Answer:
1. 1800 lb ft, 5.27 J
2. 905
Explanation:
1. Work is the change in energy.
W = mgh
W = (120 lb) (15 ft)
W = 1800 lb ft
W = mgh
W = (0.250 kg) (9.8 m/s²) (2.15 m)
W = 5.27 J
2. The IMA (ideal mechanical advantage) of a screw is the circumference of the screw divided by the pitch.
IMA = 2πr / p
The IMA of a handle (lever) is the circumference of the handle divided by the circumference of the screw.
IMA = 2πR / 2πr
The total IMA is the product:
IMA = (2πR / 2πr) (2πr / p)
IMA = 2πR / p
IMA = 2π (36) / (0.25)
IMA = 905
All objects near the earths surface - regardless of size and weight - hhave the same force of gravityvacting on them.
A. True
B. False
Answer:
B. False
Explanation:
Not all objects near the earths surface - regardless of size and weight - have the same force of gravity on them.
Haley is driving down a straight highway at 73 mph. A construction sign warns that the speed limit will drop to 55 mph in 0.50 mi.
What constant acceleration (in m/s2) will bring Haley to this lower speed in the distance available?
Answer:
0.29[tex]\text{m}\text{s}^{2}[/tex]
Explanation:
Given: Haley is driving down a straight highway at 73 mph. A construction sign warns that the speed limit will drop to 55 mph in 0.50 mi.
To Find: constant acceleration (in [tex]\text{m}\text{s}^{2}[/tex]) that will bring Haley to this lower speed in the distance available
Solution:
we know that,
[tex]1[/tex] mile=[tex]1.609[/tex]km
[tex]1[/tex] km[tex]\setminus[/tex]h = [tex]5\setminus18 m\setminus s[/tex]
Initial Speed of Haley(u)=[tex]73[/tex] mph
[tex]73\times 1.609\times 5\setminus 18[/tex]
[tex]32.63[/tex][tex]\text{m}\setminus\text{s}[/tex]
Final Speed of Haley(v)= [tex]55[/tex] mph
[tex]55\times 1.609\times 5\setminus 18[/tex]
[tex]24.58[/tex][tex]\text{m}\setminus\text{s}[/tex]
The distance to be travelled while lowering the speed(S)=[tex]0.5[/tex] mile
[tex]0.805[/tex] km
[tex]805[/tex] m
according to third equation of motion,
[tex]\text{v}^{2}-\text{u}^{2}=2\text{a}\text{S}[/tex]
as speed is lowering down the acceleration will be in the opposite direction of motion, hence acceleration will be negative, equation will become
[tex]\text{u}^{2} -\text{v}^{2}=2a\text{S}[/tex]
putting values,
[tex]32.63^{2}[/tex]-[tex]24.58^{2}[/tex]=[tex]2a\times805[/tex]
a=[tex]460.55\setminus1610[/tex]
a≅[tex]0.29[/tex][tex]\text{m}\text{s}^{2}[/tex]
the constant acceleration that will drop speed to [tex]55 \text{mph}[/tex] is [tex]0.29[/tex][tex]\text{m}\text{s}^{2}[/tex]
The constant acceleration of the driver is -0.286 m/s²
The given parameters;
initial velocity of the driver, v = 73 mph
final velocity of the driver, v = 55 mph
distance available, d = 0.5 mile
The constant acceleration of the driver is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{55^2 - 73^2}{2(0.5)} \\\\a = -2,304\ mi/h^2[/tex]
The acceleration in m/s²
1 mile = 1609 m
[tex]a = \frac{-2304 \ mi}{hr^2} \times \frac{1609 \ m}{1 \ mile} \times \frac{1 \ hr^2}{(3600)^2 \ s^2} \\\\a = -0.286 \ m/s^2[/tex]
Thus, the constant acceleration of the driver is -0.286 m/s²
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A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) calculate the average force on the person if he is stopped by a padded dashboard that compresses an average of 1.00 cm. (b) calculate the average force on the person if he is stopped by an air bag that compresses an average of 15.0 c
(a) [tex]-1.5\cdot 10^6 N[/tex]
First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:
[tex]v^2 -u ^2 = 2ad[/tex]
where
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 1.00 cm = 0.01 m is the displacement of the person
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.01)}=-20000 m/s^2[/tex]
And the average force on the person is given by
[tex]F=ma[/tex]
with m = 75.0 kg being the mass of the person. Substituting,
[tex]F=(75)(-20000)=-1.5\cdot 10^6 N[/tex]
where the negative sign means the force is opposite to the direction of motion of the person.
b) [tex]-1.0\cdot 10^5 N[/tex]
In this case,
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 15.00 cm = 0.15 m is the displacement of the person with the air bag
So the acceleration is
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.15)}=-1333 m/s^2[/tex]
So the average force on the person is
[tex]F=ma=(75)(-1333)=-1.0\cdot 10^5 N[/tex]
The average force on the person if he is stopped by a padded dashboard is -1.5 x 10⁶ N.
The average force on the person if he is stopped by an air bag is -10,000 N.
The given parameters;
mass of the person, m = 75 kgvelocity of the person, u = 20 m/sThe deceleration of the car when the padded dashboard compresses an average of 1.00 cm.
s = 0.01 m
[tex]v^2 = u^2 + 2as\\\\0 = 20^2 + (2\times 0.01)a\\\\0 = 400 + 0.02a\\\\0.02a = -400\\\\a = \frac{-400}{0.02} \\\\a = -20,000 \ m/s^2[/tex]
The average force is calculated as;
F = ma
F = 75 x (-20,000)
F = -1.5 x 10⁶ N.
The deceleration of the car when the air bag compresses an average of 15 cm.
s = 0.15 m
[tex]v^2 = u^2 + 2as\\\\0 = u^2 + 2as\\\\-2as = u^2\\\\a = \frac{u^2}{-2s} = \frac{20^2}{-2\times 0.15} = -1,333.33 \ m/s^2[/tex]
The average force is calculated as follows;
F = ma
F = 75 x (-1,333.33)
F = -10,000 N.
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The 3.6-lb rod AB is hanging in the vertical position. A 2.2-lb block, sliding on a smooth horizontal surface with a velocity of 12 ft/s, strikes the rod at its end B.
What is the velocity of the block immediately after the collision. The coefficient of restitution between the block and the rod at B is e = 0.85?
Answer:
v = 4.17 m/s
Explanation:
as we know that block is moving towards the end of the rod
so let say just after the collision the block is moving with speed v' and rod is rotating with speed "w"
so we have angular momentum conservation about the hinge point
[tex]mvL = mv'L + I\omega[/tex]
[tex](2.2)(12)L = (2.2)v'(L) + (\frac{3.6 L^2}{3}) \omega[/tex]
[tex]12 = v' + 0.545(\omega L)[/tex]
now by the equation of coefficient of restitution we can say
[tex]0.85 = \frac{(\omega L) - v'}{12}[/tex]
now we have
[tex]\omega L - v' = 10.2[/tex]
now we have
[tex]\omega L = 14.37 [/tex]
[tex]v' = 4.17 m/s[/tex]
so velocity of block just after collision is 4.17 m/s
To find the velocity of the block after the collision, we can use the principle of conservation of momentum and the coefficient of restitution.
Explanation:First, we need to calculate the velocity of the 3.6-lb rod AB after the collision with the 2.2-lb block.
To do this, we can use the principle of conservation of momentum. The momentum before the collision is given by the sum of the momentums of the rod and the block: (3.6 lb) x (0 ft/s) + (2.2 lb) x (12 ft/s) = 26.4 lb-ft/s.
The coefficient of restitution (e) is defined as the ratio of the relative velocity of separation to the relative velocity of approach. In this case, the relative velocity of approach is 12 ft/s, and the relative velocity of separation is the sought-after velocity of the block immediately after the collision.
When you touch a cold piece of ice with your finger,energy flowsA) from your finger to the ice.B) from the ice to your finger.C) actually, both ways.
Answer:
I believe the answer is C.) actually, both ways
hope this helps! (:
Explanation:
A 1024 Hz tuning fork is used to obtain a series of resonance levels in a gas column of variable length, with one end closed and the other open. The length of the column changes by 20 cm from resonance to resonance. From this data, the speed of sound in this gas is:
Answer:
The speed of sound in this gas is 409.6 m/s.
Explanation:
The length of the column changes from 20 cm from resonance to resonance. Thus,
[tex]L=\frac {(2n+1)\lambda}{4}[/tex]
The length change from one resonance to resonance. so, there is 1 loop change. So,
ΔL = 1 loop = λ/2
ΔL = 20 cm (given)
Also, 1 cm = 0.01 m
So,
ΔL = 0.2 cm (given)
The wavelength is:
λ = ΔL×2
λ = 2x0.2 = 0.4 m
Given:
Frequency (ν) = 1024 Hz
Velocity of the sound in the gas = ν×λ = 1024×0.4 m/s = 409.6 m/s
BASEBALL IN SPACE baseball in space You are observing a baseball in space, with mass m = 0.145 kg. It is moving past your spacecraft with speed v = 22.9 m/sec, to the north. Your Acme Tractor Beam can exert 1.3 Newtons of pull force on that baseball. CALCULATE: How long will it take your Acme Tractor Beam to slow down the baseball to a stop? Enter the numeric part of your answer, to the nearest hundredth of a second, i.e., 0.01 sec. E.g., if your answer is 60.5329 sec, then type in 60.53 in the answer box.
Answer:
It takes 2.55 sec
Explanation:
First we need to find the aceleration with this equation:
F = m*a
[tex]a = \frac{F}{m}[/tex]
Where: F = 1.3 Newtons
m = 0.145 kg
Then [tex]a = \frac{1.3 Newtons}{0.145 kg}[/tex]
[tex]a = 8.9655 m/sec^{2}[/tex]
Now we are ready to calculate the time with the next equation:
[tex]V_{f} = V_{o} + a*t[/tex]
[tex]V_{f} - V_{o} = a*t[/tex]
[tex]t = \frac{V_{f} - V_{o}}{a}[/tex]
Where: [tex]V_{o} = 22.9 m/sec[/tex] (at the beginning)
[tex]V_{f} = 0 m/sec[/tex] (at the end because it stops)
We must use [tex]a = - 8.9655 m/sec^{2}[/tex] because in this case speed is decreasing
Finally: [tex]t = \frac{0 m/sec - 22.9 m/sec}{- 8.9655 m/sec^{2} }[/tex]
t = 2.55 sec
By applying Newton's second law of motion and the motion equation, we find that it will take approximately 2.55 seconds for the Acme Tractor Beam to slow down the baseball to a stop in space.
Explanation:This problem is a classic example of the application of Newton's second law of motion which states: F = ma, where F is the force, m is the mass, and a is the acceleration. Here we are given the force F exerted by your Acme Tractor Beam (1.3 N) and the mass of the baseball m (0.145 kg).
First, use F=ma to calculate acceleration a. The acceleration a = F/m = 1.3 N / 0.145 kg = 8.97 m/s². This is the rate at which the baseball would slow down under the influence of the tractor beam.
Next, use the formula v = u + at, where 'v' is the final velocity, 'u' is the initial velocity, 't' is the time, and 'a' is the acceleration we just calculated. We want to know the time 't' when the baseball comes to stop (v = 0), so we rearrange this equation to solve for 't': t = (v - u) / a.
Plugging in the given numbers: t = (0 - 22.9 m/s) / -8.97 m/s², we find t = 2.55 s.
So it will take approximately 2.55 seconds for your Acme Tractor Beam to slow down the baseball to a stop.
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If the amplitude of a sound wave is tripled, by what factor will the intensity increase?
Answer:
By a factor 9
Explanation:
The intensity of a sound wave is proportional to the square of the amplitude of the wave:
[tex]I \propto A^2[/tex]
where
I is the intensity
A is the amplitude of the wave
In this problem, the amplitude of the sound wave is increased by a factor 3:
A' = 3A
So the intensity would change by
[tex]I' \propto A'^2 = (3A)^2 = 9 A^2 = 9I[/tex]
So, the intensity would increase by a factor 9.
Final answer:
Tripling the amplitude of a sound wave increases its intensity by a factor of nine, since intensity is proportional to the square of the amplitude.
Explanation:
If the amplitude of a sound wave is tripled, the intensity of the sound wave will increase by a factor of nine (3²), because intensity is proportional to the square of the amplitude.
The relationship between amplitude and intensity for sound waves tells us that if you increase the amplitude (A) of a wave, the intensity (I), which is the power per unit area (W/m²), changes according to the formula I ≈ A². Therefore, if the initial intensity is I₀ when the amplitude is A₀, tripling the amplitude to 3A₀ means the new intensity I will be (3A₀)² = 9A₀² = 9I₀.
Example:
If the original intensity of a sound wave is 1.00 W/m² and the amplitude is tripled, the new intensity would be 9.00 W/m².
Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence of dark energy. Rank each model from left to right based on the ratio of its actual mass density to the critical density, from smallest ratio (mass density much smaller than critical density) to largest ratio (mass density much greater than critical density).
(A) coasting universe
(B) critical universe
(C) recollapsing universe
Answer:
1) Recollapsing universe
2) critical universe
3) Coasting universe
Explanation:
According to the smallest ration (ratio actual mass density to current density) to largest ration, rank of models for expansion of universe are
1) Recollapsing universe -in this, metric expansion of space is reverse and universe recollapses.
2) critical universe - in this, expansion of universe is very low.
3) Coasting universe - in this, expansion of universe is steady and uniform
You recently purchased a large plot of land in the Amazon jungle at an extremely low cost. You are quite pleased with yourself until you arrive there and find that the nearest source of electricity is 1500 miles away, a fact that your brother-in-law, the real estate agent, somehow forgot to mention. Since the local hardware store does not carry 1500-mile-long extension cords, you decide to build a small hydroelectric generator under a 35.0-m high waterfall located nearby. The flow rate of the waterfall is 0.150x10^2 m^3/h, and you anticipate needing 1750 kW h/wk to run you lights, air conditioner, and television.
What is the maximum power theoretically available from the waterfall? Is the power sufficient to meet your needs?
Answer:
The maximum power available from the water fall is 239.568 kWh/week.
Explanation:
Given:
Flow rate = 0.15 × 10² m³/h
now the mass of water flowing per hour will be = flow rate × mass density of water(i.e 1000 kg/m³)
mass of water flowing per hour will be = 0.15 × 10² m³/h × 1000 kg/m³ = 15000 kg/h
The gravitational potential energy of the falling water = mgh = 15000 × 9.8 × 35 = 5145000 J/h
or
5145000/3600 = 1426.166 J/s (as 1 h = 3600 seconds)
or
1426.166 W = 1.426 kW
Now, the number of hours in a week = 7 × 24 = 168 hours
Now the energy produce in a week = 1.426 kW × 168 hr = 239.568 kWh/week.
No it is not sufficient to meet the needs
The maximum hydraulic power generated by the waterfall is around 14.35 kW, which is more than the required energy for weekly consumption (2.28 kW). Therefore, the waterfall can especially serve as a power source.
Explanation:The problem involves the calculation of hydraulic power, which is given by the formula Power = p * g * h * Q, where p is the density of water (around 1000 kg/m³ in standard conditions), g is the gravitational acceleration (approximated as 9.8 m/s²), h is the waterfall height (in meters), and Q is the flow rate (in m³/s). First, we need to convert the flow rate from m³/h to m³/s. We get Q = 0.150 * 10² m³/h = 0.04167 m³/s.
After substituting the figures into the formula, we get: Power = 1000 kg/m³ * 9.8 m/s² * 35.0 m * 0.04167 m³/s = 14350.5 Watts, or about 14.35 kW.
Given the energy requirement per week is 1750 kWh, let's convert it to the energy required per second. That is 1750 kWh/wk = (1750 * 1000) / (7 * 24 * 60 * 60) = 2.28 kW. This means, the maximum power available from the waterfall is more than enough to meet your requirements.
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A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of 0a= 36.0 degrees , the ray refracted into the water makes an angle of 49.6 degrees with the normal to the interface.What is the smallest value of the incident angle 0a for which none of the ray refracts into the water? Express your answer with the appropriate units.
Answer:
[tex]\theta = 50.5 degree[/tex]
Explanation:
As we know that the refractive index of the water is
[tex]\mu_w = \mu_1[/tex]
also we know that refractive index of the glass is
[tex]\mu_g = \mu_2[/tex]
now we know by Snell's law
[tex]\mu_1 sin\theta_1 = \mu_2 sin\theta_2[/tex]
so we have
[tex]\mu_1 sin36 = \mu_2 sin49.6[/tex]
[tex]\frac{\mu_1}{\mu_2} = 1.2956[/tex]
now if no light will refract into the water then in that case
[tex]\mu_1 sin\theta = \mu_2 sin90[/tex]
now we have
[tex](1.2956) sin\theta = 1[/tex]
[tex]sin\theta = 0.77[/tex]
[tex]\theta = 50.52 degree[/tex]
Suppose an object in free fall is dropped from a building. its starting velocity is 0 m/s. whats its speed (in m/s) of the object after falling 2 seconds?
Answer:
19.6 m/s
Explanation:
If we take down to be positive, the acceleration due to gravity is 9.8 m/s².
Acceleration = change in velocity / change in time
a = Δv / Δt
9.8 m/s² = (v − 0 m/s) / 2 s
v = 19.6 m/s
The speed after 2 seconds is 19.6 m/s.
When electrons are removed from the outermost shell of a calcium atom, the atom becomesA. an anion that has a larger radius than the atomB. an anion that has a smaller radius than the atom.C. a cation that has a larger radius than the atom.D. a cation that has a smaller radius than the atom.
Answer:
D. a cation that has a smaller radius than the atom.
Explanation:
When electrons are removed from the outermost shell of a calcium atom, the atom becomes a cation that has a smaller radius than the atom.
When electrons are removed from the outermost shell of a calcium atom, the atom becomes: D. a cation that has a smaller radius than the atom.
An ion can be defined as an atom or molecules (group of atoms) that has either lost or gained one or more of its valence electrons, thereby, making it to have a net positive or negative electrical charge respectively.
Basically, there are two (2) types of ion and these are:
Anion.Cation.A cation is a positive electrical charge which is formed when the atom of a chemical element losses an electron.
Hence, a cation is formed when an electron is removed from the atom of a chemical element.
Generally, the atomic size of a chemical element would increase when its number of electrons increase, thereby, increasing its atomic radius.
Consequently, the radius of a cation in comparison with the radius of its neutral atom becomes smaller because it losses more electron from the outermost shell of its atom.
In conclusion, when electrons are removed from the outermost shell of a calcium atom, the atom becomes a cation that has a smaller radius than its atom.
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A frog jumps for 4.0 seconds at a maximum horizontal distance of 0.8m. what is its velocity along the road?
Answer:
The frog's horizontal velocity is 0.2 m/s.
Explanation:
To solve this problem, we must first remember what velocity is and how we solve for it. Velocity can be solved for using the formula x/t, where x represents horizontal distance and t represents time (in seconds), that it takes to travel this distance. If we plug in the given numbers for these variables and solve, we get the following:
v = x/t
v = 0.8m/4s
v = 0.2 m/s
Therefore, the correct answer is 0.2 m/s. We can verify that these units are correct because the formula calls for distance divided by time, so meters per second is a sensible answer.
Hope this helps!
Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the centerline does the second bright fringe occur?
Answer: [tex]1.8\°[/tex]
Explanation:
The diffraction angles [tex]\theta_{n}[/tex] when we have a slit divided into [tex]n[/tex] parts are obtained by the following equation:
[tex]dsin\theta_{n}=n\lambda[/tex] (1)
Where:
[tex]d=3.4(10)^{-5}m[/tex] is the width of the slit
[tex]\lambda=540 nm=540(10)^{-9}m[/tex] is the wavelength of the light
[tex]n[/tex] is an integer different from zero.
Now, the second-order diffraction angle is given when [tex]n=2[/tex], hence equation (1) becomes:
[tex]dsin\theta_{2}=2\lambda[/tex] (2)
Now we have to find the value of [tex]\theta_{2}[/tex]:
[tex]sin\theta_{2}=\frac{2\lambda}{d}[/tex] (3)
Then:
[tex]\theta_{2}=arcsin(\frac{2\lambda}{d})[/tex] (4)
[tex]\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})[/tex] (5)
Finally:
[tex]\theta_{2}=1.8\°[/tex] (6)
A 10-cm-long spring is attached to the ceiling. When a 2.0 kg mass is hung from it, the spring stretches to a length of 15 cm.?A) What is the spring constant k? B) How long is the spring when a 3.0 kg mass is suspended from it?
A) 392 N/m
The spring constant can be found by applying Hooke's Law:
F = kx
where
F is the force applied to the spring
k is the spring constant
x is the stretching of the spring
Here we have:
F is the weight of the block hanging from the spring, which is
[tex]F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N[/tex]
The stretching of the spring is
[tex]x=15 cm - 10 cm = 5 cm = 0.05 m[/tex]
Therefore its spring constant is
[tex]k=\frac{F}{x}=\frac{19.6 N}{0.05 m}=392 N/m[/tex]
B) 17.5 cm
Now that we know the value of the spring constant, we can calculate the new stretching of the spring when a mass of m=3.0 kg is applied to it. In this case, the force applied on the spring is
[tex]F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N[/tex]
Therefore the stretching of the spring is
[tex]x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m = 7.5 cm[/tex]
And since the natural length of the spring is 10 cm, the new length will be
L = 10 cm + 7.5 cm = 17.5 cm
A horizontal force of 50 N is required to push a wagon across a sidewalk at a constant speed.
a. What is the net (unbalanced) force acting on the wagon?
b. What forces are acting on the wagon?
c. If the mass on the box increased, use Newton's law to explain what must change to get the object moving at constant speed.
A) The net unbalanced force acting on the wagon is; F_net = 50 N
B) The forces acting on the wagon are;
Applied Force = 50 N
Frictional Force = 50 N
C) What will happen if mass is increase is that;
frictional force will increase.
Formula for force is;
Force = mass x acceleration
F = ma
A) We are told that the speed is constant and at constant speed, acceleration is zero. Thus; F_net = 0 N
B) We are told that the force required to push the wagon at constant speed is 50 N.
Now, from newtons third law of motion, it states that to every action, there is an equal and opposite reaction.
Thus, there will be an equal opposite reaction which will be the frictional force. Thus; F_f = 50 N
C) Frictional force has the formula;
F_f = μmg
So if the mass is increased, the frictional force will also increase.
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(a) The net force acting on the wagon is zero (0).
(b) The forces acting on the wagon are, the applied horizontal force on the wagon and frictional force between the wagon and the sidewalk.
(c) As the mass increases, the applied force should increase by equal amount to keep the moving at constant speed.
The given parameters;
the horizontal force applied on the wagon, F = 50 NThe net horizontal force acting on the wagon is given as;
[tex]\Sigma F_x = 0\\\\F - F_k = ma[/tex]
where;
Fk is the frictional force between the wagon and the sidewalkm is the mass of the wagona is the acceleration of the wagonAt constant speed, the acceleration of the wagon = 0
[tex]F - F_k = m(0)\\\\F- F_k = 0\\\\[/tex]
Thus, the net force acting on the wagon is zero (0).
(b) The forces acting on the wagon are;
the applied horizontal force on the wagonfrictional force between the wagon and the sidewalk(c) The Newton's second law of motion is given as;
F = ma
[tex]F = \frac{m(\Delta v)}{t} \\\\\Delta v =\frac{Ft}{m} \\\\\frac{F_1}{m_1} = \frac{2F_1}{2m_1}[/tex]
Thus, as the mass increases, the applied force should increase by equal amount to keep the moving at constant speed.
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A 5.30 m long pole is balanced vertically on its tip. What will be the speed (in meters/second) of the tip of the pole just before it hits the ground? (Assume the lower end of the pole does not slip.)
Answer:
v = 17.66 m/s
Explanation:
As we know that the lower end of the pole is fixed in the ground and it start rotating about that end
so here we can say that the gravitational potential energy of the pole will convert into rotational kinetic energy of the pole about its one end
so we have
[tex]mgL = \frac{1}{2}(\frac{mL^2}{3})\omega^2[/tex]
so we have
[tex]\omega = \sqrt{\frac{6g}{L}}[/tex]
now we have
[tex]\omega = \sqrt{\frac{6(9.81)}{5.30}}[/tex]
[tex]\omega = 3.33 rad/s[/tex]
now the speed of the other tip of the pole is given as
[tex]v = \omega L[/tex]
[tex]v = (3.33)(5.30) = 17.66 m/s[/tex]
Review Problem. A light string with a mass per unit length of 8.20 g/m has its ends tied to two walls separated by a distance equal to three-fourths the length of the string (Fig. P16.30). An object of mass m is suspended from the center of the string, putting a tension in the string. (A) (mg)^1/2(B) 1/4(mg x 1000)^1/2 (C) 1/8(5mg x 1000)^1/2(D) None
Answer:
The answer is 'D' none
Explanation:
In the figure shown we have
[tex]2Tsin(\theta )=mg\\\\\therefore T=\frac{mg}{2sin(\theta )}[/tex]
From the figure we can see that
[tex]cos(\theta )=\frac{\frac{3L}{8}}{\frac{L}{2}}\\\\\therefore \theta =cos^{-1}(\frac{3}{4})\\\\sin(\theta )=\frac{\sqrt{7}}{4}[/tex]
Thus value of tension be will be
[tex]T=\frac{2mg}{\sqrt{7}}[/tex]
The tension in the string when an object of mass m is suspended from the center is given by T = (mg)/(2L), where T is the tension, m is the mass of the object, g is the acceleration due to gravity, and L is the length of the string. So, the correct answer is (A) (mg)^1/2.
Explanation:To find the tension in the string when an object of mass m is suspended from the center, we can use the formula for tension in a string:
T = (mg)/(2L)
Where T is the tension, m is the mass of the object, g is the acceleration due to gravity, and L is the length of the string.
Substituting the given values, the tension in the string is:
T = (m)(9.8)/(2L)
Therefore, the correct answer is (A) (mg)^1/2.
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A stone is dropped from the upper observation deck of a tower, 300 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the distance (in meters) of the stone above ground level at time t. h(t) = Correct: Your answer is correct. (b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.) 7.82 Correct: Your answer is correct. s (c) With what velocity does it strike the ground? (Round your answer to one decimal place.) -76.64 Correct: Your answer is correct. m/s (d) If the stone is thrown downward with a speed of 2 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.) 8.00 Incorrect: Your answer is incorrect. s
Answers:
(a) [tex]y(t)=300m-4.9m/s^{2}t^{2}[/tex]
(b) [tex]t=7.82s[/tex]
(c) [tex]V=-76.7 m/s[/tex]
(d) [tex]t=7.6s[/tex]
Explanation:
This problem is a good example of vertical motion, and the main equations for this situation are as follows:
[tex]y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex] (1)
[tex]V=V_{o}-gt[/tex] (2)
Where:
[tex]y[/tex] is the height of the stone at a given time
[tex]y_{o}=300m[/tex] is the initial height of the stone
[tex]V_{o}[/tex] is the initial velocity of the stone
[tex]t[/tex] is the time
[tex]g=9.8m/s^{2}[/tex] is the acceleration due to gravity on Earth
Now, for parts (a), (b) and (c) we are specifically talking about free fall (where the main condition is that the initial velocity must be zero [tex]V_{o}=0[/tex]).
How do we know this?
Because we are told "the stone is dropped".
Having this clear, let's start with the calculations:
(a) Find the distance of the stone above ground level at time t:In order to approach this, we will use equation (1), remembering the condition [tex]V_{o}=0[/tex]:
[tex]y(t)=y_{o}-\frac{1}{2}gt^{2}[/tex] (3)
[tex]y(t)=300m-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex] (4)
[tex]y(t)=300m-4.9m/s^{2}t^{2}[/tex] (5) This is the distance of the stone above ground level at time t
(b) How long does it take the stone to reach the ground?:In this case, [tex]y=0[/tex], because we have to find the time when the stone finally hits the ground.
Rewritting (1) with this condition:
[tex]0=y_{o}-\frac{1}{2}gt^{2}[/tex] (6)
Isolating [tex]t[/tex]:
[tex]t=\sqrt{\frac{2y_{o}}{g}}[/tex] (7)
[tex]t=\sqrt{\frac{2(300m)}{9.8m/s^{2}}}[/tex] (8)
Then:
[tex]t=7.82s[/tex] (9)
(c) With what velocity does it strike the ground?In this part, rewritting equation (2) will be usefull, remembering [tex]V_{o}=0[/tex] and using the timewe found in (9), which is the time when the stone strikes the ground:
[tex]V=-gt[/tex] (10)
[tex]V=-(9.8m/s^{2})(7.82s)[/tex] (11)
[tex]V=-76.68m/s \approx -76.7 m/s[/tex] (12) Note the velocity has a negative sign because its direction is downwards.
(d) Finding [tex]t[/tex] when [tex]y_{o}=0[/tex] and [tex]V_{o}\neq 0[/tex]:Now the initial velocity is [tex]V_{o}=-2m/s[/tex] (downwards). So, let's use equation (1) again an find [tex]t[/tex]:
[tex]0=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex] (13)
[tex]0=300m+(-2m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex]
[tex]0=-4.9m/s^{2}t^{2}+(-2m/s)t+300m[/tex] (14)
At this point we have a quadratic equation of the form [tex]0=at^{2}+bt+c[/tex], and we have to use the quadratic formula if we want to find [tex]t[/tex]:
[tex]t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex] (15)
Where [tex]a=-4.9[/tex], [tex]b=-2[/tex], [tex]c=300[/tex]
Substituting the known values and choosing the positive result of the equation:
[tex]t=7.623 s \approx 7.6 s[/tex] This is the time it takes to the stone to reach the ground when [tex]V_{o}=-2m/s[/tex]
A stone dropped from a height of 300m reaches the ground with a velocity of -76.64 m/s after approximately 7.82s. If it's thrown downwards initially with a speed of 2m/s, it takes about 7.774s to reach the ground.
Explanation:In this physics problem, a stone is dropped from the upper observation deck of a tower, specifically 300 m above the ground. Because we're assuming acceleration due to gravity (g) is 9.8 m/s², we can calculate necessary information about the stone's fall.
For part (a), the distance (in meters) of the stone above ground level at time t could be found using the equation h(t) = 300-9.8*t².
To determine how long it takes the stone to reach the ground (part b), we can rearrange the previous equation and solve for t, which gives us t = sqrt(300/9.8) which is approximately 7.82s.
For part (c), one can find the velocity at which the stone strikes the ground by using the formula v = gt, which gives us v = 9.8*7.82 = -76.64 m/s.
Finally for (d), if the stone is thrown downward with a speed of 2 m/s, we again adjust our distance equation and solve for t, which gives us t = (300 + v²/2*g) / v = 7.774s approximately.
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the acceleration means that for each second that passes, falling objects move 9.8 meters.
A. True
B. False
Answer:
B. False
Explanation:
An acceleration of 9.8 m/s² means the velocity increases by 9.8 m/s every second.
What is meant by constructive and destructive interference
An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s.(a) Calculate the amplitude of the motion._____ m(b) Calculate the maximum velocity attained by the object. [Hint: Use conservation of energy.]______ m/s
Answer:
Part a)
A = 0.066 m
Part b)
maximum speed = 0.58 m/s
Explanation:
As we know that angular frequency of spring block system is given as
[tex]\omega = \sqrt{\frac{k}{m}}[/tex]
here we know
m = 3.5 kg
k = 270 N/m
now we have
[tex]\omega = \sqrt{\frac{270}{3.5}}[/tex]
[tex]\omega = 8.78 rad/s[/tex]
Part a)
Speed of SHM at distance x = 0.020 m from its equilibrium position is given as
[tex]v = \omega \sqrt{A^2 - x^2}[/tex]
[tex]0.55 = 8.78 \sqrt{A^2 - 0.020^2}[/tex]
[tex]A = 0.066 m[/tex]
Part b)
Maximum speed of SHM at its mean position is given as
[tex]v_{max} = A\omega[/tex]
[tex]v_{max} = 0.066(8.78) = 0.58 m/s[/tex]
An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first [math]3.00 \mu s[/math] after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.
To determine the magnitude and direction of the electric field when an electron is released in a uniform electric field, we use the displacement formula and compare the acceleration due to the electric field with the acceleration due to gravity.
Explanation:(a) To determine the magnitude of the electric field, we can use the formula E = delta V / delta x. The electron experiences constant acceleration, so the displacement equation is given by x = (1/2)at^2, where x is the displacement, a is the acceleration, and t is the time. Plugging in the given values, we find that the acceleration is 5.00 x 10^11 m/s^2 (upward).
(b) We can justify ignoring the effects of gravity by comparing the acceleration due to the electric field with the acceleration due to gravity. For an electron in a uniform electric field, the acceleration is much greater than the acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s^2. Therefore, the effects of gravity can be ignored in this scenario.
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The magnitude and direction of the electric field can be determined by calculating the electric field strength. Gravity can be ignored in this situation because the electric field is much stronger.
Explanation:(a) To find the magnitude and direction of the electric field, we can use the equation:
Electric field strength (E) = Force (F) / Charge (q)
Given that the electron is accelerating upward, the direction of the electric field is downward. Using the formula, we can rearrange it to find the electric field strength:
Electric field strength = Force / Charge = mass x acceleration / charge = (9.11 x 10^-31 kg) x (9.8 m/s^2) / (1.6 x 10^-19 C) = 5.67 x 10^11 N/C
The magnitude of the electric field is 5.67 x 10^11 N/C, and the direction is downward.
(b) We can justify ignoring the effects of gravity by comparing the magnitude of the electric field to the gravitational field strength. The gravitational field strength near the surface of the Earth is approximately 9.8 N/kg. Comparing this to the electric field strength of 5.67 x 10^11 N/C, we can see that the electric field is much stronger than the gravitational field. Therefore, the effects of gravity can be ignored.
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Suppose the energy transferred to a dead battery during charging is W. The recharged battery is then used until fully discharged again. Is the total energy transferred out of the battery during use also W?
Answer:
No
Explanation:
The amount of energy transferred out of battery will not be the same as given during charging.
A battery has its internal resistance as well which will draw some energy, through this internal resistance energy is lost
P = I² R
where P is energy, I is current passing and R is resistance.
As per law of conservation energy will be saved as it is the sum of drawn energy and energy wasted in internal resistance. But the total energy transferred out will not be same.
On the earth, when an astronaut throws a 0.250 kg stone vertically upward, it returns to his hand a time T later. On planet X he finds that, under the same circumstances, the stone returns to his hand in 2T. In both cases, he throws the stone with the same initial velocity and it feels negligible air resistance. What is the acceleration due to gravity on planet X (in terms of g)?
Answer:[tex]\frac{g}{2}[/tex]
Explanation:
Given
mass of stone[tex]\left ( m\right )[/tex]=0.250 kg
Let initial velocity with which it is thrown upward is u
therefore after time t it's velocity is zero at highest point
t=[tex]\frac{u}{g}[/tex]
where g= gravity at earth
therefore [tex]T=2\times \frac{u}{g}[/tex]-------1
Now same thing is done in Planet X where gravity is g'
therefore time taken by stone to reach surface is
[tex]T'=T=2\times \frac{u}{g'}[/tex]-------2
Divide 1 & 2
[tex]\frac{T}{T'}[/tex]=[tex]\frac{g'}{g}[/tex]
[tex]\frac{1}{2}[/tex]=[tex]\frac{g'}{g}[/tex]
g'=[tex]\frac{g}{2}[/tex]
A campground official wants to measure the distance across an irregularly-shaped lake, but can't do so directly. He picks a point south of the lake. From that point to the left side of the lake, the distance is 105 meters. From that point to the right side of the lake is 119 meters. The angle between these two measurements is 83 degrees. How far is it across the lake?
Answer:
149 m
Explanation:
The distances across the lake is forming a triangle.
let the distance between the point and the left side be 'x'
and the distance between the point and the right be 'y'
and the distance across the lake be 'z' and the angle opposite to 'z' be 'Z' given:
∠Z = 83°
x = 105 m
y = 119 m
Now, applying the Law of Cosines, we get
z² = x² + y² - 2xycos(Z)
Substituting the values in the above equation, we get
z² = 105² + 119² - 2×105×119×cos(83°)
or
z = √22140.48
or
z = 148.796 m ≈ 149 m
The point is 149 m across the lake
Which of the following statements are true?
(A)At any junction point in a circuit, the sum of all the currents entering the junction must equal the sum of all the currents leaving the junction.
(B)Kirchhoff's junction rule is based on the conservation of energy.
(C)The sum of the changes in potential around any closed loop of a circuit must equal zero.
(D)Kirchhoff's loop rule is based on the conservation of charge.
The true statements are: (A) At any junction point in a circuit, the sum of all the currents entering the junction must equal the sum of all the currents leaving the junction. (C) The sum of the changes in potential around any closed loop of a circuit must equal zero. (D) Kirchhoff's loop rule is based on the conservation of energy.
Explanation:(A) At any junction point in a circuit, the sum of all the currents entering the junction must equal the sum of all the currents leaving the junction. This is known as Kirchhoff's junction rule or the law of conservation of charge.
(B) Kirchhoff's junction rule is based on the conservation of charge, not the conservation of energy.
(C) The sum of the changes in potential around any closed loop of a circuit must equal zero. This is known as Kirchhoff's loop rule or the law of conservation of energy.
(D) Kirchhoff's loop rule is based on the conservation of energy, not the conservation of charge.
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On July 15, 2004, NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit 705 km above the earth's surface, and we shall assume a circular orbit.
(a) How many hours does it take this satellite to make one orbit?
h
(b) How fast (in km/s) is the Aura spacecraft moving?
km/s
Answer:
a). T = 1.64 hr
b). v = 7.503 km/s
Explanation:
Given :
NASA launched the Aura spacecraft to study the earth's climate and atmosphere.
Height of the satellite orbit from the earth's surface, h = 705 km
= 705000 m
Therefore we know that,
a).Time period of the space craft is
[tex]T = 2\pi \sqrt{\frac{(R+h)^{3}}{G\times M}}[/tex]
where, G = Universal Gravitational constant
= [tex]6.67 \times 10^{-11} N-m^{2}/kg^{2}[/tex]
M = Mass of the earth
= 5.98 x [tex]10^{24}[/tex] kg
R = Radius of the earth
= 6.38 x [tex]10^{6}[/tex] m
∴[tex]T = 2\pi \sqrt{\frac{(R+h)^{3}}{G\times M}}[/tex]
[tex]T = 2\pi \sqrt{\frac{((6.36\times 10^{6})+705000)^{3}}{6.67\times 10^{-11}\times 5398\times 10^{24}}}[/tex]
[tex]T = 5933[/tex] s
= 1.64 hr
Thus, the satellite will take 1.64 hr to make one orbit.
b). We know velocity of the spacecraft is given by
[tex]v=\sqrt{\frac{G\times M}{R+h}}[/tex]
[tex]v=\sqrt{\frac{6.67\times 10^{-11}\times 5.98\times 10^{24}}{(6.38\times 10^{6})+705000}}[/tex]
v = 7503 m/s
= 7.503 km/s
Thus, the Aura satellite is moving with velocity v = 7.503 km/s
A) The number of hours it takes for the satellite to make one orbit ( h ) = 1.64 hours
b) The speed of the Aura spacecraft = 7.5 Km/s
Given data
Height of satellite Orbit ( h ) = 705 km ≈ 705000 m
A) Determine the time taken in hours for the satellite to make a single orbit
T = [tex]2\pi \sqrt{\frac{(R+h)^3}{GM} }[/tex] -------- ( 1 )
where ; G = 6.67 * 10⁻¹¹ N-m²/kg², M ( mass of earth ) = 5.98 * 10²⁴,
R = 6.38 * 10⁶ m , h = 705000 m
Insert the values into equation ( 1 )
T = 5933 secs
= 1.64 hours ( time taken to complete one orbit )
B) Determine how fast Aura spacecraft is moving
V = [tex]\sqrt{\frac{GM}{R+h} }[/tex] -------- ( 2 )
where ; G = 6.67 * 10⁻¹¹ N-m²/kg², M ( mass of earth ) = 5.98 * 10²⁴,
R = 6.38 * 10⁶ m , h = 705000 m
Insert values into equation ( 2 )
∴ V = 7503 m/s
= 7.5 km/s
Hence we can conclude that the number of hours it takes for the satellite to make one orbit ( h ) = 1.64 hours and The speed of the Aura spacecraft = 7.5 Km/s
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easy A solid disk rotates in the horizontal plane at an angular velocity of 0.067 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.10 kg · m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?
Answer:
[tex]\omega_f = 0.0067\ rad/s[/tex]
Explanation:
Given:
Initial angular velocity of the disc, [tex]\omega_o[/tex] = 0.067 rad/s
Initial moment of inertia of the disc, [tex]I_o[/tex] = 0.10 kg.m²
Distance of sand ring from the axis, r = 0.40m
Mass of the sand ring = 0.50 kg
Now, no external torque is applied to the rotating disk.
Thus, the angular momentum of the system will remain conserved.
Also,from the properties of moment of inertia, the addition of the sand ring will increase the initial moment of inertia by an amount Mr²
thus, we have
Initial Angular Momentum = Final Angular Momentum
or
[tex]I_o\omega_o = I_f\times\omega_f[/tex]
[tex]I_o\omega_o = (I_o + Mr^2)\times\omega_f[/tex]
Where,
[tex]I_f[/tex] = Final moment of inertia
[tex]\omega_f[/tex] = Final angular velocity
substituting the values, we get
[tex]0.10\times0.067 = (0.10 + 0.50\times 0.40^2)\times\omega_f[/tex]
or
[tex]0.0067 = (0.18)\times\omega_f[/tex]
or
[tex]\omega_f = 0.0067\ rad/s[/tex]
The rate of change of angular displacement is defined as angular velocity. After all the sand is in place the angular velocity of the disk will be 0.067 rad/sec.
What is the definition of Angular velocity?The rate of change of angular displacement is defined as angular velocity, and it is stated as follows:
ω = θ t
Where,
θ is the angle of rotation,
t is the time
ω is the angular speed
ω₀ is the initial angular velocity of the disc = 0.067 rad/s
I₀ is the initial moment of inertia of the disc, = 0.10 kg.m²
r is the distance of sand ring from the axis = 0.40m
m is the mass of the sand ring = 0.50 kg
If the net external torque is applied to the rotating disk is zero
According to the angular momentum conservation principle;
Initial Angular Momentum = Final Angular Momentum
[tex]\rm I_0 \omega_0= I_f \omega_f \\\\[/tex]
According to parallel axis theorem ;
[tex]\\\\ I_f = I_0 + mr^2[/tex]
[tex]\rm I_0 \omega_0= (I_0 + mr^2 ) \omega_f[/tex]
[tex]\rm 0.10 \times 0.067= (0.10 + 1020 \times 0.40^2 ) \omega_f \\\\ \rm \omega_f=0.0067\ rad/sec[/tex]
Hence the angular velocity of the disk will be 0.067 rad/sec.
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