Answer:
0.12 mm ; 140.50 rad/m ; 628.32 rad/sec ; +
Explanation:
Given the wave equation of the form :
y(x, t) = ym sin(kx ± ωt)
Mas per unit length (u) = 5 g/cm = (5÷1000)kg / 0.01m) = 0.005kg/0.01m = 0.5kg/m
Tension, T = 10 N
Amplitude, A = 0.12 mm
Frequency, F = 100 Hz
Comparing with the general wave equation :
y = Asin(kx ± ωt)
A = amplitude = ym = 0.12 mm
2.) k = 2π / λ
Recall :
v = fλ
v = sqrt(T/u) = sqrt(10/0.5) = sqrt(20) = 4.472
λ = v/ f = 4.472 / 100 = 0.04472
Hence,
k = (2 * π) / 0.04472
k = 140.50 rad/m
3.) Angular frequency, ω
ω = 2πf = 2 * 3.14 * 100 = 628.32 rad/sec
4.) sign is +ve
Direction of wave propagation as given is in the negative x axis
g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal radiation from the filament becomes visible. One bulb filament has a surface area of 30 mm2 and emits 60 W when operating. If the bulb filament has an emissivity of 0.8, what is the operating temperature of the filament
Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K
Lenz’s Law allows us to find _______.
the direction of the induced current.
the magnitude of the induced emf.
the direction of the induced emf.
the magnitude of the induced current.
Answer:
Explanation:
a
2.19 The drag characteristics of a blimp traveling at 4 m/s are to be studied by experiments in a water tunnel. The prototype is 20 m in diameter and 110 m long. The model is one-twentieth scale. What velocity must the model have for dynamic similarity
Answer:
[tex]Vm=0.894m/s[/tex]
Explanation:
From the question we are told that
Velocity if travel [tex]v=4m/s[/tex]
Diameter of prototype [tex]d_1=20m[/tex] and [tex]d_2=110m[/tex]
Scale ratio=[tex]\frac{1}{20}[/tex]
Generally Velocity of of the model using Froud's model is mathematically given as
[tex]Fm=Fp[/tex]
[tex]\frac{Vm}{\sqrt{Lmg}} =\frac{Vp}{\sqrt{Lpg}}[/tex]
[tex]Vm=Vp*\frac{Vp}{\sqrt{Lpg} }[/tex]
[tex]Vm=4*\frac{1}{\sqrt{20}}[/tex]
[tex]Vm=0.894m/s[/tex]
A constant force FA is applied to an object of mass M, initially at rest. The object moves in the horizontal x-direction, and the force is applied in the same direction. After the force has been applied, the object has a speed of vf. Which mathematical routines can be used to determine the time in which the force is applied to the object of mas
Answer:
t = [tex]\frac{ v \ F}{ m}[/tex]
Explanation:
The question is a bit strange, for this exercise we must use the mathematical relationship of Newton's second law to find the acceleration of the body
F = m a
a = F / m (1)
with this acceleration the mathematical relations of kinematics of accelerated motion must be used
v = v₀ + a t
with the body starting from rest its initial velocity is zero
v = a t
t = v / a (2)
if we substitute the equation 1 in 2
t = [tex]\frac{ v \ F}{ m}[/tex]
this is the final mathematical expression that allows to find the time based on the data of the problem
If you carry a box across the floor, you don't do work. Why not?
The force exerted by the person is an upward force equal to the weight of the box, and that force is perpendicular to the motion. If there is no motion in the direction of the force, then no work is done by that force. Yet you certainly feel like you are doing work if you carry a heavy box.
In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force. Energy is required to do work and the basic SI unit of energy is the joule, the amount of energy required to exert a force of 1 Newton through a distance of 1 meter (1 joule = 1 newton meter).*picture 1*For the special case of a constant force, the work may be calculated by multiplying the distance times the component of force which acts in the direction of motion.*picture 2*
In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force.*picture 3*For a constant force F which moves an object in a straight line from x1 to x2 , the work done by the force can be visualized as the area enclosed under the force line below*picture 4*For the more general case of a variable force F(x) which is a function of x, the work is still the area under the force curve, and the work expression becomes an integral. *picture 5*credits-goo.gle,hyper physics
greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but
3 leaps of the hound are equal to 5 leaps of the hare. Compare the speed of the
hound and the hare,
need full solution:-
[tex]{\large{\bold{\rm{\underline{Given \; that}}}}}[/tex]
★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
[tex]{\large{\bold{\rm{\underline{To\; find}}}}}[/tex]
★ The speed of the hound and the hare
[tex]{\large{\bold{\rm{\underline{Solution}}}}}[/tex]
★ The speed of the hound and the hare = 25:18
[tex]{\large{\bold{\rm{\underline{Full \; Solution}}}}}[/tex]
[tex]\dashrightarrow[/tex] As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
So firstly let us assume a metres as the distance covered by the hare in one leap.
Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.
But 3 leaps of the hound are equal to 5 leaps of the hare.
Henceforth, (5/3)a meters is the distance that is covered by the hound.
Now according to the question,
Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)
Now the distance travelled by the hound in it's 5 leaps..!
(5/3)a × 525/3a metresNow the distance travelled by the hare in it's 6 leaps..!
6a metresNow let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!
25/3a = 6a25/3 = 625:18If a person weighs 140 lb'on Earth, their mass in kilograms is
Answer:
70 kg
Explanation:
divide it by 2
Hope this helped!
Answer:
63.502932 Kilograms
Explanation:
Which landform is produced at location E where the Mississippi River enters the Gulf of
Mexico?
a delta a drumlin an out wash an escarpment
Answer:
a delta
Explanation:
The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.
A delta is a depositional landform where a smaller body of water enters into a larger one.
The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.
So, this feature is a delta.
A iron block with a mass of 4.8 kg initially slides over a rough horizontal surface with a speed of 1.2 m/s. Friction slows the block to rest. While slowing to rest, 85.0% of the kinetic energy of the block is absorbed by the block itself as internal energy. What is the temperature increase of the block
Answer:
Explanation:
Kinetic energy of block will be converted into heat energy by friction .
Heat energy produced = 1/2 m v²
= .5 x 4.8 x 1.2²
= 3.456 J
85% of energy is converted into heat energy , so heat energy produced
= .85 x 3.456 = 2.9376 J .
If Q heat is given to m mass of object having s as specific heat and Δt is increase in temperature
Q = msΔt
specific heat of iron s = 462 J / kg C
Putting the values ,
2.9376 = 4.8 x 462 x Δt
Δt = 13.24 x 10⁻⁴ ⁰C.
A box has a mass of 150 kg. If a net force of 3,000 N acts
on the box, what is the box’s acceleration?
A truck travelling down the street suddenly brakes, applying a 14 N force over 3.5 seconds. What was the impulse over the given time.
Answer:
49 Ns
Explanation:
Given data
Force= 14N
time = 3.5seconds
Applying the expression for impulse
P= Ft
substitute
P=14*3.5
P=49 Ns
Hence the impulse is 49 Ns
A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 4.00 kg particle such that the center of mass of the three-particle system has the coordinates (-0.500 m, -0.700 m)
Answer:
a) The x coordinate of the third mass is -1.562 meters.
b) The y coordinate of the third mass is -0.944 meters.
Explanation:
The center of mass of a system of particles ([tex]\vec r_{cm}[/tex]), measured in meters, is defined by this weighted average:
[tex]\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}[/tex] (1)
Where:
[tex]m_{i}[/tex] - Mass of the i-th particle, measured in kilograms.
[tex]\vec r_{i}[/tex] - Location of the i-th particle with respect to origin, measured in meters.
If we know that [tex]\vec r_{cm} = (-0.500\,m,-0.700\,m)[/tex], [tex]m_{1} = 1\,kg[/tex], [tex]\vec r_{1} = (-1.20\,m, 0.500\,m)[/tex], [tex]m_{2} = 4.50\,kg[/tex], [tex]\vec r_{2} = (0.600\,m, -0.750\,m)[/tex] and [tex]m_{3} = 4\,kg[/tex], then the coordinates of the third particle are:
[tex](-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}[/tex]
[tex](-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})[/tex]
[tex](4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)[/tex]
[tex](x_{3},y_{3}) = (-1.562\,m,-0.944\,m)[/tex]
a) The x coordinate of the third mass is -1.562 meters.
b) The y coordinate of the third mass is -0.944 meters.
a lamp with flux 1400 lm, has an intensity of
a.) 1400 cd
b.) 100 cd
c.) 1000 cd
d.) 111 cd
Leticia leaves the grocery store And walks 150 M’s to parking lot then she turns 90° to the right and walks an additional 70 M’s to her car what is the magnitude of displacement of her car from the grocery store at exit
Answer:
Explanation:
Its 165.5m
A magnifying glass uses a convex lens of focal length 6.25 cm. When it is held 5.20 cm in front of an object, what is the image distance?
(Mind your minus signs)
(Unit=cm)
Answer:
The answer is -30.95.
Explanation:
Use the lens equation: 1/focal length = 1/object distance + 1/image distance. The answer comes out to -30.95. This is correct on Acellus.
When it is held 5.20 cm in front of an object, the image distance will be "-30.95 cm". To understand the calcultaion, check below.
Convex lensAccording to the question,
Object distance, u = -5.20 cm
Focal length, f = 6.25 cm
By using the Lens formula, we get
→ [tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u}[/tex]
or,
→ [tex]\frac{1}{v} = \frac{1}{f} + \frac{1}{u}[/tex]
By substituting the values, we get
[tex]= \frac{1}{6.25} - \frac{1}{5.20}[/tex]
[tex]\frac{1}{v} = -\frac{21}{650}[/tex]
By applying cross-multiplication,
v = -30.95 cm
Thus the above answer is correct.
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Curtis, a student in our class, makes the following statement: The puck reached a slightly higher location on the ramp than I predicted. This is because I used the wrong mass for the puck when I did all my calculations. I accidentally used the mass of the smaller puck rather than the mass of the larger puck in my video." Is this a plausible explanation? Would the using the wrong mass for the puck during the calculations mean the puck would reach a greater height? Explain your reasoning.
Answer and Explanation: No, the explanation is not plausible. The puck sliding on the ice is an example of the Principle of Conservation of Energy, which can be enunciated as "total energy of a system is constant. It can be changed or transferred but the total is always the same".
When a player hit the pluck, it starts to move, gaining kinetic energy (K). As it goes up a ramp, kinetic energy decreases and potential energy (P) increases until it reaches its maximum. When potential energy is maximum, kinetic energy is zero and vice-versa.
So, at the beginning of the movement the puck only has kinetic energy. At the end, it gains potential energy until its maximum.
The representation is as followed:
[tex]K_{i}+P_{i}=K_{f}+P_{f}[/tex]
[tex]K_{i}+0=0+P_{f}[/tex]
[tex]\frac{1}{2}mv^{2} = mgh[/tex]
As we noticed, mass of the object can be cancelled from the equation, making height be:
[tex]h=\frac{v^{2}}{2g}[/tex]
So, the height the puck reaches depends on velocity and acceleration due to gravity, not mass of the puck.
It would be impossible to build a microscope that could use visible light to see the molecular structure of a crystal because. It would be impossible to build a microscope that could use visible light to see the molecular structure of a crystal because. lenses cannot be ground with fine enough precision. diffraction limits the resolving power to about the size of the wavelength of the light used. lenses with enough magnification cannot be made. lenses cannot be placed in the correct place with enough precision. More than one of the above is correct.
Answer:
the correct one is: a diffraction limits the resolving power to approximately the size of the wavelength of the light used
Explanation:
To be able to solve two structures with a light source, the Rayleigh criterion must be met that stable the two structures are solved when the first minimum of diffraction at one point is in the code of the first maximum of the other point
Using this criterion we can find an expression for the first minimization of the diffraction spectrum m = 1
sin θ tea = λ / a
now the structure of the comatose has a separation of around 1 nm and the wavelength of visible light ranges from 400 to 700 nm, when substituting we find
sin θ = 400/1 10
sin θ = 400
sin θ = 700/1
sin θ = 700
These values are neither impossible since the sin function is bounded between -1 to 1, so we cannot see the diffraction
When reviewing the different statements, the correct one is: a diffraction limits the resolving power to approximately the size of the wavelength of the light used:
It would be impossible to build a microscope that could use visible light to
see the molecular structure of a crystal because diffraction limits the
resolving power to about the size of the wavelength of the light used.
Diffraction occurs when waves encounter an obstacle which leads to it
bending around it. Visible light when diffracted results in the overlapping of
the patterns thereby formation of images as one instead of in different parts.
This thereby leads to the resolving power being limited to the size of the
wavelength of light used and images won't be able to be identified
separately.
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A 40 kg boy standing on a skateboard throws a 2 kg ball 20 m/s to the left.
a. What is the ball's momentum?
O 10 kg m/s
O 20 kg mis
O 40 kg m/s
O 1 kg m/s
Answer:
40 kg m/s
Explanation:
Given the following data;
Mass of boy = 40kg
Mass of ball = 2kg
Velocity = 20m/s
To find the momentum;
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
Momentum =mass * velocity
Substituting into the equation, we have
Momentum = 2 * 20
Momentum = 40 kg m/s
A beaker is filled with water to the rim. Gently you place a toy duck in the beaker and some of the water spills out. The duck floats. The weight of the beaker with the duck floating in it is Group of answer choices less than its weight before adding the duck. the same as its weight before adding the duck. greater or less than its weight before adding the duck, depending on the weight of the duck. greater than its weight before adding the duck.
Answer:
Explanation:
The same as its weight before adding the duck
Does changing the height of point C affect the speed of the coaster car at point D?
Without friction, NO.
The speed at D depends only on the difference in height between A and D. Whatever happens between them doesn't matter.
The speed of the coaster car at point D will be affected if the height of point C is changed.
Potencial Energy:
It is the enrgy in a body due to the position of differnt part of the object or system.
As we increase the the hight of the car the potetial enrgy increase, the gravitational acceleration on car will be more due to the high of the point C.
Therefore, the speed of the coaster car at point D will be affected if the height of point C is changed.
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A level test track has a coefficient of road adhesion of 0.80, and a car being tested has a coefficient of rolling friction that is approximated as 0.018 for all speeds. The vehicle is tested unloaded and achieves the theoretical minimum stop in 180 ft (from brake application). The initial speed was 60 mi/h. Ignoring aerodynamic resistance, what is the unloaded braking efficiency
Answer:
the unloaded braking efficiency is 84.6 %
Explanation:
Given the data in the question;
by Ignoring aerodynamic resistance; we can find the theoretical stopping distance using the following formula
S = (Y[tex]_{b}[/tex]( V₁² - V₂²)) / ( 2g( ηbμ + [tex]f_{rl}[/tex] ± sin∅[tex]_{g}[/tex]))
now given that the tracked is levelled, ∅[tex]_{g}[/tex] = 0, also Y[tex]_{b}[/tex] = 1.04 for level or flat road
Speed V₁ = 60mil/hr = (60×5280)/(1×60×60) = 316800ft/3600s = 88ft/s
now, we substitute in our values to get the braking efficiency;
180ft = (1.04( (88ft/s)² - 0²)) / ( 2(32.2( (ηb/100)(0.80) + (0.018) ± sin(0°)))
180ft = 8053.76 / ( 64.4)(0.008ηb + 0.018)
180ft = 8053.76 / ( 0.5152ηb + 1.1592)
180( 0.5152ηb + 1.1592) = 8053.76
( 0.5152ηb + 1.1592) = 8053.76 /180
0.5152ηb + 1.1592 = 44.7431
0.5152ηb = 44.7431 - 1.1592
0.5152ηb = 43.5839
ηb = 43.5839 / 0.5152
ηb = 84.596 ≈ 84.6 %
Therefore, the unloaded braking efficiency is 84.6 %
A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 8.11 m below its highest point?
Answer:
a) the kinetic energy of the ball at its highest point is 69.58 J
b) its speed when it is 8.11 m below its highest point is 55.97 m/s
Explanation:
Given that;
mass of golf ball m = 46.8 g = 0.0468 kg
initial speed of the ball v₁ = 58.8 m/s
height h = 24.7 m
acceleration due to gravity = 9.8 m/s²
the kinetic energy of the ball at its highest point = ?
from the conservation of energy;
Kinetic energy at the highest point will be;
K.Ei + P.Ei = KEf + PEf
now the Initial potential energy of the ball P.Ei = 0 J
so
1/2mv² + 0 J = KEf + mgh
K.Ef = 1/2mv² - mgh
we substitute
K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]
K.Ef = 80.904 - 11.3284
K.Ef = 69.58 J
Therefore, the kinetic energy of the ball at its highest point is 69.58 J
b) when the ball is 8.11 m below the highest point, speed = ?
so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m
so our velocity will be v₂
also using the principle of energy conservation;
K.Ei + P.Ei = KEh + PEh
1/2mv² + 0 J = 1/2mv₂² + mgh'
1/2mv₂² = 1/2mv² - mgh'
multiply through by 2/m
v₂² = v² - 2gh'
v₂ = √( v² - 2gh' )
we substitute
v₂ = √( (58.8)² - 2×9.8×16.59 )
v₂ = √( 3457.44 - 325.164 )
v₂ = √( 3132.276 )
v₂ = 55.97 m/s
Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s
"45 meters north" is an example of
Answer:
Displacement
Explanation:
The quantity 45m north is a typical example of displacement.
Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.
When we are specifying the displacement of a body, the direction must be indicated accurately. Therefore, the quantity given is displacementTwo trains, each having a speed of 22 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 51 km apart and heads directly for the other train. On reaching the other train it flies directly back to the first train, and so forth. (We have no idea why a bird would behave in this way.) What is the total distance the bird travels before the trains collide
Answer:
the total distance the bird travels before the trains collide is 69.54 km
Explanation:
Given the data in the question,
dA = 22km/h × t
dB = 51km - 22km/h × t
now we find the time, when the two trains collide
dA = Db
22km/h × t = 51km - 22km/h × t
44km/h × t = 51km
t = 51km / 44km/h
t = 1.159 hrs
so the bird can fly back and forth for 1.159 hrs before the train collide.
hence, distance travelled by the bird in total will be;
d = v × t
we substitute
d = 60 km/h × 1.159 h
d = 69.54 km
Therefore, the total distance the bird travels before the trains collide is 69.54 km
Match each letter to the description
A woman walks in a
straight line with the sun
to her right at six o'clock in
the morning. Toward
which pole does she walk?
A line of latitude.
A man walks in a straight
line with the sun to his
right at six o'clock in the
evening. Toward which
pole does he walk?
A line of longitude.
A woman walks in a straight line with the sun to her right at six o'clock in the morning.
The sun rises East of her, so the woman is walking toward the North pole.
A man walks in a straight line with the sun to his right at six o'clock in the evening.
The sun sets West of him, so the man is walking toward the South pole.
The woman and the man are both walking along lines of constant longitude.
A 56-kg skater initially at rest throws a 5.0-kg medicine ball horizontally to the left. Suppose the ball is accelerated through a distance of 1.0 m before leaving the skater's hand at a speed of 7.0 m/s. Assume the skater and the ball to be point-like and the surface to be frictionless and ignore air resistance. Use a vertical y-axis with the positive direction pointing up and a horizontal x-axis with the positive direction pointing to the right. What will happen to the skater and the ball after the ball is thrown
Answer:
[tex]a_2=24.5\ \text{m/s}^2[/tex] towards right
[tex]a_1=2.1875\ \text{m/s}^2[/tex] towards left
Explanation:
[tex]m_1[/tex] = Mass of skater = 56 kg
[tex]m_2[/tex] = Mass of ball = 5 kg
v = Final velocity of ball = 7 m/s
u = Initial velocity of ball = 0
s = Distance the ball moved in the hand of the skater = 1 m
Moving left is considered and moving right is considered positive.
From kinematic equations of motion we have
[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{7^2-0^2}{2\times 1}\\\Rightarrow a=24.5\ \text{m/s}^2[/tex]
So, the ball will move towards right with a magnitude of acceleration [tex]a_2=24.5\ \text{m/s}^2[/tex].
The force on the ball will be
[tex]F_2=m_2a_2\\\Rightarrow F_2=5\times 24.5\\\Rightarrow F_2=122.5\ \text{N}[/tex]
The force on the ball is [tex]122.5\ \text{N}[/tex]
The reaction force on the skater will be equal to the force on the ball but will have opposite direction.
[tex]-F_1=F_2\\\Rightarrow -m_1a_1=F_2\\\Rightarrow a_1=-\dfrac{122.5}{56}\\\Rightarrow a_1=-2.1875\ \text{m/s}^2[/tex]
So, the skater will move towards left with a magnitude of acceleration [tex]2.1875\ \text{m/s}^2[/tex]
After the ball is thrown, the skater will accelerate to the left at 2.19 m/s² while the ball will accelerate to the right at 24.5 m/s².
The given parameters
Mass of the skater, m1 = 56 kgMass of the ball, m2 = 5 kgDistance traveled by the ball, d = 1 mSpeed of the ball, v = 7 m/sThe acceleration of the ball as it leaves the skaters hand is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\ v^2 = 0 + 2as\\\\ a = \frac{v^2}{2s} \\\\ a = \frac{7^2}{2(1)} \\\\ a = 24.5 \ m/s^2[/tex]
The force experienced by the ball is calculated as follows;
[tex]F = ma\\\\ F_b = 5 \times 24.5\\\\ F_b = 122.5 \ N[/tex]
The force experienced by the skater is equal in magnitude to that of the ball but opposite in direction.
[tex]F_s = -F_b\\\\ F_s = -122.5 \ N[/tex]
The acceleration of the skater after the ball was thrown is calculated as follows;
[tex]F = ma \\\\ a = \frac{F}{m} \\\\ a = \frac{-122.5 }{56} \\\\ a =-2.19 \ m/s^2[/tex]
Thus, after the ball is thrown, the skater will accelerate to the left at 2.19 m/s² while the ball will accelerate to the right at 24.5 m/s².
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A boat is cruising in a straight line at a constant speed of 2.5 m/s when it is shifted into neutral. After coasting 14 mm the engine is engaged again, and the boat resumes cruising at the reduced constant speed of 1.5 m/s. Assuming constant acceleration while coasting,
a. How much time did it take for the boat to coast the 12 m?
b. What was the boat’s acceleration while it was coasting?
c. What was the speed of the boat when it had coasted for 6.0 m? Explain.
Answer:
Explanation:
a ) distance s = 12 m .
constant speed = 1.5 m/s
time taken to coast 12 m
= distance / speed
= 12 / 1.5 = 8 s
b ) initial velocity u = 2.5 m /s
final velocity v = 1.5 m /s
displacement s = 14 m
acceleration a = ?
v² = u² + 2 as
1.5² = 2.5² + 2 a x 14
a = - .1428 m /s²
= - 14.28 cm / s²
c )
v = ?
u = 2.5 m /s
s = 6 m
a = - .1428 m /s²
v² = u² - 2 as
= 2.5² - 2 x 6 x .1428
= 6.25 - 1.71
= 4.54
v = 2.13 m /s .
In the graph, which two regions show the particle undergoing zero acceleration and negative acceleration respectively?
A.
BC shows zero acceleration, and AB shows negative acceleration.
OB.
AB shows zero acceleration, and CD shows negative acceleration.
O C.
BC shows zero acceleration, and CD shows negative acceleration.
D.
AB shows zero acceleration, and BC shows negative acceleration.
Answer:
c
Explanation:
Answer:
c
Explanation:
True or False. The larger a waves wavelength, the more energy it carries. (1 Point) True O False O Maybe
Explanation:
False, it is oppisite the shorter the wavelength the more energy it carries.
A circus tightrope walker weighing 800 N is standing in the middle of a 15 meter long cable stretched between two posts. The cable was originally horizontal. The lowest point of the cable is now at his feet and is 30 cm below the horizontal. Assume the cable is massless. What is the tension in the cable
Answer:
T = 10010 N
Explanation:
To solve this problem we must use the translational equilibrium relation, let's set a reference frame
X axis
Fₓ-Fₓ = 0
Fₓ = Fₓ
whereby the horizontal components of the tension in the cable cancel
Y Axis
[tex]F_{y} + F_{y} - W =0[/tex]
2[tex]F_{y}[/tex] = W
let's use trigonometry to find the angles
tan θ = y / x
θ = tan⁻¹ (0.30 / 0.50 L)
θ = tan⁻¹ (0.30 / 0.50 15)
θ = 2.29º
the components of stress are
F_{y} = T sin θ
we substitute
2 T sin θ = W
T = W / 2sin θ
T = [tex]\frac{ 800}{ 2sin 2.29}[/tex]
T = 10010 N