Answer:
h '= [ ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]² h
Explanation:
Let's analyze this problem, first the two bodies travel together, second the superball bounces, third it collides with the marble and fourth the marble rises to a height h ’
let's start by finding the velocity of the two bodies just before the collision, let's use the concepts of energy
starting point. Starting point
Em₀ = U = m g h
final point. Just before the crash
Em_f = K = ½ m v²
as there is no friction the mechanical energy is conserved
Em₀ = Em_f
mg h = ½ m v²
v = √2gh
this speed is the same for the two bodies.
Second point. The superball collides with the ground, this process is very fast, so we will assume that the marble has not collided, let's use the concept of conservation of moment
initial instant. Just when the superball starts contacting the ground
p₀ = M v
this velocity is negative because it points down
final instant. Just as the superball comes up from the floor
p_f = M v '
the other body does not move
p₀ = p_f
- m v = M v '
v ’= -v
Therefore, the speed of the asuperbola is the same speed with which it arrived, but in the opposite direction, that is, upwards.
Let's use the subscript 1 for the marble and the subscript 2 for the superball
Third part. The superball and the marble collide
the system is formed by the two bodies, so that the forces during the collision are internal and the moment is conserved
initial instant. Moment of shock
p₀ = M [tex]v_{1'}[/tex]+ m v_2
final instant. When the marble shoots out.
P_f = Mv_{1f'}+ m v_{2f}
p₀ = p_
M v_{1'}+ m v_2 = M v_{1f'} + m v_{2f}
M (v_{1'} - v_{1f'}) = -m (v_2 - v_{2f})
in this expression we look for the exit velocity of the marble (v2f), as they indicate that the collision is elastic the kinetic nerve is also conserved
K₀ = K_f
½ M v_{1'}² + m v₂² = M v_{1f'}² + ½ m v_{2f}²
M (v_{1'}² - v_{1f'}²) = - m (v₂² - v_{2f}²)
Let's set the relation (a + b) (a-b) = a² - b²
M (v_{1'} + v_{1f'}) (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})
let's write our two equations
M ( v_{1'} - v_{1f'}) = -m (v₂ - v_{2f}) (1)
M (v_{1'} + v_{1f'}) (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})
if we divide these two expressions
(v_{1'}+ v_{1f'}) = (v₂ + v_{2f} )
we substitute this result in equation 1 and solve
v_{1f'}= (v₂ + v_{2f}) - v_{1'}
M (v_{1'} - [(v₂ + v_{2f}) - v_{1'}] = -m (v₂ - v_{2f})
-M v₂ - M v_{2f1'} + 2M v_{1'} = m v₂ - m v_{2f}
-M v_{2f} -m v_{2f} = m v₂ -M v₂ + 2M v_{1'}
v_{2f} (M + m) = - v₂ (M-m) + 2 M v_{1'}
v_{2f} = - [tex]( \frac{ M-m}{M +m } )[/tex]) v₂ + 2 [tex](\frac{M}{M+m})[/tex] v_{1'}
now we can substitute the velocity values found in the first two parts
[tex]v_{2f}[/tex] = - ( \frac{ M-m}{M +m } ) √2gh + 2(\frac{M}{M+m}) √2gh
we simplify
v_{2f} = [( \frac{ M-m}{M +m } ) + 2 (\frac{M}{M+m})] [tex]\sqrt{2gh}[/tex]
let's call the quantity in brackets that only depends on the masses
A = ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]
v_{2f}= A \sqrt{2gh}
in general, the marble is much lighter than the superball, so its speed is much higher than the speed of the superball
finally with the conservation of energy we find the height that the marble reaches
Starting point
Emo = K = ½ mv_{2f}²
Final point
Emf = U = m g h'
Em₀ = Em_f
½ m v_{2f}² = m g h ’
h ’= ½ v_{2f}² / g
h ’= ½ (A \sqrt{2gh})² / g
h ’= A² h
h '= [ ( \frac{ M-m}{M +m } )+ 2 (\frac{M}{M+m})]² h
A certain compact disc (CD) contains 783.216 megabytes of digital information. Each byte consists of exactly 8 bits. When played, a CD player reads the CD's information at a constant rate of 1.5 megabits per second. How many minutes does it take the player to read the entire CD? Express your answer using two significant figures
Answer:
69.62 minutes
Explanation:
From the information we have here,
1 byte is = 8 bits
So 1 megabyte = 8 megabits
Then
783.216 x 8 megabits = 6265.728 megabits
This player has its reading capacity at 1.5 megabits / second
So 1 minute = 60x1.5
= 90 megabits / min
Then we have the entire reading time of this CD player to be =
6265.728/90
= 69.62 minutes.
This answers the question
A torsional pendulum is formed by attaching a wire to the center of a meter stick with a mass of 5.00 kg. If the resulting period is 4.00 min, what is the torsion constant for the wire
Answer:
The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].
Explanation:
The angular frequency of the torsional pendulum ([tex]\omega[/tex]), measured in radians per second, is defined by the following expression:
[tex]\omega = \sqrt{\frac{\kappa}{I} }[/tex] (1)
Where:
[tex]\kappa[/tex] - Torsional constant, measured in newton-meters.
[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.
The angular frequency and the moment of inertia are represented by the following formulas:
[tex]\omega = \frac{2\pi}{T}[/tex] (2)
[tex]I = \frac{m\cdot L^{2}}{12}[/tex] (3)
Where:
[tex]T[/tex] - Period, measured in seconds.
[tex]m[/tex] - Mass of the stick, measured in kilograms.
[tex]L[/tex] - Length of the stick, measured in meters.
By (2) and (3), (1) is now expanded:
[tex]\frac{2\pi}{T} = \sqrt{\frac{12\cdot \kappa}{m\cdot L^{2}} }[/tex]
[tex]\frac{2\pi}{T} = \frac{2}{L}\cdot \sqrt{\frac{3\cdot \kappa}{m} }[/tex]
[tex]\frac{\pi\cdot L}{T} = \sqrt{\frac{3\cdot \kappa}{m} }[/tex]
[tex]\frac{\pi^{2}\cdot L^{2}}{T^{2}} = \frac{3\cdot \kappa}{m}[/tex]
[tex]\kappa = \frac{\pi^{2}\cdot m\cdot L^{2}}{3\cdot T^{2}}[/tex]
If we know that [tex]m = 5\,kg[/tex], [tex]L = 1\,m[/tex] and [tex]T = 240\,s[/tex], then the torsion constant for the wire is:
[tex]\kappa = \frac{\pi^{2}\cdot (5\,kg)\cdot (1\,m)^{2}}{3\cdot (240\,s)^{2}}[/tex]
[tex]\kappa = 2.856\times 10^{-4}\,N\cdot m[/tex]
The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].
An ambitious physics major decides to check out the Uncertainty Principle for macroscopic systems. She goes to the top of the UD tower and drops a marble of mass m to the ground, trying to hit one of the cracks between bricks on the mall. To aim her marble, she teeters precariously directly over the desired crack and uses a very sophisticated apparatus of the highest possible precision, which she has borrowed from the General Physics Lab. Alas, try as she might, she cannot hit the crack.
Required:
Prove that the marble will inevitably miss the crack.
Answer:
The order = [tex]\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}[/tex]
Explanation:
To miss the crack at a given distance is apparently not the same as the uncertainty that occurred in the distance while falling from the tower. However, it is believed that the uncertainties in both cases appear to be the same.
So, let's work it out together
According to Heisenberg's uncertainty principle:
[tex]\Delta s. \Delta p =\dfrac{h}{2} =\dfrac{h}{4 \pi}[/tex]
Also; if we recall from the equation of motion that:
[tex]v = u + at ---(1) \\ \\ v^2 - u^2 = 2as --- (2) \\ \\ s = ut + \dfrac{1}{2}at^2 --- (3)[/tex]
So, if u = 0 and a = g
Then;
[tex]v = gt --- (1) \\ \\ v^2 = 2gs - - - ( 2) \\ \\ s = \dfrac{1}{2}gt^2 --- (3)[/tex]
From (2)
Making (s) the subject, we have:
[tex]s = \dfrac{v^2}{2g}[/tex]
[tex]s = \dfrac{p^2}{2gm^2}[/tex]
By differentiation;
[tex]ds = d (\dfrac{p^2}{2gm^2})[/tex]
[tex]ds = \dfrac{2pdp}{2gm^2}[/tex]
[tex]\Delta \ s = \dfrac{p \Delta p}{gm^2 }[/tex]
where;
[tex]\Delta p = \dfrac{h}{4 \pi \Delta \ s}[/tex] from uncertainty principle
This implies that:
[tex]\Delta s = \dfrac{p(\dfrac{h}{4 \pi \Delta s }) }{gm^2}[/tex]
[tex]\Delta s = p(\dfrac{h}{4 \pi gm^2 }) \times \dfrac{1}{ \Delta s}}[/tex]
[tex](\Delta s)^2 = \dfrac{hmv} {4 \pi gm^2 }[/tex]
here;
v = 2gH
So;
[tex](\Delta s)^2 = \dfrac {h \sqrt{2gH} }{4 \pi gm }[/tex]
[tex]\mathbf{(\Delta s)^2 = \sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }[/tex]
Thus, the order = [tex]\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}[/tex]
Mary is trying to pull Julie on a sled across a flat snowy field. Mary pulls on the rope attached to the sled. Her pulling force is directed horizontally. Julie weighs 109 pounds. The sled weights 12 pounds. If the coefficient of static friction between the sled runners and the snow is 0.42, how much force must Mary pull with (in lbs) to start moving the sled
Answer: F = 498.04 lbs
Explanation: The forces acting on the sled and Julie are show in the figure below. In it, we notice that, for the sled and Julie to go accross the field, they only need force of friction, because, force of friction is a force that resists the relative motion of surfaces.
Force of friction is given by the formula
[tex]F_{f}=\mu.F_{N}[/tex]
where
μ is coefficient of friction
[tex]F_{N}[/tex] is normal force
Normal force is the force the surface exerts on the object. It is always perpendicular and a force of contact.
In the case of the sled, since it is on a horizontal plane, Normal Force has the same magnitude of Gravitational Force. So
[tex]F_{N}=m.g[/tex]
Coefficient of friction is how much friction exists between two surfaces.
Rearraging friction force is
[tex]F_{f}=\mu.m.g[/tex]
Mass for this system is the sum of Julie and the sled, therefore
m = 109 + 12
m = 121 lb
Calculating Friction Force:
[tex]F_{f}=0.42.121.9.8[/tex]
[tex]F_{f}=[/tex] 498.04 lbs
LBS is a unit of measurement referred as pound by weight.
In conclusion, force Mary needs to start moving the sled is 498.04 lbs
Suppose that 6 J of work is needed to stretch a spring from its natural length of 26 cm to a length of 39 cm. (a) How much work (in J) is needed to stretch the spring from 30 cm to 35 cm
Answer:
Workdone = 0.89 Joules
Explanation:
Given the following data;
Workdone = 6J
Extension = 39 - 26 = 13cm to meters = 13/100 = 0.13m
The workdone to stretch a string is given by the formula;
Workdone = ½ke²
Where;
k is the constant of elasticity.
e is the extension of the string.
We would solve for string constant, k;
6 = ½*k*0.13²
6 = ½*k*0.0169
Cross-multiplying, we have;
12 = 0.0169k
k = 12/0.0169
k = 710.06
a. To find the workdone when e = 30, 35.
Extension = 35 - 30 = 5 to meters = 5/100 = 0.05m
Workdone = ½*710.06*0.05²
Workdone = 355.03*0.0025
Workdone = 0.89 Joules
Therefore, the amount of work (in J) needed to stretch the spring from 30 cm to 35 cm is 0.89.
A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water-ice mixture. Fine thermocouple wires suspend the sphere, and the temperature is observed to change from 15 to 14C in 6.35 s. The experiment is repeated with a metallic sphere of the same diameter, but of unknown composition with a mass of 1.263 kg. If the same observed temperature change occurs in 4.59 s, what is the specific heat of the unknown material
Answer:
The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
Explanation:
Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ([tex]\dot Q[/tex]), measured in watts, that is, joules per second, by the following formula:
[tex]\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}[/tex] (1)
Where:
[tex]m[/tex] - Mass of the sphere, measured in kilograms.
[tex]c[/tex] - Specific heat of the material, measured in joules per kilogram-degree Celsius.
[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the sphere, measured in degrees Celsius.
[tex]\Delta t[/tex] - Time, measured in seconds.
In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:
[tex]\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}[/tex] (2)
Where:
[tex]m_{I}[/tex], [tex]m_{X}[/tex] - Masses of the iron and unknown spheres, measured in kilograms.
[tex]\Delta t_{I}[/tex], [tex]\Delta t_{X}[/tex] - Times of the iron and unknown spheres, measured in seconds.
[tex]c_{I}[/tex], [tex]c_{X}[/tex] - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.
[tex]c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}[/tex]
If we know that [tex]\Delta t_{I} = 6.35\,s[/tex], [tex]\Delta t_{X} = 4.59\,s[/tex], [tex]m_{I} = 0.515\,kg[/tex], [tex]m_{X} = 1.263\,kg[/tex] and [tex]c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the specific heat of the unknown material is:
[tex]c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)[/tex]
[tex]c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}[/tex]
Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.
Two steamrollers begin 105 mm apart and head toward each other, each at a constant speed of 1.20 m/s. At the same instant, a fly that travels at a constant speed of 2.50 m/s starts from the front roller of the southbound steamroller and flies to the front roller of the northbound one, then turns around and flies to the front roller of the southbound once again, and continues in this way until it is crushed between the steamrollers in a collision.
Required:
What distance does the fly travel?
Answer: 109.4 mm
Explanation: Distance is a scalar quantity and it is the measure of how much path there are between two locations. It can be calculated as the product of velocity and time: d = vt
The separation between the two steamrollers is 105 mm or 0.105 m. They collide to each other at the middle of the separation:
location of collision = [tex]\frac{0.105}{2}[/tex] = 0.0525 m
To reach that point, both steamrollers will have spent
[tex]v=\frac{\Delta x}{t}[/tex]
[tex]t=\frac{\Delta x}{v}[/tex]
[tex]t=\frac{0.0525}{1.2}[/tex]
t = 0.04375 s
The fly is travelling with speed of 2.5 m/s. So, at t = 0.04375 s:
d = 2.5*0.04375
d = 0.109375 m
Until it is crushed, the fly will have traveled 109.4 mm.
The Burj Khalifa is the tallest building in the world at 828 m. How much work would a man with a weight of 700 N do if he climbed to the top of the building
it would help if work was shown :,)
Answer:
579600J
Explanation:
Given parameters:
Height of the building = 828m
Weight of the man = 700N
Unknown:
Work done by the man = ?
Solution:
The work done by the man is the same as the potential energy expended.
Work done:
Work done = Weight x height = 700 x 828
Work done = 579600J
In addition to the sources of error mentioned above, since the caliper jaws squeezed the flexible, rubber surface of the ball slightly, the measured diameters were slightly __________________ compared to what a non-contact method of measuring would provide. This represents an ________________ ________________ error in ________________. This additional source of error ________________ ________________ significant. When the caliper jaws closed, the zero mark on the sliding Vernier scale, ________________ line up with the zero mark on the measuring scale. This means the caliper ________________ calibrated correctly.
Answer:
MINORS, SYSTEMATIC, STATISTICAL, BOTOM LINE, ZERO MATCHES
Explanation:
In general the sources of error or uncertainty can be classified
* Statistics. Which are those that describe the statistical formulas, for example: average, standard deviation, absolute error, etc.
* Systematic. That they occur due to an inappropriate measurement or to an interaction between the system and the instrument that cannot be quantified, in general this error shifts the measurements towards an explicit side
* Random, so errors that sometimes occur in the measurement and sometimes not, for example temperature changes during the medical process
In this case, you are asked to complete the sentences with the appropriate word
the measured diameters were slightly ___ MINORS________ compared to what a non-contact method of measuring would provide
. This represents an ____SYSTEMATIC_______ error in ________________.
This additional source of error ________STATISTICAL________ significant.
When the caliper jaws closed, the zero mark on the sliding Vernier scale, BOTOM LINE AND THE __________ line up with the zero mark on the measuring scale.
This means the caliper ___ZERO MATCHES_____________ calibrated correctly.
In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of 12,000 yards/s. the Columbiad is 900 ft long, but part of it is packed with poweder, so the spaceship accelerates over a distance of only 700 ft. Estimate the acceleration experienced by the occupants of the spaceship during launch. Give your answer in m/s2. (Verne realized that the "travelers would...encounter a violent recoil," but he probably didn't know that people generally lose consciousness if they experience accelerations greater than about 7g ~70 m/s2.)
Answer:
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
Explanation:
Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ([tex]a[/tex]), measured in meters per square second, is estimated by this kinematic formula:
[tex]a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }[/tex] (1)
Where:
[tex]\Delta s[/tex] - Travelled distance, measured in meters.
[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speeds of the spaceship, measured in meters.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v = 10968\,\frac{m}{s}[/tex] and [tex]\Delta s = 212.8\,m[/tex], then the acceleration experimented by the spaceship is:
[tex]a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}[/tex]
[tex]a = 282652.782\,\frac{m}{s^{2}}[/tex]
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
a) What magnitude point charge creates a 12596.37 N/C electric
held at a distance of 0.593 m?
Answer:
[tex]Q = 4.9216 * 10^{-7}C[/tex]
Explanation:
Given
[tex]E = 12596.37 N/C[/tex]
[tex]r = 0.593m[/tex]
Required
Determine the magnitude point charge (Q)
This question will be solved using [tex]the\ magnitude[/tex] of the electric field formula
[tex]E = \frac{kQ}{r^2}[/tex]
Where
[tex]k = 9 * 10^9\ Nm^2 / C^2[/tex]
Make Q the subject in [tex]E = \frac{kQ}{r^2}[/tex]
[tex]E * r^2 = kQ[/tex]
[tex]Q = \frac{E * r^2}{k}[/tex]
Substitute values for E, r and k
[tex]Q = \frac{12596.37 * 0.593^2}{9 * 10^9}[/tex]
[tex]Q = \frac{4429.50}{9 * 10^9}[/tex]
[tex]Q = \frac{492.16}{10^9}[/tex]
[tex]Q = 492.16 * 10^{-9}[/tex]
Express in standard form
[tex]Q = 4.9216 * 10^2 * 10^{-9}[/tex]
[tex]Q = 4.9216 * 10^{2-9}[/tex]
[tex]Q = 4.9216 * 10^{-7}C[/tex]
Required
Momentum
The magnitude of the momentum of an object is 64 kg*m/s. If the velocity of the
object is doubled, what will be the magnitude of the momentum of the object? *
32 kg*m/s
64 kg*m/s
128 kg*m/s
256 kg*m/s
Answer:
C) 128 kg*m/s
Explanation:
When you double something you multiply it by 2 most of the time. 64*2=128 or you can add it 64+64=128. Hope this helps.
F=9 N, a=3 m/s², m=?
Answer:
3kg
Explanation:
Given parameters:
Force = 9N
Acceleration = 3m/s²
Unknown:
Mass = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
So;
9 = mass x 3
mass = 3kg
A student is measuring the mass of 20 paper clips using an electronic balance that measures to the thousandths of a gram. The balance displays the value for the mass of the paper clips: 20.120 g. Which of the values would be acceptable ways to record this mass in a lab notebook
The question is incomplete, the complete question is;
A student is measuring the mass of 20 paper clips using an electric balance that measures to the thousandths of a gram. The balance displays the following values for the mass of the paper clip: 20.120 g.
Which of the following choices would be acceptable ways to record this mass on a lab report? Select all that apply.
a) 20g
b) 20.1 g
c) 20.12 g
d) 20.120g
e) 20.1200g
Answer:
d) 20.120g
Explanation:
We have been told in the question that the electronic balance measures the mass of the paper clips to thousandths of a gram.
This implies that the value of mass is measured to the third decimal place.
If we look at the options, 20.120 g is the measurement of the mass to thousandths of a gram hence that is the correct answer to the question.
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 3.1 ss for the boat to travel from its highest point to its lowest, a total distance of 0.55 mm. The fisherman sees that the wave crests are spaced 4.2 mm apart.
A. How fast are the waves traveling?
B. What is the amplitude of each wave?
C. If the total vertical distance traveled by the boat were 0.500, but the other data remained the same, how fast are the waves traveling?
D. If the total vertical distance traveled by the boat were 0.500, but the other data remained the same, what is the amplitude of each wave?
Answer:
Explanation:
It takes 3.1 s for the boat to travel from its highest point to its lowest, so the period of oscillation
T = 2 x 3.1 = 6.2 s
frequency of wave n = 1 / T = .1613 per sec
Amplitude of oscillation = .55/2 = .275 mm
The fisherman sees that the wave crests are spaced 4.2 mm apart. so wavelength of wave λ = 4.2 mm .
A ) velocity of wave v = n λ
.1613 x 4.2 = .677 mm /s
B ) Amplitude of wave = .275 mm
C ) The vertical distance determines only the amplitude which does not affect the velocity , so velocity will remain unchanged .
D ) Amplitude of wave depends only on the vertical displacement .
The amplitude will become .5 / 2 = .25 mm .
3. What is the SI unit of force? What is this unit equivalent to in terms of fundamental units?
4. Why is force a vector quantity?
Answer:
force = mass * acceleration
therefore the SI unit is kg*m/s2 or newton's
it's a vector quantity because it has both direction(acceleration) and size (mass)
A hazard sign has 3 identical
parallelogram-shaped stripes as shown.
Charles must outline each stripe with
reflective tape. Is one roll of 144 inches
of tape enough to finish the job?
Answer and Explanation: To know how much tape he will need, we have to calculate the perimeter of each parallelogram-shaped stripe.
Perimeter is the sum of all the sides of a figure.
For a parallelogram:
P = 2*length + 2*width
So, we need to determine width and length of the stripe.
Width is 3 inches. Length is the hypotenuse of the right triangle, whose sides are 6 and 18 inches. Then, length is
[tex]h=\sqrt{18^{2}+6^{2}}[/tex]
[tex]h=\sqrt{360}[/tex]
h = 19 in
Perimeter of the first stripe is
P = (2*19) + (2*3)
P = 44 inches
The hazard sign has 3 stripes. So total perimeter is
[tex]P_{t}=[/tex] 44 + 44 + 44
[tex]P_{t}=[/tex] 132 inches
To outline the parallelogram-shaped stripes, Charles need a total of 132 inches of tape. Since one roll has 144 inches, he will have enough tape to finish the job.
A particle has a velocity that is 90.% of the speed of light. If the wavelength of the particle is 1.5 x 10^-15 m, calculate the mass of the particle
Answer:
[tex]m=1.63\times 10^{-27}\ kg[/tex]
Explanation:
The velocity of a particle is 90% of the speed of light.
The wavelength of the particle is [tex]1.5\times 10^{-15}\ m[/tex]
We need to find the mass of the particle.
The formula for the wavelength of a particle is given by :
[tex]\lambda=\dfrac{h}{mv}[/tex]
h is Planck's constant
v is 90% of speed of light
m is mass of the particle
[tex]m=\dfrac{h}{\lambda v}\\\\m=\dfrac{6.63\times 10^{-34}}{1.5\times 10^{-15}\times 0.9\times 3\times 10^8}\\\\m=1.63\times 10^{-27}\ kg[/tex]
So, the mass of the particle is [tex]1.63\times 10^{-27}\ kg[/tex].
Which of the following do not involve a direction?
Check all that apply.
A. Velocity
B. Distance
C. Time
D. Acceleration
Distance and time I think
A trolley of mass 5.0 kg is moving at 1.0 ms to the right. A constant force of 25 N acts to the left for 0.75 seconds.
Calculate the change of kinetic energy of the trolley.
(Show Work)
Answer:
change in kinetic energy of the trolley is 53.91 J.
Explanation:
mass of the trolley, m = 5.0 kg
initial velocity of the trolley, u = 1.0 m/s
external force on the trolley, F = 25 N
time of force action, t = 0.75 s
The final velocity of the trolley at the end of 0.75 s is calculated as follows;
[tex]F = \frac{m(v-u)}{t} \\\\25 = \frac{5(v-1)}{0.75}\\\\5(v-1) = 18.75\\\\v-1 = \frac{18.75}{5} \\\\v-1 = 3.75\\\\v = 4.75 \ m/s \ in \ the \ direction \ of \ the \ applied \ force[/tex]
The change in kinetic energy of the trolley is calculated as;
Δ K.E = ¹/₂m(v² - u²)
Δ K.E = ¹/₂ x 5(4.75² - 1²)
Δ K.E = 53.91 J.
Therefore, change in kinetic energy of the trolley is 53.91 J.
Suppose you were digging a well into saturated sediments. Why is the sediment’s permeability an important factor in deciding where to put your well?
Answer:
The importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.
Explanation:
When digging a well into saturated sediments, the possibility of the sediment with either little saturation or full saturation being able to provide steady water supply will be limited by how permeable it is. Now, the importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.
what is the weight in Newtons of a ball with a mass of 7.77 kg?
Answer:
76.1N
Explanation:
Given parameters:
Mass of the ball = 7.77kg
Unknow:
Weight of balloon = ?
Solution:
Weight is the vertical force applied on a body.
Weight = mass x acceleration due gravity
So;
Weight = mass x acceleration due to gravity
So;
Weight = 7.77 x 9.8 = 76.1N
When a drag strip vehicle reaches a velocity of 60 m/s, it begins a negative acceleration by releasing a drag chute and applying its brakes. While reducing its velocity back to zero, its acceleration along a straight line path is a constant -7.5 m/s2 . What displacement does it undergo during this deceleration period
Answer:
240 meters
Explanation:
The distance traveled by the vehicle can be calculated using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex] (1)
Where:
x: is the displacement
[tex]v_{f}[/tex]: is the final speed = 0 (reduces its velocity back to zero)
[tex]v_{0}[/tex]: is the initial speed = 60 m/s
a: is the acceleration = -7.5 m/s²
By solving equation (1) for x we have:
[tex] x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m [/tex]
Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.
I hope it helps you!
What Coulombs discovered almost 300
years ago
Answer:
ummm hehe this is my time to shine
Explanation:
MERICIA!!!!!!!!!!!!!!!!!!!!!!!
A block is released from rest at the top of a hill of height h. If there is negligible friction between the block and the hill, the block arrives at the bottom of the hill with speed v. The block is released from rest at the top of another hill with a rough surface and height h. If one-half of the initial mechanical energy of the block-Earth system is lost due to friction as the block descends the hill, the block will reach the bottom of the hill with a speed of
Answer:
v₁ =√2gh, v₂ = v₁ /√2
Explanation:
Let's use the concepts of energy and work to analyze each case
hill without rubbing. Energy is conserved
starting point. Highest part
Em₀ = U = mg h
final point. Lower part
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v₁ =√2gh
rubbing hill
in this case the energy is not conserved because it is converted into work of the friction force, therefore the variation of the energy is the work of the friction
W = Em_{f} - Em₀
they indicate half of the initial mechanical energy is lost due to friction
W = ½ Em₀
we substitute
- ½ Em₀ = Em_{f} - Em₀
The negative sign is because the friction work always opposes the movement
Em_{f} = ½ Em₀
½ m v₂² = ½ m g h
v₂ = √½ √2gh
v₂ = v₁ /√2
Two spherical objects are separated by a distance that is 1.08 x 10-3 m. The objects are initially electrically neutral and are very small compared to the distance between them. Each object acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of 3.89 x 10-21 N. How many electrons did it take to produce the charge on one of the objects
Answer:
the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
Explanation:
from coulumbs law, The force that is acting over both charge can be computed as
F=( kq1q2)/r^2..............eqn(1)
Where
F=electrostatic force= 3.89 x 10-21 N
k= column constant= 9 x 10^9 Nm^2/C^2
q1 and q2= magnitude of the charges
r= distance between the charges= 1.08 x 10-3 m.
Since both charges are experiencing the same force, eqn(1) can be written as
F=( kq^2)/r^2.
We can make q subject of the formula
q= √(Fr^2)/k
= √[(3.89 x 10^-21× (1.08 x 10^-3)^2]/8.99 x 10^9
q= 71.043×10^-20 C
Hence, the charge is 71.043×10^-20 C
From quantization law, the number of electron can be computed as
N=q/e
Where
N= number of electron
q= charges
=1.6×10^-19C
N=71.043×10^-20/1.6×10^-19
=4.44
Hence, the charge on the object is 71.043×10^-20 C and the number of electron is 4.44
25 points!
A 6 kg object accelerates from 5 m•s to 25 m•s in 30 seconds. What was the net force acting on the
object? Give your answer in Newtons to one significant figure and without a unit.
(Show Work)
Answer:
6N
Explanation:
Given parameters:
Mass of object = 6kg
Initial velocity = 5m/s
Final velocity = 25m/s
Time = 30s
Unknown:
Net force acting on the object = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
Acceleration is the rate of change of velocity with time
Acceleration = [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
Force = mass x [tex]\frac{Final velocity - Initial velocity }{time}[/tex]
So;
Force = 6 x [tex]\frac{25 - 5}{30}[/tex] = 6N
effects of heat on matter
Answer:
it can melt orcan put them past their boiling point
Explanation:
A cylindrical tank of radius R, filled to the top with a liquid, has a small hole in the side, of radius r, at distance d below the surface. Find an expression for the volume flow rate through the hole. A 4.0-mm-diameter hole is 1.0 m below the surface of a 2.0-m-diameter tank of water. What is the rate, in mm/min, at which the water level will initially drop if the water is not replenished?
Answer:
a)[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]
b)[tex]dh / dt = 0.2658 mm / min[/tex]
Explanation:
From the question we are told that
Diameter of hole [tex]d_h=4mm=>0.004m[/tex]
Depth of hole [tex]D=0mm=>0.001m[/tex]
Diameter of tank [tex]d_t=2mm=>0.002m[/tex]
Generally the equation for pressure is mathematically given as
[tex]Pressure P= \rho*g*d[/tex]
[tex]P= 1/2*\rho *v^2[/tex]
Where
[tex]v = \sqrt {2gd}[/tex]
[tex]V = Area*v[/tex]
[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]
Generally the level at which the water level will initially drop if the water is not replenished is mathematically given by
[tex]dh / dt = (r/R)^2 *sqrt{2gd}\\dh / dt = (2/2000)^2 *sqrt(2*9.81*1) \\dh / dt = 4.429*10^-3 mm/s \\[/tex]
Therefore the level at which the water level will initially drop if the water is not replenished
[tex]dh / dt = 0.2658 mm / min[/tex]
The rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.
Given data:
The diameter of hole is, d = 4.0 mm = 0.004 m.
The depth of hole is, h = 1.0 m.
The diameter of tank is, d' = 2.0 m.
The given problem is based on the flow rate, which is defined as the flow of liquid through a given section per unit time.
Let us first obtain the equation of pressure as,
[tex]P=\dfrac{1}{2} \times \rho \times v^{2}[/tex]
Here, v is the velocity of efflux and its value is,
[tex]v=\sqrt{2gh} \\\\v^{2}=2gh[/tex]
And the level at which the water level will initially drop if the water is not replenished is mathematically given by,
[tex]\dfrac{dH}{dt}=(r/R)^{2} \times v[/tex]
Here,
r is the radius of hole.
R is the radius of tank.
Solving as,
[tex]\dfrac{dH}{dt}=((d/2) /(d'/2))^{2} \times \sqrt{2gh} \\\\\dfrac{dH}{dt}=((0.004/2) /(2/2))^{2} \times \sqrt{2 \times 9.8 \times 1}\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \;\rm m/s\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \times 6 \times 10^{4} \;\rm mm/min\\\\\dfrac{dH}{dt}=1.0625 \;\rm mm/min[/tex]
Thus, we can conclude that the rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.
Learn more about the flow rate here:
https://brainly.com/question/11816739
I REALLY NEED HELP!!!!
Running at 3.0 m/s, Burce, the 50.0 kg quarterback, collides with Max, the 100.0 kg tackle, who is traveling at 6.0 m/s in the other direction. Upon collision, Max continues to travel forward at 2.0 m/s.How much impulse does Max experience as a result of the collision? (list unknown variable and known variables, write an equation, plug in numbers, and get answer with unit.)
Answer:
400 Ns
Explanation:
Impulse = Change in momentum
i.e I = ΔP
So that,
Impulse experienced by Max = Change in Max's momentum
Change in Max's momentum = m(v - u)
Where m is the mass, v is the velocity after collision, and u is the velocity before collision.
m = 100.0 kg, v = 2.0 m/s, u = 6.0 m/s
Change in Max's momentum = 100 x (2 -6)
= -400 kg m/s
The negative sign shows that the change in momentum was against his direction of motion.
Impulse experienced by Max = 400 Ns.
Thus,
Max experienced an impulse of 400 Ns as a result of the collision.