Answer:
a) v = 25.54 ft/s
b) Xmax = 0.4180
c) Xmax = 0.0048 ft
Explanation:
a) determine the speed v at which the amplitude of the boat and the trailer will be a maximum;
to calculate the distance travelled
S = vt
given an expression of wavelength
l = vt
l = 2πv / ω
ω = 2πv / l
equation for counter of the road
y = Y sin ωt
= Y sin ( (2πv / l)t)
next we calculate the angular frequency
ω = √ ( k/m)
= √ ( g / Sst )
= √( 32.2 / ( 1.5/12))
= 16.05 rad/s
Now we calculate the speed v at which amplitude of the boat will be maximum
ω = 2πv / l
16.05 = (2 × π × v ) / 10
160.5 = 2πv
v = 160.5 / 2π
v = 25.54 ft/s
b)
we calculate the maximum amplitude using the following expression;
X/Y = [ √( 1 + ( 2Sr)²)] / [ √(( 1 - r²)² + ( 2Sr)²))]
Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×1)²)] / [ √(( 1 - 1²)² + ( 2×0.05×r)²))]
Xmax / 0.0416 = 1.004987 / 0.1
Xmax 0.1 = 0.0418
Xmax = 0.0418 / 0.1
Xmax = 0.4180
c)
we convert speed of the boat from mph velocity m/s
v = 55 mph × 1.46667ft/s / 1mph
v = 80.66 ft/s
next we calculate angular velocity
ω = 2πv / l
= 2π × 80.66 / 10
= 50.68 rad/s
next is the frequency ratio
r = 50.68 / 16.05
= 3.16
finally we find the amplitude of the boat
X/Y = [ √( 1 + ( 2Sr)²)] / [ √(( 1 - r²)² + ( 2Sr)²))]
Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×3.16)²)] / [ √(( 1 - 3.16²)² + ( 2×0.05×3.16)²))]
Xmax / 0.0416 = √1.0998 / √ ( 80.74 + 0.0998)
Xmax / 0.0416 = 1.0487 / 8.9910
Xmax 8.9910 = 0.0436
Xmax = 0.0436 / 8.9910
Xmax = 0.0048 ft
2. The weight of a basketball player is 6 ft and 1 12 inches. Change his height to:
a. Feet
b. Inches
C. Centimeters
d. Meters
Answer:
a feat
Explanation:
cause the mf already 7ft on the dot there is no such thing as 6 ft 12
i need to know the angle of incidence, reflection, and refraction
Answer:
incident 50°
reflection 90°
refraction 20°
Two hockey pucks are moving toward each other. The mass of the first puck is half the mass of the second puck. Initially, the first puck is going at a speed of 5.0 m/s to the right, while the second puck is going at a speed of 2.0 m/s. The first puck rebounds with a speed of 9.0 m/s. What is the final speed of the second puck
The final speed of the second puck is 5 m/s.
What is speed?Speed can be defined as the rate of change of distance with respect to time.
To calculate the final speed of the second puck, we use the formula below.
Formula:
mu+m'u' = mv+m'v'................ Equation 1Where:
m = mass of the first puckm' = mass of the second pucku = initial velocity of the first pucku = initial velocity of the second puckv = final velocity of the first puckv' = final velocity of the second puck.
Make v' the subject of the equation
v' = (mu+m'u'-mv)/m'.............. Equation 2Assuming,
The mass of the first puck is y The left direction is negative and the right direction is positiveFrom the question,
Given:
m = ym' = 2yu = 5 m/su' = -2 m/sv = -9 m/s ( rebounds)Substitute these values into equation 2
v' = [(y×5)+(2y×(-2))-(y×(-9))]/2yv' = (5y-4y+9y)/2yv' = 10y/2yv' = 5 m/sHence, The final speed of the second puck is 5 m/s.
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Name the charcteristic which help us distinguish berween aman's voice and awamen's seeing them
Answer:
Asalamalikum maryam
Following are the characteristics that help to distinguish between man's and women's voice
Mans have low pitched sound while as women have hight pitched sound.Mans voice is thick while as women's voice is thin.Explain why we need to study the climate.
Answer:The current focus of climate science involves carbon dioxide (CO2)emissions. Carbon dioxide in the atmosphere acts as a blanket over the planet by trapping long wave radiation, which would otherwise radiate heat away from the planet. As the amount of carbon dioxide increases, so will its warming effect.
Explanation:
If, according to Newton’s Third Law, every action force has an equal reaction
force that acts in the opposite direction, why do these forces not just cancel each
other out, resulting in no net force and therefore no motion? Explain.
Answer:
They act on different objects
Explanation:
Let's say that I push a cart, the cart moves because the force is also going into the ground. Hope this makes sense.
The wavelengths of visible light vary from about 300 nm to 700 nm. What is the range of frequencies of visible light in a vacuum
The range of frequencies of visible light in a vacuum is mathematically given as
Fmin=4.19*10^14Hz to Fmax=1*10^15Hz
What is the range of frequencies of visible light in a vacuum?Question Parameters:
The wavelengths of visible light vary from about 300 nm to 700 nm.
Generally, the equation for the frequency is mathematically given as
F=C/\lambda
Therefore
For Fmax
[tex]Fmax=\frac{300*10^8}{3*10^9}[/tex]
Fmax=1*10^15Hz
Where
[tex]Fmin=\frac{3*10^8}{700*10^9}[/tex]
Fmin=4.19*10^14Hz
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a 2 kg ball traveling at 5m/s collides with a 1 kg ball at rest. After the collision the 1 kg ball moves off with a speed of 7 m/s. Find the final speed of the 2 kg ball.
A series circuit contains both a 6Ω and a 3Ω resistor. Through the 6Ω resistor, the current is 0.5 amps. What will the current be through the 3Ω resistor?
Answer:
0.5 amps
Explanation:
the amps will be constant no matter what
Which type of energy increases when you compress a spring?
Question 8 options:
radiant energy
kinetic energy
elastic potential energy
sound energy
Answer:
Elastic potential energy
Answer:
Elastic potential energy: increases when you compress a spring.
Explanation:
what is acceleration due to gravity?
The acceleration due to gravity is 9.8 m/s^2.
How many atoms of each element are present in the formula for bromoethane?
CH3CH2BT
Answer:
Explanation:
How many atoms of each element are present in the formula for bromoethane
What is the mass of a child that has a KE of 400 J who is riding her bike at 3.9 m/s?
Answer:
52.6 kg
Explanation:
KE = 1/2mv^2
400 = 1/2 x m x 3.9^2
m = 2(400 J)/ ( 3.9 m/s)^2
m = 52.6 kg
A 21.2 kg mass falls from a height of 4.000m. The momentum of the mass just before it hits the ground is
A. 144.2
B. 187.8
C. 320.0
D. 442.4
E. 502.1
By third equation of motion -
[tex]\green{ \underline { \boxed{ \sf{v^2-u^2=2aS}}}}[/tex]
where
v= final velocityu = initial velocitya = accelerationS = distance travelledPutting Values to find final velocity of mass before hitting the ground-
[tex]\begin{gathered}\\\implies\quad \sf v^2-(0)^2=2\times g \times 4 \quad (g = acceleration \: due \:to \: gravity) \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf v^2=2\times 9.8 \times 4\quad (g= 9.8 \:m/s) \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf v=\sqrt{78.6} \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf v= 8.86 \:m/s \\\end{gathered} [/tex]
Now finding the momentum of the mass at that moment -
[tex]\green{ \underline { \boxed{ \sf{Momentum= mass \times velocity}}}}[/tex]
[tex]\begin{gathered}\\\implies\quad \sf Momentum= 21.2 \times 8.86 \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\implies\quad \sf Momentum= 187.8 \:kgms^{-1} \\\end{gathered} [/tex]
[tex]\longrightarrow[/tex]The momentum of the mass just before it hits the ground is 187.8 kgm/s
[tex]\\[/tex]
[tex]\therefore \sf Option \: B) \: is \:correct [/tex]✔️
(based on 15-103 in the text) At an initial instant, a 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a speed (vB)1 = 4.8 ft/s. The attached cord is then pulled down through the hole with a constant speed vr = 2.2 ft/s. a. Determine the ball's speed at the instant r2 = 2 ft. Neglect friction and the size of the ball. Note that particle path is no longer of constant radius, and the particle has velocity components in both tangential and radial directions. b. How much work is done to pull down the cord from the initial instant to the instant when r2 = 2 ft? Neglect friction and the size o
Answer:
a
[tex]v_r =8.65 \ ft/s [/tex]
b
[tex] W_{1-2} = 3.24 \ ft \cdot lb[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 4 \ lb[/tex]
The radius is [tex]r= 3 \ ft[/tex]
The speed is [tex]v_B_1 = 4.8 \ ft /s[/tex]
The speed of the attached cord is [tex]v_c =2.2 \ ft[/tex]
The position that is been considered is [tex]r_1 = 2 \ ft[/tex]h
Generally according to the law of angular momentum conservation
[tex]L_a = L_b[/tex]
Here [tex]L_a[/tex] is the initial momentum of the ball which is mathematically represented as
[tex]L_a = m* v_B_1 * r[/tex]
while
[tex]L_b[/tex] is the momentum of the ball at r = 2 ft which is mathematically represented as
[tex]L_a = m* v_B_2 * r_1[/tex]
So
[tex]m* v_B_1 * r = m* v_B_2 * r_1[/tex]
=> [tex] 4.8 * 3 = v_B_2 * 2[/tex]
=> [tex] v_B_2 = 7.2 \ ft/s [/tex]
Generally the resultant velocity of the ball is
[tex]v_r = \sqrt{v_B_2^2 + v_B_1^2 }[/tex]
=> [tex]v_r = \sqrt{7.2^2 + 4.8^2 }[/tex]
=> [tex]v_r =8.65 \ ft/s [/tex]
Generally according to equation for principle of work and energy we have that
[tex]K_1 + \sum W_{1-2} = K_2[/tex]
Here [tex]K_1[/tex] is the initial kinetic energy of the ball which is mathematically represented as
[tex]K_1 = \frac{1}{2} * m* v_B_1^2[/tex]
While [tex]\sum W_{1-2}[/tex] is the sum of the total workdone by the ball
and [tex]K_2[/tex] is the final kinetic energy of the ball which is mathematically represented as [tex]K_2 = \frac{1}{2} * m* v_r^2[/tex]
So
[tex]\sum W_{1-2} = \frac{1}{2} * m (v_r^2 - v_B_1^2)[/tex]
Here m is the mass which is mathematically represented as
[tex]m = \frac{W}{g}[/tex] here W is the weight in lb and g is the acceleration due to gravity which is [tex]g = 32 \ ft/s^2[/tex]
So
[tex]\sum W_{1-2} = \frac{1}{2} * \frac{4}{32} * (8.65^2 - 4.8^2)[/tex]
=> [tex] W_{1-2} = 3.24 \ ft \cdot lb[/tex]
g A 38-g ball at the end of a string is swung in a vertical circle with a radius of 21 cm. The tangential velocity is 200.0 cm/s. Find the tension in the string:
Answer:
It depends on the location of the ball during the motion. The string tension are approximately 3.82 N (at the lowest point), 3.06 N (at the highest point), and 3.44 N (at the horizontal point).
Explanation:
Tension in the String can be determined by the Newton's 1st Law of Motion (The ball shouldn't be escaped from the trajectory). The value of [tex]\theta[/tex] indicates the angle that is measured from the vertical lines and the rope of length R)
[tex]\sum F=0\rightarrow \frac{mv^{2}}{R}+mg\cos\theta-T=0[/tex]
[tex]T=mg\cos\theta+\frac{mv^{2}}{R}[/tex]
[tex]T=(38\times10^{-3})(10)(\cos\theta)+(38\times10^{-3})\frac{2^{2}}{0.21^{2}}=0.38\cos\theta+3.44[/tex]
When the ball is at the lowest point, the value of angle [tex]\theta=0[/tex], so the string tension is approximately 3.82 N. If the ball is at the highest point the value of [tex]\theta=180^{0}[/tex], so the string tension is approximately 3.06 N, and at the horizontal point [tex]\theta=90^{0}[/tex], so the string tension is approximately 3.44 N.
What is the mass of a toy car if it has 5 J of potential energy and is sitting on top of a track that has a height of 2m?
(PE= m x g x h) (hint g=9.8 m/s2)
Explanation:
PE=mgh
5=m(9.8)(2)
m=5/19.6
m=0.2251 kg
m=225.1 grams
this diagram shows the sectionary box sitting on the floor which statement about other force exerted on the box are true?
questions:
it's normal force exerted by the floor. it balances the gravitational force.
it's friction force exerted by the floor. it balances the gravitational force.
it's normal for exerted by the floor. does not balance the gravitational force.
it's an applied force exerted from the left. It doesn't balance the gravitational force.
Answer: A) It's a normal force exerted by the floor. It balances out the gravitational force.
The normal force pushes up which means the box doesn't go through the floor. The up and down forces are balance. If the normal force was larger than the gravitational force, then the box would be pushed up and float/jump in the air. But the box is stationary and not moving, so that's why we don't have any acceleration and the forces effectively cancel each other out. In a way you can think of it like a see-saw.
Friction would only come into play if you applied a force and were trying to move the box. Friction counteracts the applied force. If you push the box to the left, then the friction would push to the right. Friction slows down the box and it allows it to not slide forever.
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Is normal force equal to gravity?
The normal force on an object at rest on a flat surface is equal to the gravitational force on that object.
▀▄▀▄Is friction a gravitational force?
Friction is the resistance to motion of one object moving relative to another. It is not a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.
▀▄▀▄Can normal force be greater than gravity?
Can you imagine the situation when Normal force is greater than mg ? Yes. When an additional downward force F is applied to a mass m resting on a horizontal surface, the normal force is FN=F+mg. ... An example is the normal force on an incline plane with an angle of θ due to a mass m.▄▀▄▀▄▀▄
What is the direction of the force you applied?
An applied force is an interaction of one object on another that causes the second object to accelerate or change velocity or direction. The force can be a push, pull, or drag. The resulting direction of an object depends on the relative direction of the force on the object. A force equation is F = ma.
A plane maintains constant acceleration and its velocity increases
from 40m/s to 100m/s over a 10 second interval. What is the plane's
average Distance ?
First calculate a, the acceleration of the plane, given by the equation;
[tex] \alpha = \frac{vf - vi}{t} = \frac{100 - 40}{10} = \frac{60}{10} = 6[/tex]
The use the relation given by
Vf²-Vi²=2ad
[tex]d = \frac{100 {}^{2} - 60 {}^{2} }{2 \times 6} = \frac{10000 - 3600}{12} = \frac{6400}{12} = 533.33 \: meters[/tex]
Problem 3
A car starts from rest at constant acceleration of 2.0 m/s2. At the same instant a truck travelling with a constant
speed of 10 m/s overtakes and passes the car.
(a) How far beyond the starting point will the car overtake the truck?
Answer:
100m
Explanation:
Equation of motion for he truck: s=ut
Equation of motion for the car: s=1/2at^2
the second solution gives , s=2u^2/a = 2*10^2/2 = 100m
Which material rises from cracks in oceanic crust?
dense crust
molten rock
continental crust
pillow-shaped rocks
Answer:
molten rocks is the correct answerThe molten rock rises from cracks in the oceanic crust, so option B is correct.
What is crust?The thinnest and most significant layer on our planet is its crust, which is made up of rocks and forms the planet's outermost layer. Despite making up less than 0.5 percent of the globe's total volume, it is essential to the majority of the natural cycles that take place all over the planet. The crust of the Earth is divided into layers, each of which contains a unique element.
Metamorphic rocks are defined as rocks that change into another rock. Sedimentary rocks are created by the deposition of material at the Earth's surface within water bodies, followed by cementation.
The crust that lies beneath land masses is known as the continental crust, whereas the crust that lies beneath the ocean's surface is known as the oceanic crust. The oceanic crust is younger and much thinner than the continental crust.
Therefore, the molten rock rises from cracks in the oceanic crust.
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1 point
An object has 50 J of kinetic energy and 20 J of potential energy. What is
the total energy possessed by the object?
30 J
35 J
50 J
70 J
Answer:
30 j
Explanation:
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The kinetic energy of an object is 50 J, and the potential energy is 20 J. The total energy of the object is 30J.
What is potential energy?In physics, potential energy is the energy that an object retains due to its position in relation to other objects, internal tensions, electric charge, or other elements.Potential energy is a type of stored energy that is determined by the interactions of various system components. When a spring is crushed or stretched, its potential energy increases. A steel ball has more potential energy if it is raised above the ground rather than falling to the ground.Potential energy, which is the latent energy in an object at rest, is one of the two types of energy. Kinetic energy, on the other hand, is the energy expressed by a moving object.To learn more about potential energy refer to:
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An electric light bulb has an efficiency of 18%.
400J of energy are supplied to the light bulb by electricity.
a Calculate the amount of energy transferred by light
Answer:
The answer that we got will be 22.22....
During which time Interval does the object travel approximately 10 meters?
OA. O seconds to 3 seconds
OB. 3 seconds to 5 seconds
OC. 5 seconds to 7 seconds
OD. 7 seconds to 8 seconds
OE. 8 seconds to 10 seconds
Answer:
[tex]OA. \: O \: seconds \: to \: 3 \: seconds[/tex]
A 50 kg cart is currently in static equilibrium. Which of the following claims is true? *
A.)The cart is experiencing unbalanced forces B.)The cart is at rest
C.)The cart is accelerating
D.)The cart is moving at a constant speed or velocity
Which statements describe kinetic and potential energy? Check all that apply.
Energy can be stored in the position of an object.
Energy is not present in a moving object.
Energy can be stored in the position of the particles that make up a substance.
Energy exists as movement of the particles of a substance.
Energy is greater in faster-moving particles than in slower-moving particles.
Energy is lower in objects with greater mass than in objects with less mass.
Answer:
First option, third option, fourth option, and the fifth option.
Explanation:
Kinetic energy is energy an object has when it's motion, the greater the speed the greater the kinetic energy. For example, a car moving and increasing in speed is kinetic energy since the object is in motion. If the car stops and parks in a parking lot that is potential energy. Potential energy is the amount of energy an object has when it's at rest or not in motion.
So, the answer for this question is as followed first option or "energy can be stored in the position of an object." Third option or "Energy can be stored in the position of the particles that make up a substance." Fourth option or "Energy exists as movement of the particles of a substance." The last answer will be the fifth option or "Energy is greater in faster-moving particles than in slower-moving particles."
Hope this helps.
Notice that the speed halfway down is not half the final speed. Another interesting point is that the final answer doesn't depend on the mass. That is really a consequence of neglecting the change in kinetic energy of Earth, which is valid when the mass of the object, the diver in this case, is much smaller than the mass of Earth. In reality, Earth also falls towards the diver, reducing the final speed, but the reduction is so minuscule it could never be measured. QUESTION Qualitatively, how will the answers change if the diver takes a running dive off the end of the board
Answer:
its speed is insignificant before the diver's speed change, so the result does not change
Explanation:
In this exercise of conservation of the momentum, the system is formed by the diver and the Earth
initial instant (before jumping)
p₀ = 0
final instant (after jumping)
[tex]p_{f}[/tex] = m v + M v²
how momentum is conserved
p₀ = p_{f}
0 = m v + M v²
v² = m / M v
since the mass of the Earth is M = 10²⁴ kg
its speed is insignificant before the diver's speed change, so the result does not change
You measured the length, diameter and mass of two different cylinders. In both cases, you found that the length had 3 significant figures and that length was the measurement with the fewest number of significant digits. If you found the weight densities to be 38119 N/m^3 and 38081 N/m^3 and you round these values to the correct number of significant figures, can you conclude the two cylinders are made of the same material (do they have the same weight density)?
a. Not enough information given.
b. Yes.
c. No.
Answer:
Weight Density 1 = 38100 N/m³
Weight Density 2 = 38100 N/m³
b. Yes
Explanation:
The formula for volume of cylinder is:
V = πr²l
where,
V = Volume
r = radius
l = length of cylinder
So, if length has the 3 significant figures which is least in all values, Then the volume must also be in 3 significant figures. The formula for weight density is:
Weight Density = Weight/Volume
Here, the volume has the least significant figures of 3, therefore, the weight densities must also have 3 significant figures:
Weight Density 1 = 38119 N/m³
Weight Density 1 = 38120 N/m³
Weight Density 1 = 38100 N/m³
Weight Density 2 = 38119 N/m³
Weight Density 2 = 38120 N/m³
Weight Density 2 = 38100 N/m³
Hence, the answer is:
b. Yes
If 600 N of force is used to move a car 4 meters, how much work is done?
Answer:
The answer is 2400 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distanceFrom the question
force = 600 N
distance = 4 m
We have
work done = 600 × 4
We have the final answer as
2400 JHope this helps you
A cheetah starts from rest and accelerates after a gazelle at a rate of 6.5 meters per second2for 3.0 seconds. Calculate the cheetah’s speed at the end of these 3.0 seconds.
Answer:
the speed of the cheetah at the end of the 3 seconds is: 19.5 m/s
Explanation:
Let's use the equation that relates speed with acceleration:
vf = vi + a * t
where vf stands for final velocity, vi stands for initial velocity, a for acceleration, and t for the time acceleration is applied. Then, in our case we have:
vf = 0 + 6.5 (3)
vf = 19.5 m/s