Answer;
-54ft/s
y = -4/5 x + 10
Explanation
Given the position of an object expressed by the function
s(t)=−16t²+10t+100
Velocity is the change in position with respect to time
v(t) = ds(t)/dt
v(t) = -32t + 10
When t = 2
v(2) = -32(2)+10
v(2) = -64+10
v(2) = -54
Hence the velocity of the object is -54ft/s
b) The standard equation of a line in point slope form is expressed as;
y = MX+c
M is the slope
c is the y-intercept
Given the coordinate (5, 6) and (10, 2)
M = 2-6/10-5
M = -4/5
Get the y-intercept
Substitute m = -4/5 and any point say (5,6) into the expression y = mx+c
6 = -4/5 (5) + c
6 = -4+c
c = 6+4
c = 10
Get the required equation
Recall that: y = mx+c
y = -4/5 x + 10
Hence the equation of a line, in slope-intercept form is y = -4/5 x + 10
Why can ultraviolet waves be dangerous
Answer:
They can cause sunburn. Exposure to UV rays can cause premature aging of the skin and signs of sun damage such as wrinkles, leathery skin, liver spots, actinic keratosis, and solar elastosis. UV rays can also cause eye problems.
Explanation:
A block of mass m = 4.4 kg slides from left to right across a frictionless surface with a speed vi= 8.4 m/s It collides in a perfectly elastic collision with a second block of mass M that is at rest. After the collision, the 4.4-kg block reverses direction, and its new speed is 2.5 m/s What is V, the speed of the second block after the collision?
Answer:
[tex]v_{2'}=8.1\:\mathrm{m/s}[/tex]
Explanation:
In a perfectly elastic collision, the total kinetic energy of the system is maintained. Therefore, we can set up the following equation:
[tex]\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2[/tex]
Since the second block was initially at rest, [tex]\frac{1}{2}m_2{v_2}^2=0[/tex].
Plugging in all given values, we have:
[tex]\frac{1}{2}m_1{v_1}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2,\\\\\frac{1}{2}\cdot4.4\cdot8.4^2=\frac{1}{2}\cdot 4.4 \cdot (-2.5)^2+\frac{1}{2}\cdot 4.4\cdot {v_{2'}}^2,\\\\{v_{2'}}=\sqrt{64.31},\\\\{v_{2'}}\approx\fbox{$8.1\:\mathrm{m/s}$}[/tex]..
A ball has a velocity of 11 m/s and a momentum of 47 kgm/s, what is its mass? Show
your given, required, and solutions.
Explanation:
p=m x v
to find the mass:
m= p/v
=47/11
=4.27 kg
A particular satellite with a mass of m is put into orbit around Ganymede (the largest moon of Jupiter) at a distance 300 km from the surface. What is the gravitational force of attraction between the satellite and the moon? (Ganymede has a mass of 1.48x1023 kg and a radius of 2631 km.) mass of satellite =5×10^8 kg.
Answer:
F = 402.18 N
Explanation:
Given that,
A particular satellite with a mass of m is put into orbit around Ganymede (the largest moon of Jupiter) at a distance 300 km from the surface. Let the mass of the satellite is 350 kg.
We need to find the gravitational force of attraction between the satellite and the moon.
The formula for the gravitational force is given by :
[tex]F=G\dfrac{Mm}{(R+h)^2}[/tex]
M is mass of Ganymede
m is mass of satellite
R is Radius of Ganymede
h is distance = 300 km
Putting all the values,
[tex]F=6.67\times 10^{-11}\times \dfrac{1.48\times 10^{23}\times 350}{(2631\times 10^{3}+300\times 10^3)^2}\\F=402.18\ N[/tex]
So, the required force of attraction between the satellite and the moon is 402.18 N.
Find the momentum of a 500,000 kg train that is stopped on the tracks?
a. O kg m/s
b. 250,000 kg m/s
c. 500,000 kg m/s
d. 16,000,000 kg m/s
Answer:
The answer should be A) 0m/s
Explanation:
It is stopped on the train tracks therefore it is not moving.
Please tell me if I am wrong because I'm not 100% sure on this. Hope it's right and that it helped you.
Describe the effect of the amplitude on the velocity of the pulse!???
Answer:
Amplitude increases with decreasing velocity.
Explanation:
At the same time, an increase in attention takes place
can lamp that works on a 2.5 v work on a 1.12 v ?
Answer:
Explanation:
Thinking about the logics it can but it may be dim because 1.12 is lower than 2,5v so this will mean u lamp may not work or may work very dimely due to the low voltage it is receiving.
What happens to the sum of the ball's kinetic energy and potential energy as the ball rolls from point A to point E? Assume there's no friction between the ball and the ground.
А. The sum decreases.
В. The sum increases.
C. The sum remains the same.
D. The sum always equals zero.
Answer:
C. The sum remains the same.
Explanation:
The sum of the kinetic and potential energy remains the same as the all rolls from point A to E.
We know this based on the law of conservation of energy that is in play within the system.
The law of conservation of energy states that "energy is neither created nor destroyed within a system but transformed from one form to another".
At the top of the potential energy is maximum As the ball rolls down, the potential energy is converted to kinetic energy. Potential energy is due to the position of a bodyKinetic energy is due to the the motion of the body10) A soccer player kicks a soccer ball (m = 0.42 kg) accelerating from rest to 32.5m/s in 0.21s. Determine the force that sends soccer ball towards the goal.
G
U
E
S
S
Formula
11) Small rockets are fired to make small adjustments in the speed of a satellite. A certain small rocket can change the velocity of a 72,000kg satellite from 0m/s to 0.63m/s in 1296s. What force is exerted by the rocket on the satellite?
G
U
E
S
S
Formula
please I need help I don't understand it and I had to deliver it yesterday helpp:(
Explanation:
(10) Mass of a soccer player, m = 0.42 kg
Initial speed, u = 0
Final speed, v = 32.5 m/s
Time, t = 0.21 s
We need to find the force that sends soccer ball towards the goal.
Force, F = ma
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.42 \times (32.5-0)}{0.21}\\\\F=65\ N[/tex]
So, 65 N of force soccer ball sends towards the goal.
(11) Mass of the satellite, m = 72,000 kg
Initial speed, u = 0 m/s
Final speed, v = 0.63 m/s
Time, t = 1296 s
We need to find the force is exerted by the rocket on the satellite.
Force, F = ma
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{72,000\times (0.63-0)}{1296}\\\\F=35\ N[/tex]
So, 35 N of the force is exerted by the rocket on the satellite.
Hence, this is the required solution.
which changes will increase the rate of reaction during combustion
Answer:
reducing temperature of the surrounding
Explanation:
combustion reactions are exothermic so they give off heat. reducing the temperature of the surrounding will enable more efficient energy transfer
A 10kg toy truck has a 5kg toy car at rest. If the toy truck was moving at 3 m/s before the collision and carries that car with it, what is the
Final velocity of the car and truck.
A. 15 m/ s
B. 30 m/ s
C 2 m/ s
D. 18 m/ sc free
Answer:
[tex]v_f=2\:\mathrm{m/s}[/tex]
Explanation:
From the Law of Conservation of Momentum, momentum is conserved. Therefore, we can set up the following equation:
[tex]m_1v_1+m_2v_1=m_fv_f[/tex]
Since the 5kg toy car was initially and rest, [tex]m_2v_2=0[/tex].
Therefore, plugging in our values, we have:
[tex]10\cdot 3=(10+5)v_f,\\v_f=\frac{30}{15},\\v_f=\fbox{$2\:\mathrm{m/s}$}[/tex].
Answer:
Let m1 = mass of big toy car=10kg
m2= mass of small toy car= 5kg
U1= initial velocity of big toy car= 3m/s
U2= initial velocity of small toy car=0
Since the big toy car moved the small one after the collision, their final velocity will be the same.
m1u1 + m2u2= (m1+m2)v
10(3)+(5)(0)=(10+5)v
30=15v
Divide both sides by 15
V=2
The final velocity is 2m/s
Explanation:
_____________________health is how you feel and how you react to situations bases on how you feel.
Group of answer choices
Mental
Physical
Emotional
Social
Answer: Emotional
Explanation:
Answer:
Physical
Explanation:
I love going going for a jog or a run so does my dog
I NEED HELP ASAPPPP PLEASE
Two ships are docked next to each other. Their centers of mass are 39m apart. One ship’s mass is 9.2 *10^7 kg and the other ship’s mass is 1.84*10^8 kg. What gravitational force exists between them?
Please help!
Answer:
742.3N
Explanation:
Given parameters:
Distance = 39m
Mass 1 = 9.2 x 10⁷kg
Mass 2 = 1.84 x 10⁸kg
Unknown:
Gravitational force between the ships = ?
Solution:
To solve this problem, we apply the newton's law of universal gravitation:
Fg = [tex]\frac{G x mass 1 x mass 2}{r^{2} }[/tex]
G is the universal gravitation constant = 6.67 x 10⁻¹¹
r is the distance or separation
Fg = [tex]\frac{6.67 x 10^{-11} x 9.2 x 10^{7} x 1.84 x 10^{8} }{39^{2} }[/tex] = 742.3N
Megan walks 1100\,\text m1100m1100, start text, m, end text to the left in 330\,\text s330s330, start text, s, end text. What was her average speed in \dfrac{\text m}{\text s} s m start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?
Answer:
v = 3.34 m/s
Explanation:
Given that,
Distance, d = 1100 m
Time, t = 330 s
We need to find the average speed of the Megan. It is equal to the total distance divided by total time taken.
[tex]v=\dfrac{1100\ m}{330\ s}\\\\v=3.34\ m/s[/tex]
So, the average speed of Megan is 3.34 m/s.
Answer:
33.3
Explanation:
PLEASEEEEEE HELPPPPPPP
Define resistance and discuss how it affects current.
Answer:
Resistance is the opposing of the flow of current through a conductor.
What gases can CFC and HCFC refrigerants decompose into at high temperatures
Answer:
Hydrochloric and Hydrofluoric Acids.
Two 2.1-cm-diameter electrodes with a 0.20-mm-thick sheet of Teflon between them are attached to a 9.0 V battery. Without disconnecting the battery, the Teflon is removed.
Required:
a. What is the charge before the Teflon is removed?
b. What is the potential difference before the Teflon is removed?
c. What is the electric field before the Teflon is removed?
d. What is the charge after the Teflon is removed?
e. What is the potential difference after the Teflon is removed?
f. What are the electric field after the Teflon is removed?
Answer:
a. Q = 1881.73 x [tex]10^{-13}[/tex] C
b. As battery is not removed so, potential difference will remain same.
c. E = 21.42 x [tex]10^{3}[/tex] V/m
d. Q = 895.5 x [tex]10^{-13}[/tex] C
e. Again the potential difference will not change it will remain same as 9 V
f. E = 45 x [tex]10^{3}[/tex] V/m
Explanation:
Solution:
Here, Teflon is used so, the dielectric constant of the Teflon K = 2.1
Diameter = 2.1 cm
Radius = 2.1/2 cm
Radius = 1.05 cm
Radius = 0.015 m
Now, we need to find the area of each plate:
A = [tex]\pi r^{2}[/tex]
A = (3.14) ([tex]0.015^{2}[/tex])
A = 0.000225 [tex]m^{2}[/tex]
A = 2.25 x [tex]10^{-4}[/tex] [tex]m^{2}[/tex]
We are given the thickness of the plate which equal to the distance between the two plates.
d = 0.20 mm = 0.2 x [tex]10^{-3}[/tex] m
d = 0.2 x [tex]10^{-3}[/tex] m = distance between two plates.
Hence, the capacitance of the dielectric without the dielectric
C = [tex]\frac{E.A}{d}[/tex]
Putting up the values we get,
E = 8.85 x [tex]10^{-12}[/tex]
C = [tex]\frac{8.85 . 10^{-12} x 2.25 . 10^{-4} }{0.002}[/tex]
C = 99.5 [tex]10^{-13}[/tex]
If dielectric is included then,
[tex]C^{'}[/tex] = K C
[tex]C^{'}[/tex] = (2.1) ( 99.5 x [tex]10^{-13}[/tex])
[tex]C^{'}[/tex] = 209.08 x [tex]10^{-13}[/tex] F
As we know the voltage of the battery V = 9V So,
a) Charge before the Teflon is removed:
Q = CV
Q = [tex]C^{'}[/tex]V
Q = (209.08 x [tex]10^{-13}[/tex] F) (9V)
Q = 1881.73 x [tex]10^{-13}[/tex] C
b) Potential Difference before the Teflon is removed = ?
As battery is not removed so, potential difference will remain same.
c) Electric Field =?
As we know,
E = V/(K.d)
E = 9V/(2.1 x 0.2 x [tex]10^{-3}[/tex])
E = 21.42 x [tex]10^{3}[/tex] V/m
d) After the Teflon is removed
Q = CV
Q = (99.5 [tex]10^{-13}[/tex] ) ( 9)
Q = 895.5 x [tex]10^{-13}[/tex] C
e) Again the potential difference will not change it will remain same as 9 V
f) Electric Field = ?
E = [tex]\frac{V}{d}[/tex] (Teflon is removed)
E = 9/0.2 x [tex]10^{-3}[/tex]
E = 45 x [tex]10^{3}[/tex] V/m
In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular orbit with a period of 7.08 × 103 s and orbital speed of 3.40 × 103 m/s . The mass of the GS is 930 kg and the radius of Mars is 3.43 × 106 m. Calculate the mass of Mars.
Answer: [tex]5.944\times 10^{23}\ kg[/tex]
Explanation:
Given
Time period [tex]T=7.08\times 10^3\ s[/tex]
Orbital speed [tex]v=3.40\times 10^3\ m/s[/tex]
mass of GS [tex]m_{GS}=930\ kg[/tex]
Radius of Mars [tex]r=3.43\times 10^6\ m[/tex]
Consider the mass of mars is M
Here, Gravitational pull will provide the centripetal force
[tex]F_G=F_c[/tex]
[tex]\dfrac{GMm_{GS}}{r^2}=\dfrac{m_{GS}v^2}{r}\\M=\dfrac{v^2\cdot r}{G}\\M=\dfrac{(3.43\times 10^3)^2\cdot 3.43\times 10^6}{6.67\times 10^{-11}}[/tex]
[tex]M=5.944\times 10^{23}\ kg[/tex]
In March 1999 the Mars Global Surveyor (GS) entered its final orbit on Mars, sending data back to Earth. The mass of Mars is approximately 6.419 × 10²³ kg.
Kepler's Third Law states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (a) of the orbit:
T² = (4π² / GM) × a³
In a circular orbit, the semi-major axis is equal to the radius of the orbit (r).
Given:
Orbital period (T) = 7.08 × 10³ s
Orbital speed (v) = 3.40 × 10³ m/s
Mass of GS (m) = 930 kg
Radius of Mars (r) = 3.43 × 10⁶ m
The orbital speed (v) is related to the radius (r) and the gravitational constant (G) by:
v = √(GM / r)
v² = GM / r
G = (v² × r) / M
T² = (4π² / [(v² × r) / M]) × r³
T² = (4π² × M × r²) / v²
M = (T² × v²) / (4π² × r²)
M = ( (7.08 × 10³)² × (3.40 × 10³)² ) / (4π² × (3.43 × 10⁶)²)
M = 6.419 × 10²³ kg
Therefore, the mass of Mars is approximately 6.419 × 10²³ kg.
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The second law of thermodynamics imposes what limit on the efficiency of a heat engine?
A. The energy a heat engine must deposit in a cold reservoir is greater than or equal to the energy it extracts from a hot reservoir.
B. The energy a heat engine must deposit in a cold reservoir is greater than or equal to the energy extracted as useful work.
C. A heat engine must deposit some energy in a cold reservoir.
Answer:
C. A heat engine must deposit some energy in a cold reservoir.
Explanation:
The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir."
This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir.
Then we have the equation:
Q = W + q
From this we can conclude that the correct option is:
C. A heat engine must deposit some energy in a cold reservoir.
There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.
C. A heat engine must deposit some energy in a cold reservoir.
The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir". This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir. Then we have the equation: Q = W + q There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.Therefore, option C is correct.
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Particle A with charge q and mass ma and particle B with charge 2q and mass
mb, are accelerated from rest by a potential difference AV and subsequently
deflected by a uniform magnetic field into semicircular paths. The radii of the
trajectories by particle A and B are R and 3R, respectively. The direction of
the magnetic field is perpendicular to the velocity of the particle. Determine
their mass ratio?
An electric heater draws a steady 15.0A on a 120-V
line. How much power does it require and how
much does it cost per month (31days) if it operates
2.0 h per day and the electric company charges 15.5
cents per kWh?
Answer:
1. 1800 W
2. $ 17.3
Explanation:
From the question given above, the following data were obtained:
Current (I) = 15 A
Voltage (V) = 120 V
Time (t) = 20 h per day
Duration = 31 days
Cost = 15.5 cents per kWh
1. Determination of the power.
Current (I) = 15 A
Voltage (V) = 120 V
Power (P) =?
P = IV
P = 15 × 120
P = 1800 W
Thus, 1800 W of power is required.
2. Determination of the cost per month (31 days).
We'll begin by converting 1800 W to KW.
1000 W = 1 KW
Therefore,
1800 W = 1800 W × 1 KW / 1000 W
1800 W = 1.8 KW
Next, we shall determine the energy consumption for 31 days. This can be obtained as follow:
Power (P) = 1.8 KW
Time (t) = 2 h per day
Time (t) for 31 days = 2 × 31 = 62 h
Energy (E) =?
E = Pt
E = 1.8 × 62
E = 111.6 KWh
Finally, we shall determine the cost of consumption. This can be obtained as follow:
1 KWh = 15.5 cents
Therefore,
111.6 KWh = 111.6 KWh × 15.5 cents / 1 KWh
111.6 KWh = 1729.8 cents
Converting 1729.8 cents to dollar, we have:
100 cents = $ 1
Therefore,
1729.8 cents = 1729.8 cents × $ 1 / 100 cents
1729.8 cents = $ 17.3
Thus, it will cost $ 17.3 per month to run the electric heater.
The amount of power it requires is 180 Watt, and the cost per month to operate the electric heater for 2.0 h is 5.58 cent
To calculate the electric power required by the electric heater, we use the formula below.
⇒ Formula:
P = Vi................ Equation 1⇒ Where:
P = PowerV = Voltagei = currentFrom the question,
⇒ Given:
V = 120 VI = 15.0 A⇒ Substitute these values into equation 1
P = 120(1.5)P = 180 Watt.
For the cost of running the electric heater per month, if it operates 2.0 h per day. We use the relation below
C = Pt×C'/1000............... Equation 2⇒ Where:
C = The cost per month to operate the electric heater for 2.0ht = timeC' = Cost per kWh charge by the electric company.From the question,
⇒ Given:
C' = 15.5 cent per kWh t = 2.0 hP = 180 Watt⇒ Substitute these values into equation 2
C = 180(15.5)(2)/1000C = 5.58 cent.Hence, The amount of power it requires is 180 Watt, and the cost per month to operate the electric heater for 2.0h is 5.58 cent
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h. The length of the shadow is different in evening and in the day. Justify
the shadows are exactly the same length in the morning as they are in the evening.
is so obvious it’s that when the sun is low you get long shadows and when the sun is up in the sky like in the noon the shadow is shorter.
what do scientists call a substance that forms during a chemical reaction
Answer:
A chemical reaction is a process in which one or more substances, also called reactants, are converted to one or more different substances, known as products.
Answer:
the answer is reactants i guess
What can we conclude from the attractive nature of the force between a positively charged rod and an object?
a. the object is positively charged
b. cannot determine
c. the object is a conductor
d. the object is an insulator
e. the object is negatively charged
Answer:
E; The object is negatively charged
Explanation:
Here, we want to state the conclusion that can be drawn from a positively charged rod being attracted to an object.
Generally as we know, oppositely charged materials attract while the ones with same charges repel each other.
Thus, in this case, for the rod to attract the object, there must have been an opposite charge of negativity on the object
So we conclude that the reason why the rod attracted the object was because of the presence of opposing charges on both of them. And since the rod has taken the positive charge, it is only correct to state that the object is negatively charged
14 J of heat are removed from a gas sample while it is being compressed by a piston that does 28 J of work.
What is the change in the thermal energy of the gas?
How does change the temperature of the gas?
The increase in thermal energy of the gas sample is +14 J, and the temperature is increased.
Given information:
The heat removed from the gas sample is [tex]Q=-14\rm\;J[/tex]. The negative sign represents the heat removal.
The work done on the gas sample is [tex]W=28\rm\; J[/tex].
Work is done on the gas. So, it will be taken as positive.
According to the first law of thermodynamics, the change in thermal energy of the gas or system will be calculated as,
[tex]\Delta E=Q+W\\\Delta E=-14+28\\\Delta E=14\rm\;J[/tex]
The change in thermal energy of the system will be 14 J. It is positive. So, the thermal energy is increased. It implies the temperature of the system or gas is also increased.
Therefore, the increase in thermal energy of the gas sample is +14 J, and the temperature is increased.
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Two balls of unequal mass are hung from two springs that are not identical. The springs stretch the same distance as the two systems reach equilibrium. Then both springs are compressed and released. Which one oscillates faster?
a) The spring with the heavier ball,
b) Springs oscillate with the same frequency,
c) The spring with the light ball.
Answer:
b) Springs oscillate with the same frequency,
Explanation:
expression for frequency of vibration of mass hanging from a spring is given as follows
f = [tex]\frac{1}{2\pi} \times \sqrt{\frac{k}{m} }[/tex]
k is force constant of spring and m is mass vibrating .
In the present case, if mass stretches the spring by x and remains balanced
mg = k x
[tex]\frac{k}{m} =\frac{g}{x}[/tex]
g and x are same for both cases
[tex]\frac{k}{m}[/tex] will also be same for both cases .
Hence frequency of vibration will also be same for both the balls .
To understand and apply the formula τ=Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle of mass m, we use Newton's second law: F net=ma , where F net is the net force acting on the particle.To find the angular acceleration α of a rigid object rotating about a fixed axis, we can use a similar formula: τnet=Iα, where τnet=∑τ is the net torque acting on the object and I is its moment of inertia.
Part A:
Assume that the mass of the swing bar, is negligible. Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, l, as well as the acceleration due to gravity g.
Part B:
Now consider a similar situation, except that now the swing bar itself has mass mbar.Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, mbar, l, as well as the acceleration due to gravity g.
Answer:
Hello your question is incomplete attached below is the missing part of the question
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.
You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I
answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L
part A = attached below
Explanation:
Part A :
Assuming that mass of swing is negligible
α = T/I
where ; T = torque, I = inertia,
hence T = L/2*9*(M1 - M2)
also; I = [tex]M1*(L/2)^2 + M2*(L/2)^2[/tex]= ( M1 + M2) * (L/2)^2
Finally the magnitude of the angular acceleration α
α = 2*[(M1 - M2)/(M1 + M2)]*g/L
Part B attached below
A typical ceiling fan running at high speed has an airflow of about 1.85 ✕ 10^3 ft^3/min, meaning that about 1.85 ✕ 10^3 cubic feet of air move over the fan blades each minute.
Determine the fan's airflow in m^3/s.
Answer:
0.83 m³/s
Explanation:
The speed of the airflow is given as;
1.85 x 10³ ft³/min
Now we are to express this unit in m³/s
1ft = 0.3m
60s = 1 min
So;
1.85 x 10³ x ft³ x [tex]\frac{1}{min}[/tex] x [tex]\frac{(0.3m)^{3} }{ft^{3} }[/tex] x [tex]\frac{1min}{60s}[/tex]
= 0.83 m³/s
The force of gravity acting on an object is directed through this
center of gravity and toward the center of the
Answer:
Earth.
Explanation:
Center of gravity can be defined as the specific point where all of the weight of an object is concentrated.
Generally, all the objects found around the world all have a center of gravity.
When an object is balanced so that a displacement lowers its center of gravity, the object is said to be in stable equilibrium.
Hence, the force of gravity acting on an object is directed through this center of gravity and toward the center of the earth.
Weight can be defined as the force acting on a body or an object as a result of gravity.
Mathematically, weight is given by the formula;
[tex] Weight, W = mg [/tex]
Where;
m is the mass of an object.
g is acceleration due to gravity.