to the mass and the acceleration of the body.
F = ma
Where F is force, m is mass and acceleration.
In the attachment we see a diagram of the forces in the system. Let's look for the acceleration of the body
fr = m a
a =[tex]\frac{fr}{m}[/tex]
a = [tex]\frac{7.5 \ 10^3}{2.0 \ 10^3 }[/tex]
a = 3.75 m / s²
This acceleration is in the opposite direction to the speed.
Let's find the distance needed to stop, the final speed is zero.
0 = v₀² - 2 ax
x = [tex]\frac{v_o^2 }{ 2a}[/tex]
Let's calculate.
x = [tex]\frac{45^2 }{2 3.75}[/tex]
x = 270 m
This is the minimum distance that the two vehicles must separate to avoid a collision.
b) We look for speed.
v₀ = [tex]\sqrt{2ax}[/tex]
v₀ = [tex]\sqrt{2 \ 3.75 \ 40.0}[/tex]
v₀ = 17.3 m / s
c) The concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.
2) They indicate that the initial velocity is v and the distance traveled to stop is d, let's find the acceleration.
0 = v₀² - 2ax
Let's substitute.
a = [tex]\frac{v^2}{2d}[/tex]
They ask the distance traveled if this car traveled from an initial speed 2v.
0 = v² - 2 a x
x = [tex]\frac{v^2}{2a}[/tex]
We substitute
x = [tex]\frac{(2v)^2 }{2} \ (\frac{2d}{v^2})[/tex]
x = 4 d
In conclusion, using the kinematic relations and Newton's second law we can find the results for the questions about the braking movement of the car are;
Question 1
a) The stopping distance is: x = 270 m
b) The initial velocity is: v₀ = 17.3 m / s
c) concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.
Question 2.
The stopping distance is x = 4d
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Explain why the temperature is not changing at X
Answer:
The temperature is constant at 'X'. No increase or decrease. Would be same answer if there were a 'X' at temperature 'D'.
Slope of graph is zero so no change in the temperature.
What is temperature?
Temperature is a physical quantity that expresses quantitatively the perceptions of hotness and coldness. Temperature is measured with a thermometer.
Given graph of temperature and time, at position x there in no change in temperature with time so temperature is not changing.
Slope of graph is zero so no change in the temperature.
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Please respect! Can you answer this for me?
Answer:
ıf u find the answer can u comment me to notification thank u
Explanation:
A block of mass m = 3.0 kg is pushed a distance d = 2.0 m along a frictionless horizontal table by
a constant applied force of magnitude F= 20.0 N directed at an angle 0= 30.0° below the horizontal
as shown in Figure. Determine the work done by (a) the applied force, (b) the normal force exerted
by the table, and (d) the net force on the block.
Explanation:
We apply the definition of work by a constant force in the first three parts, but then in the fourth part we add up the answers. The total (net) work is the sum of the amounts of work done by the individual forces, and is the work done by the total (net) force. This identification is not represented by an equation in the chapter text, but is something you know by thinking about it, without relying on an equation in a list.
The definition of work by a constant force is W=FΔrcosθ.
(a) The applied force does work given by
W=FΔrcosθ=(16.0N)(2.20m)cos25.00=31.9J
(b), (c) The normal force and the weight are both at 900 to the displacement in any time interval. Both do 0 work.
(d) ∑W=31.9J+0+0=31.9J
A ball is launched as a projectile with initial speed v at an angle θ above the horizontal. using conservation of energy, find the maximum height hmax of the ball's flight. express your answer in terms of v, g, and θθ.
Answer:
Explanation:
I will us θ as the angle above horizontal as I don't know how to make your chosen symbol for angle.
KEy = PE
½m(vsinθ)² = mgh
h = (vsinθ)²/2g
at room tmeperature none of the metals are
What metal is not at room temperature?
Mercury is a metallic chemical element that is an only metal that is not a solid at room temperature.
In order to create a charged object, you need to ________.
If the velocity and frequency of a wave are both doubled, how does the wavelength change?
The wavelength will remain unchanged.
Explanation:
The velocity [tex]v[/tex] of a wave in terms of its wavelength [tex]\lambda[/tex] and frequency [tex]\nu[/tex] is
[tex]v = \lambda\nu[/tex] (1)
so if we double both the velocity and the frequency, the equation above becomes
[tex]2v = \lambda(2\nu)[/tex] (2)
Solving for the wavelength from Eqn(2), we get
[tex]\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}[/tex]
We would have gotten the same result had we used Eqn(1) instead.
Answer:
the wavelength increases
Explanation:
A 30-cm-tall, 4.0-cm-diameter plastic tube has a sealed bottom. 250 g of lead pellets are poured into the bottom of the tube, whose mass is 30 g, then the tube is lowered into a liquid. The tube floats with 5.0 cm extending above the surface. What is the density of the liquid
The density of the liquid will be equal to [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]
What is density?The density of an object is defined as the ratio of the mass of an object to the volume of the object.
Volume of tube = 2^2 * pi * 30 = 377 cm^3
Volume of tube submerged = 25* 377 / 30 = 314 cm^3
Buoyancy = weight of liquid displaced
Volume of liquid displaced = 314 cm^3
Mass of tube and lead = 250 + 30 = 280 g
Now from the mass density by definition
[tex]\rho = \dfrac{m}{v}[/tex]
[tex]m=\rho \times v[/tex]
Mass of liquid displaced = Mass being supported
[tex]314 \times \rho = 280 g[/tex]
[tex]\rho= \dfrac{280}{ 314 } = .892 \frac{g}{cm^3}[/tex]
Thus the density of the liquid will be equal to [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]
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The density of the liquid is 1.67 g/[tex]cm^3[/tex].
The volume of the tube is
30 * 4 * 3.14 * 0.25 = 94.2 [tex]cm^3[/tex].
The mass of the lead pellets and the plastic tube is
30 + 250 = 280 g.
The volume of the lead pellets is
250 / 11.34 = 22 [tex]cm^3[/tex].
The volume of the liquid that the tube displaces is
94.2 - 22 = 72 [tex]cm^3[/tex].
The density of the liquid is
280 / 72 = 1.67 g/ [tex]cm^3[/tex].
Therefore, the density of the liquid is 1.67 g/ [tex]cm^3[/tex].
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During take-off a 8kg model rocket is burning fuel causing its speed to increase
at a rate of 4m/s2 despite experiencing a 90N drag.
What’s is the strength of the thrust?
(Answer unit is in N)( and the answer isn’t 212)
The strength of the thrust is 122 newtons.
The motion of the rocket is described by the second Newton's law, whose model is shown below:
[tex]\Sigma F = F - D = m\cdot a[/tex] (1)
Where:
[tex]F[/tex] - Thrust, in newtons[tex]D[/tex] - Drag, in newtons[tex]m[/tex] - Mass of the rocket, in kilograms[tex]a[/tex] - Net acceleration of the rocket, in meters per square secondIf we know that [tex]D = 90\,N[/tex], [tex]m = 8\,kg[/tex] and [tex]a = 4\,\frac{m}{s^{2}}[/tex], then the strength of the thrust is:
[tex]F = D + m\cdot a[/tex]
[tex]F = 90\,N + (8\,kg)\cdot \left(4\,\frac{m}{s^{2}} \right)[/tex]
[tex]F = 122\,N[/tex]
The strength of the thrust is 122 newtons.
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Answer This!!!!!!
I'll give brainliest to whoever gets it right.
Answer:
yes the answer is C. 32.16
which statement about metals is true?
All metals are solid at (our) living temperatures except for mercury which is a liquid. Metals are shiny for a while at least when properly finished. Metals are good conductors of both heat and electricity. This is due to the excess electrons in their valence clouds that facilitate energy transfer.What are 5 characteristics of metal?
Metals are lustrous, malleable, ductile, good conductors of heat and electricity
Silver conducts electricity better than any other metal.
Brass is an alloy made from zinc and copper.
Pure gold is too soft for many things so most gold is combined with other metals to make it stronger. ...
World time periods are often listed by the metal used.
which of the following are ways in which energy can be transferred to or from a substance?
a.) heat and internal energy
b.) work and internal energy
c.) heat and work
d.) heat and kinetic energy
Answer:
a
Explanation:
the answer is a because it is feasible
Which statement best describes wavelength?
wavelength is the speed at which a wave travels
wavelength is the height vertically of a wave
wavelength is the length of wave per sec
wavelength is the horizontal length of one complete wave cycle
Answer:
B
Explanation:
Wave length is the height perpendicular verically of a wave
Answer:
b
Explanation:
makes the most sense to me
How much energy does a 150 N child have sitting on a 35M high cliff
Answer:
150+35=185 Plsss Brainliest plssss
You are angry at Dr. Anderson for this exam, so you throw a 0.30-kg stone at his car with a speed of 44 m/s. How much kinetic energy does the stone have
Answer:
Explanation:
KE = ½mv²
KE = ½(0.30)44²
KE = 290 J rounded to 2 s.d.
if the momentum of a 1,400 kg car is the same as the truck in question 17, what is the velocity of the car?
Answer:
Explanation:
momentum is mass times velocity
p = mv
so take the momentum of the truck in question 17 and divide by the mass of this car
v = p/m = p / 1400
why does gas have the most energy but moves the slowest
Gases have heavier molecules. Since all gases have the same average kinetic energy at the same temperature, lighter molecules move faster and heavier molecules move slower on average.
write 2 situations in which the energy changes mentioned occur
Answer:
The types of energy is bond breaking and bond forming in chemical energy.
Explanation:
During Chemical reaction energy is required either for breaking up bonds in case of reactants and building bonds to form products.
The chemical reaction in which energy is released is called exothermic reactions, which is released due to making up the bonds.
The chemical reaction in which energy is absorbed is called endothermic reactions, in which energy is absorbed for breaking up the bonds.
A punter wants to kick a football so that the football has a total flight time of 4.70s and lands 56.0m away (measured along the ground). Neglect drag and the initial height of the football.
How long does the football need to rise?
What height will the football reach?
With what speed does the punter need to kick the football?
At what angle (θ), with the horizontal, does the punter need to kick the football?
Answer:
Explanation:
How long does the football need to rise?
4.70/3 = 2.35 s
What height will the football reach?
h = ½(9.81)2.35² = 27.1 m
With what speed does the punter need to kick the football?
vy = g•t = 9.81(2.35) = 23.1 m/s
vx = d/t = 56.0/4.70 = 11.9 m/s
v = √(vx²+vy²) = 26.0 m/s
At what angle (θ), with the horizontal, does the punter need to kick the football?
θ = arctan(vy/vx) = 62.7°
what is the value of acceleration due to gravity g at the pole of earth
Answer:
The value of Acceleration due to gravity at the pole of the earth is 9.870m/s^2 .Explanation:
I hope this helps you !!2 difference between calorimetry and calorimeter
Calorimetry :
the process of measuring the amount of heat released or absorbed during a chemical reaction.
Calorimeter :
device for measuring the heat developed during a mechanical, electrical, or chemical reaction, and for calculating the heat capacity of materials.
Can someone help me with this problem?
Answer:
no
Explanation:
u flipped it i cant see
