Answer:
Using multiple pulleys decreases the amount of force necessary to move an object by increasing the amount of rope used to raise the object. The mechanical advantage (MA) of a pulley system is equal to the number of ropes supporting the movable load.The object has a mass of 100kg. The Tension is 200N[U]. What is the acceleration of this elevator? *
A) 8m/s/s
B) 8m/s/s[D]
C) 9.8m/s/s[D]
D) 0.5m/s/s[D]
Answer:
So the answer is B. A is wrong because negative answer = deceleration
With what speed must a ball be thrown directly upward so that it remains in the air for 10 seconds?
a) What will be its speed when it hits the ground?
b) How high does the ball rise?
Answer:
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To accurately describe the wind, the measurement should include
A) a direction, but not a speed
B)a speed, but not a direction
C) both a speed and a direction
D) neither a speed nor a direction
Answer:
C. both a speed and a direction
A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s. (a) What is the moment of inertia of the system
Complete Question
A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s.
(a) What is the moment of inertia of the system
(b) What is its rotational kinetic energy? axis 50 cm 30 cm 20 cm
Answer:
a) [tex]M.I=0.32kg.m^2[/tex]
b) [tex]K.E=621.8J[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=2.0kg[/tex]
Disk Radius [tex]r_d=50cm=0.5m[/tex]
Mass of annular cylinder [tex]M_c=1.0kg[/tex]
Inner Radius of cylinder [tex]R_i=20cm=0.2m[/tex]
Outer Radius of cylinder [tex]R_o=0.3m[/tex]
Angular Velocity [tex]\omega=10rev/s=2 \pi*10=62.83[/tex]
Generally the equation for moment of inertia is mathematically given by
[tex]M.I=0.5M r_d^2+0.5M_c(R_i^2+R_o^2)[/tex]
[tex]M.I=0.5(2*0.50)^2)+0.5*1*(0.2^2+0.30^)[/tex]
[tex]M.I=0.32kg.m^2[/tex]
Generally the equation for Rotational Kinetic Energy is mathematically given by
[tex]K.E=0.5M.I \omega^2[/tex]
[tex]K.E=0.5 *0.32*62.83[/tex]
[tex]K.E=621.8J[/tex]
A 25.0 kg child slides down a long slide in a playground. She starts from rest at a height h1 of 21.00 m. When she is partway down the slide, at a height h2 of 8.00 m, she is moving at a speed of 7.80 m/s. Calculate the mechanical energy lost due to friction (as heat, etc.).
Answer:
The mechanical energy lost due to friction is 2,424.5 J
Explanation:
Given;
mass of the child, m = 25 kg
intial velocity of the child, u = 0
final velocity of the child, v = 7.8 m/s
initial position of the child, h₁ = 21 m
final position of the child, h₂ = 8 m
Let the energy lost due to heat = ΔE
ΔE + ΔK.E + ΔP.E = 0
ΔE + ¹/₂m(v² - u²) + mg(h₂ - h₁) = 0
ΔE + ¹/₂ x 25(7.8² - 0) + 25 x 9.8(8 - 21) = 0
ΔE + 760.5 J - 3185 J =
ΔE - 2,424.5 J = 0
ΔE = 2,424.5 J
Therefore, the mechanical energy lost due to friction is 2,424.5 J
plsss help me
thank you very much
When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3m3 of water flows through the dam each second. The water is released 220 mm below the top of the reservoir. If the generators that the dam employs are 90% efficient, what is the maximum possible electric power output?
Answer:
The output electric power is 1338876 W.
Explanation:
Volume, V = 690 cubic meter
height, h = 220 mm = 0.22 m
efficiency = 90 %
time , t = 1 s
Let the mass is m.
m = volume x density
m = 690 x 1000 = 690000 kg
The input power is
P = m g h = 690000 x 9.8 x 0.22 = 1497640 W
The electric power out put is
[tex]P' = 90 % of 1487640\\\\\\P' = 1338876 W[/tex]
Speed of light in water
Answer:
225,000 kilometers per second
Explanation:
Have a nice day
A major artery with a 1.3 cm^2 cross-sectional area branches into 18 smaller arteries, each with an average cross-sectional area of 0.6 cm^2. By what factor is the average velocity of the blood reduced when it passes into these branches?
Answer:
When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12
Explanation:
Given;
initial area of the artery, A₁ = 1.3 cm²
Area of each smaller 18 arteries, a₂ = 0.6 cm²
Total area of the smaller 18 arteries, A₂ = 18 x 0.6 cm²
Apply flow rate equation;
Q = AV
where;
Q is the flow rate of the blood
V is the average velocity of the blood
If the flow rate is constant, then;
A₁V₁ = A₂V₂
[tex]V_2 = \frac{A_1V_1}{A_2} = \frac{1.3\times V_1}{18\times 0.6} \\\\V_2 = 0.12 \ V_1[/tex]
When the blood passes into the smaller branches, its average velocity reduces by a factor of 0.12
A student on a new planet wants to determine the value of gravity on that planet. Luckily for them they brought equipment that they can use to set up an oscillating spring or an oscillating pendulum. Which procedure would allow the student to determine the value of gravity on the new planet
Answer:
By measure the effective length and the time period of the pendulum.
Explanation:
Let the student take the oscillating pendulum at the planet.
He measure the time period of the pendulum by using the stop watch or the ordinary watch.
Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.
Now, use the formula of the time period of the pendulum,
[tex]T =2\pi\sqrt\frac{L}{g}[/tex]
Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.
By rearranging the terms, we get
[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]
Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 7.56 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.22 rev/s.
Required:
a. Which rate of rotation gives the greater speed for the ball?
b. What is the centripetal acceleration of the ball at 8.16 rev/s?
c. What is the centripetal acceleration at 6.35 rev/s?
Answer:
a) [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b) [tex]a=30.7[/tex]
c) [tex]a=35.91[/tex]
Explanation:
From the question we are told that:
Initial angular velocity [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
Initial Length [tex]L_1=0.600m[/tex]
Final angular velocity [tex]\omega _2=6.22rev/s=39rad/s[/tex]
Final Length [tex]L_2=0.900m[/tex]
a)
Generally the rotation with the greater speed is
[tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b)
Generally the equation for centripetal acceleration at 8.16 is mathematically given by
[tex]a=\omega_1^2*L_1[/tex]
[tex]a=8.16 rev/s*0.6[/tex]
[tex]a=30.7[/tex]
c)
At 6.35 rev/s
[tex]a=6.35 rev/s*0.9[/tex]
[tex]a=35.91[/tex]
Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N, this time on a frictionless floor. The velocity of the crate is in the direction of the force Steve is applying to the crate. What is the net work done on the crate
Answer:
The correct answer is "4000 J".
Explanation:
Given that,
Force,
= 200 N
Displacement,
= 20 m
Now,
The work done will be:
⇒ [tex]Work=Force\times displacement[/tex]
By putting the values, we get
[tex]=200\times 20[/tex]
[tex]=4000 \ J[/tex]
two factor of a number are 5 and 6 .what is the number show working
Answer:
30
Explanation:
since [tex]\frac{30}{5}[/tex]=6
[tex]\frac{30}{6}[/tex]=5
then both 5 and 6 are factors of 30
Have a nice day
The motion that is not repeated in regular interval of time is called....?
please help me!
Answer:
Motion which repeats itself after regular intervals of time is known as periodic motion.Có bao nhiêu đặc trưng sinh lý của âm
Answer:
Ba đặc trưng sinh lí của âm là độ cao, độ to và âm sắc.
Explanation:
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down? Assume the speed of sound is 343 m/s.
Answer:
The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
Explanation:
We can calculate the speed of the train using the Doppler equation:
[tex] f = f_{0}\frac{v + v_{o}}{v - v_{s}} [/tex]
Where:
f₀: is the emitted frequency
f: is the frequency heard by the observer
v: is the speed of the sound = 343 m/s
[tex] v_{o}[/tex]: is the speed of the observer = 0 (it is heard in the town)
[tex] v_{s}[/tex]: is the speed of the source =?
The frequency of the train before slowing down is given by:
[tex] f_{b} = f_{0}\frac{v}{v - v_{s_{b}}} [/tex] (1)
Now, the frequency of the train after slowing down is:
[tex] f_{a} = f_{0}\frac{v}{v - v_{s_{a}}} [/tex] (2)
Dividing equation (1) by (2) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}} [/tex]
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}} [/tex] (3)
Also, we know that the speed of the train when it is slowing down is half the initial speed so:
[tex] v_{s_{b}} = 2v_{s_{a}} [/tex] (4)
Now, by entering equation (4) into (3) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}} [/tex]
[tex] \frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}} [/tex]
By solving the above equation for [tex]v_{s_{a}}[/tex] we can find the speed of the train after slowing down:
[tex] v_{s_{a}} = 11.06 m/s [/tex]
Finally, the speed of the train before slowing down is:
[tex] v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s [/tex]
Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
I hope it helps you!
3) show thal escape veloerty ve where symbols have their usual meaning?
Answer:
[tex]v = \sqrt{\frac{2GM}{r} }[/tex]
The formula for escape velocity where:
G - Gravitational constant (9.81 etc.)
M - the mass of the object the escape should be made from
r - distance to the centre of that mass
How does the magnitude of the normal force exerted by the ramp in the figure compare to the weight of the static block? The normal force is:______ a. greater than the weight of the block. b. possibly greater than or less than the weight of the block, depending on whether or not the ramp surface is smooth. c. equal to the weight of the block. d. possibly greater than or equal to the weight of the block, depending on whether or not the ramp surface is smooth. less than the weight of the block.
Answer:
less than the weight of the block.
Explanation:
From the free body diagram, we get.
The normal force is N = Mg cosθ
The tension in the string is T = Mg sinθ
Wight of the block when the block is static, W = Mg
Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,
therefore, the normal force is less than the weight of the static block.
You are driving home from school steadily at 97 km/h for 190 km . It then begins to rain and you slow to 60 km/h instantly. You arrive home after driving 4.0 hours.
how far is your hometown from school?
Please delete my answer. I made a mistake
What is the efficiency of a ramp that is 5.5 m long when used to move a 66 kg object to a height of 110 cm when the object is pushed by a 150 N force .
Answer and I will give you brainiliest
Explanation:
Energy input = F×d = (150 N)(5.5 m) = 825 J
Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J
efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%
You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 13-m-high hill, then descends 20 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.4 m and that a loaded car will have a maximum mass of 430 kg. For safety reasons, the spring constant should be 13 % larger than the minimum needed for the car to just make it over the top. Part A What spring constant should you specify
Answer:
22.15 N/m
Explanation:
As we know potential energy = m*g*h
Potential energy of spring = (1/2)kx^2
m*g*h = (1/2)kx^2
Substituting the given values, we get -
(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)
k = 39200/2.645
k = 19600 N/m
For safety reasons, this spring constant is increased by 13 % So the new spring constant is
k = 19600 * 1.13 = 22148 N/m = 22.15 N/m
If the temperature stays constant, which change would decrease the amount
of thermal energy in an object?
A. Decreasing its density
B. Increasing its velocity
c. Decreasing its mass
D. Increasing its mass
Two stars that are in the same constellation:
Answer:
A binary star is a star system consisting of two stars orbiting around their common barycenter.
❣️(◍Jess bregoli◍)❣️#keep learning
An object is positively charged if it has more what
Answer:
An Object is positively charged if it has more Positive Electrons in that object
How do solar panels work with conduction, convection and radiation?
Answer:
In the case of a solar thermal panel we are trying to heat above the ambient temperature so conduction and convection will work against us by taking heat from the panel to the out- side world. ... The sun (at 6000 C surface temperature) is hotter than the solar panel so the panel will get hot due to the solar radiation.
Explanation:
An object produces a sound wave with a wavelength 75.0cm. if the speed of sound is 350.0m/s, the frequency of the sound is
Answer:
wavelength=75.0
speed of sound(v)=350 .0m/s
frequency(f)=?
we know,
v=f*wavelengh
350.0 =f*750
f. =350/75
=4.667
pls mark me brainlest
Can someone help me
Btw the last one say current
Question 9 of 10
What causes the different seasons on Earth?
A. The angles at which the suns rays strike the Earth
Ο Ο Ο
B. The distance between Earth and the sun
C. The speed at which the Earth rotates on its axis
O
D. Increasing levels of carbon dioxide in the atmosphere.
SUBMIT
Answer:
B
Explanation:
The seasons are measured in how far or close the earth is to the sun.
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h, and g : Are any of the answers changed if the initial angle is changed?
Complete question is;
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:
(a) the work done by the force of gravity on the projectile,
(b) the change in kinetic energy of the projectile since it was fired, and
(c) the final kinetic energy of the projectile.
(d) Are any of the answers changed if the initial angle is changed?
Answer:
A) W = mgh
B) ΔKE = mgh
C) K2 = mgh + ½mv_o²
D) No they wouldn't change
Explanation:
We are expressing in terms of m, v0, h, and g. They are;
m is mass
v0 is initial velocity
h is height of projectile fired
g is acceleration due to gravity
A) Now, the formula for workdone by force of gravity on projectile is;
W = F × h
Now, Force(F) can be expressed as mg since it is force of gravity.
Thus; W = mgh
Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.
Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.
B) Change in kinetic energy is simply;
ΔKE = K2 - K1
Where K2 is final kinetic energy and K1 is initial kinetic energy.
However, from conservation of energy, we now that change in kinetic energy = change in potential energy.
Thus;
ΔKE = ΔPE
ΔPE = U2 - U1
U2 is final potential energy = mgh
U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.
Thus;
ΔKE = ΔPE = mgh
Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.
C) As seen in B above,
ΔKE = ΔPE
Thus;
½mv² - ½mv_o² = mgh
Where final kinetic energy, K2 = ½mv²
And initial kinetic energy = ½mv_o²
Thus;
K2 = mgh + ½mv_o²
Similar to a and B above, this will not change even if initial angle is changed
D) All of the answers wouldn't change because their equations don't depend on the angle.
I HAVE A PHYSICS LOCKDOWN EXAM TODAY, THEY ARE 25 QUESTIONS AND I HAVE ABOUT AN HOUR TO SOLVE IT, I NEED HELP WITH THEM ASAP. PLEASE IF YOU'RE GOOD AT PHYSICS LET ME KNOW ILL BE SO GRATEFUL.
Answer:
I’ll try my best!
