Answers
The final speed of the skier at the top mountain is determined as 9.27 m/s.
What is the change in the energy of the skier?
The change in the energy of the skier due to frictional force is calculated as follows;
ΔP.E = Pi + Ef
where;
Pi is the initial potential at the topEf is the energy lost to frictionThe distance of the plane travelled is calculated as;
sin35 = 2.5/L
L = 2.5 / sin35
L = 4.36 m
ΔP.E = mghi - μmgcosθ(L)
where;
m is the masshi is the initial heightg is acceleration due to gravityμ is coefficient of frictionΔP.E = (60 x 9.8 x 2.5) - (0.38)(60)(9.8) cos(35) x (4.36)
ΔP.E = 671.98 1 J
The final speed of the skier at the top of the plane;
P.E = K.E
P.E = ¹/₂mv²
v² = 2P.E /m
v = √(2P.E /m)
v = √(2 x 671.98) / 60)
v = 4.73 m/s
Total speed = -4.73 m/s + 14 m/s = 9.27 m/s
Thus, due to frictional force opposing the upward motion of the skier, the final speed at the top will be smaller than the initial speed.
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Related Questions
27. Scientists have observed an increase in global temperatures over the past 100 years. Which phenomena do scientists believe contributes to the increase in temperatures? A. an increase in undersea volcanic activity B. a decrease in the distance between Earth and the Sun C. an increase in certain gases released during the use of fossil fuels D. a decrease in the amount of water on Earth due to overconsumption
Answers
The answer is letter C) An increase in certain gases released during the use of fossil fuels. Although the others do cause an increase in temperature, their scale cannot be compared to the one caused by fossil fuels.
Question 10 of 10If one of two interacting charges is doubled, the force between the chargeswillO A. decrease by 4 timesO B. increaseC. decreaseD. stay the sameSUBMIT
Answers
The force between the two charges is
[tex]Force\text{ }\propto\text{ charge}[/tex]Thus, if the charge is doubled, then its force will also be doubled.
Hence the correct option is increase.
some more disconnectiob
Answers
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A rock sample has a mass of 6 kg and a volume of 0.002 m3. Calculate the density of this rock sample.
Answers
So 6/0.002=3,000 kg/m3.
The density of the rock sample is 3000 kg/m³.
Density is an important property of matter because it can be used to determine the weight of an object. For example, if you know the density of a rock sample and its volume, you can calculate its weight by multiplying the density by the volume. Density can also be used to determine the composition of a material. For example, if you know the densities of different materials, you can identify the material of a rock sample by measuring its density.
Density is a measure of how much mass is contained in a given volume.. The calculation of the density of the rock sample:
Density = Mass / Volume
Density = 6 kg / 0.002 m³
Density = 3000 kg/m³
As a result, the rock sample has a density of 3000 kg/m³.
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A convex spherical mirror has a radius of curvatureof 9 40 cm. A) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 17.5 cmCalculate the size of the imageC) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 10.0cmE) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 2.65cmG) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 9.60m
Answers
0.We are asked to determine the location of an image formed by an 7.75mm tall object that is located a distance of 17.5 cm from a convex mirror.
First, we will calculate the focal length using the following formula:
[tex]f=-\frac{R}{2}[/tex]Where:
[tex]\begin{gathered} f=\text{ focal length} \\ R=\text{ radius} \end{gathered}[/tex]Substituting the values we get:
[tex]f=-\frac{9.40cm}{2}[/tex]Solving the operations:
[tex]f=-4.7cm[/tex]Now, we use the following formula:
[tex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/tex]Where:
[tex]\begin{gathered} d_0=\text{ distance of the object} \\ d_i=\text{ distance of the image} \end{gathered}[/tex]Now, we substitute the known values:
[tex]\frac{1}{17.5cm}+\frac{1}{d_i}=-\frac{1}{4.7cm}[/tex]Now, we solve for the distance of the image. First, we subtract 1/17.5 from both sides:
[tex]\frac{1}{d_i}=-\frac{1}{4.7cm}-\frac{1}{17.5cm}[/tex]Solving the operation:
[tex]\frac{1}{d_i}=-0.27\frac{1}{cm}[/tex]Now, we invert both sides:
[tex]d_i=\frac{1}{-0.27}cm=-3.7cm[/tex]Therefore. the location of the image is -3.7 centimeters.
The other parts are solved using the same procedure.
Part B. To calculate the size of the image we will use the following relationship:
[tex]\frac{h_i}{h_o}=-\frac{d_i}{d_0}[/tex]Where:
[tex]h_i,h_0=\text{ height of the image and height of the object}[/tex]Substituting we get:
[tex]\frac{h_i}{7.75mm}=-\frac{-3.7cm}{17.5cm}[/tex]Solving the operations on the right side:
[tex]\frac{h_i}{7.75mm}=0.21[/tex]Now, we multiply both sides by 7.75:
[tex]h_i=(7.75mm)(0.21)[/tex]Solving the operations:
[tex]h_i=1.64mm[/tex]Therefore, the height of the iamge is 1.64 mm.
A bus is travelling along straight road at 100km/hr and the bus conductor walks a 6km/hr on the floor of the bus and in the same direction as the bus. Find the speed of the conductor relative to the road and relative to the bus. If the bus conductor now walks at the same rate but in opposite direction as the bus, find his new speed relative to the road.
Answers
Given data:
* The speed of the bus is 100 km/hr.
* The speed of the conductor is 6 km/hr.
Solution:
(a). If the bus and conductor are traveling in the same direction.
The net speed of the conductor relative to the road is,
[tex]v_1=v_b+v_c_{}[/tex]where v_1 is the velocity of the conductor relative to the road, v_b is the velocity of the bus, and v_c is the velocity of the conductor inside the bus,
Substituting the known values,
[tex]\begin{gathered} v_1=100+6 \\ v_1=106\text{ km/hr} \end{gathered}[/tex]Thus, the speed of the conductor relative to the road is 106 km/hr.
(b). The speed of the conductor relative to the bus is the speed of the conductor on the bus,
Thus, the speed of the conductor relative to the bus is 6 km/hr.
What is the momentum of a 934 kg car moving 10 m/s?
Answers
ANSWER:
9340 kg*m/s
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 934 kg
Speed (v) = 10 m/s
The formula to calculate the momentum is as follows:
[tex]\begin{gathered} p=m\cdot v \\ \text{ we replacing} \\ p=934\cdot10 \\ p=9340\text{ kg*m/s} \end{gathered}[/tex]The momentum of the car is 9340 kg*m/s.
A Tour de France cyclist at a rate of 6.5 m/s and is 333 m
ahead of a crew van with powerdrink refills. The van is
traveling at 15 m/s and is accelerating at a constant rate of
0.4 m/s2
How much time will it take for the crew van to catch up
with the cyclist?
Answers
Given data:
* The speed of the cyclist is 6.5 m/s.
* The initial distance of the cyclist from the van is 333 m.
* The initial velocity of the van is 15 m/s.
* The acceleration of the van is,
[tex]a=0.4ms^{-2}[/tex]Solution:
Let x be the distance from the van initial position at which the cyclist and van meet.
Let the cyclist meet the van at time t.
By the kinematics equation, the position of the cyclist at time t is,
[tex]x-333=u_ct+\frac{1}{2}a_ct^2[/tex]where u_c is the speed of the cyclist, a_c is the acceleration of the cyclist, t is the time taken and 333-x is the distance traveled by the cyclist at time t,
The acceleration of the cyclist is zero.
Substituting the known values,
[tex]\begin{gathered} x-333=6.5t+0 \\ x-333=6.5t \end{gathered}[/tex]By the kinematics equation, the position of the van after time t is,
[tex]x=u_vt+\frac{1}{2}a_vt^2[/tex]where u_v is the velocity of the van, a_v is the acceleration of the van, and t is the time taken,
Substituting the known values,
[tex]\begin{gathered} x=15t+\frac{1}{2}\times0.4\times t^2 \\ x=15t+0.2t^2 \end{gathered}[/tex]Substituting this value of x in the kinematics equation of the cyclist,
[tex]\begin{gathered} (15t+0.2t^2)-333=6.5t \\ 15t+0.2t^2-6.5t-333=0 \\ 0.2t^2+8.5t-333=0 \end{gathered}[/tex]By solving the quadratic equation,
[tex]\begin{gathered} t=\frac{-8.5\pm\sqrt[]{8.5^2-(4\times0.2\times(-333))}}{2\times0.2} \\ t=\frac{-8.5\pm\sqrt[]{^{}8.5^2+(4\times0.2\times333)}}{2\times0.2} \\ t=\frac{-8.5\pm18.4}{0.4} \\ t=24.8\text{ s or-67.25 s} \end{gathered}[/tex]As the value of time cannot be negative.
Thus, the time at which the cyclist and van meet is 24.8 seconds.
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A pair of fuzzy dice is hanging by a string from your rearview mirror. While you are accelerating from a stop-light to 28 m/s in 6.0 s, what angle theta does the string make with the vertical?
Answers
For the pair of fuzzy dice hanging by a string from the rearview mirror, while accelerating from a stop-light to 28 m/s in 6.0 s, the string makes an angle of 25.5 degrees with the vertical.
What is an angle?An angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle.
Parameters given include:
velocity: 28,/s time= 6.0 seconds
a = v/t = 28m/s / 6.0s = 4.667 m/s²
Taking into account the forces acting on the dice...there is a force of gravity acting straight down with a magnitude of mg.
so we have that
Angle = tan-1(a/g)
Angle = tan-1(4.667 / 9.8) = 25.5 degrees.
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upon leaving her club, the golf ball moved upward to a height above the surrounding trees. is the ke and pe increasing, decreasing, or staying the same?
Answers
ANSWER
PE increases and KE decreases
EXPLANATION
As described, the golf ball is moving and changing its height, like in the following diagram,
By the law of conservation of energy, the total energy when the ball starts moving and during the whole motion until it stops, must be the same. This total energy is the sum of the potential energy and the kinetic energy.
When the club hits the ball, it gives it a certain amount of kinetic energy but no potential energy. As the ball starts going uphill, the potential energy starts to increase, since it depends on the height of the object. Therefore, to maintain the total energy constant, the kinetic energy must decrease.
Hi can you help me understand how to do this?
Answers
From the information given,
initial velocity of parachute = 198 ft/s
We want to convert 198 ft/s to m/s
Recall,
1 m = 3.3 ft
x m = 198 ft
By crossmultiplying, we have
3.3x = 198
x = 198/3.3
x = 60
Thus,
198 ft/s = 60 m/s
Rate = 60 m/s
To determine the distance that the parachute will fall in 10 seconds, we would apply one of Newton's equations of motion which is expressed as
s = ut + 1/2gt^2
where
s = distance covered
u = initial velocity
t = time
g = acceleration due to gravity and its value is - 9.8m/s^2
From the information given,
t = 10
u = 60
By substituting these values into the formula, we have
s = 60 x 10 - 1/2 x 9.8 x 10^2
s = 600 - 490
s = 110
The distance covered by the parachute in 10s is 110 m
Name the instrument which is made on the basis of expansion of heat.
Answers
The instrument is the Thermometer
Answer:
It is a thermometer and it helps to see the temperature
I dont understand this formula I need helpF = 6.67408 * 10^-11 * (1.5 * 10^5) (8.5 * 10^2) ------------------------------ 2500^2
Answers
Answer:
F = 1.36*10^-9
Explanation:
The given equation is
[tex]F=6.67\ast10^{-11}\ast\frac{(1.5\ast10^5)(8.5\ast10^2)}{2500^2}[/tex]To calculate the value of F, you need to multiply 1.5*10^5 by 8.5*10^2 as follows:
[tex]F=6.67\ast10^{-11}\ast\frac{1.275\times10^8}{2500^2}[/tex]Then, 2500² = 2500 x 2500 = 6.25 * 10^6, so replacing the value, we get:
[tex]F=6.67\ast10^{-11}\ast\frac{1.275\ast10^8}{6.25\ast10^6}[/tex]Now, we need to divide 1.275*10^7 by 6.25*10^6, so
[tex]F=6.67\ast10^{-11}\ast(20.4)[/tex]Finally, multiply 6.67*10^-11 by 20.4, so
[tex]F=1.36\ast10^{-9}[/tex]A sled of mass 26 kg has an 18 kg child on it. If big brother is pulling with a 30 N force to the right and 10 N up, and big sister is pushing with a 40 N force to the right and 16 N down, what is the normal force?
Answers
Given data:
* The mass of the sled is m_1 = 26 kg.
* The mass of the child is m_2 = 18 kg.
* The force in the upwards direction by the big brother is F_1 = 10 N.
* The force in the downwards direction by the big sister is F_2 = 16 N.
Solution:
The net mass on the sled along with the child is,
[tex]\begin{gathered} m=m_1+m_2 \\ m=26+18 \\ m=44\text{ kg} \end{gathered}[/tex]The net weight of the sled along with the child is,
[tex]\begin{gathered} w=mg \\ w=44\times9.8 \\ w=431.2\text{ N} \end{gathered}[/tex]The weight of the sled along the child is acting on the sled in the downwards direction.
Thus, the normal force acting on the sled (taking upward force as negative and downward force as positive) is,
[tex]\begin{gathered} N=w+F_2-F_1 \\ N=431.2+16-10 \\ N=437.2\text{ newton} \end{gathered}[/tex]Thus, the normal force acting on the sled is 437.2 N.
A student pushes a baseball of m = 0.13 kg down onto the top of a vertical spring that has its lower end fixed to a table, compressing the spring a distance of d = 0.18 meters from its original equilibrium point. The spring constant of the spring is k = 970 N/m. Let the gravitational potential energy be zero at the position of the baseball in the compressed spring.
A. What is the maximum height, h, in meters, that the ball reaches above the equilibrium point?
B. What is the ball’s velocity, in meters per second, at half of the maximum height relative to the equilibrium point?
Answers
The maximum height reached by the baseball above the equilibrium point is 12.14 m.
The ball’s velocity, in meters per second, at half of the maximum height relative to the equilibrium point is 458.8 m/s.
What is the maximum height attained by the ball?The maximum height reached by the ball is determined as follows:
Data given:
compression of the spring is d = 0.18 m.mass of the baseball is m = 0.13 kgthe spring constant, K = 970 N/mThe gravitational potential energy at the compressed position is zero.
Based on the law of conservation of energy, the total energy of the system is conserved.
Let the final height from the bottom be h
m * g* h = ¹/₂ * K * d²
h = (k * d²) / (2 * m * g)
h = 970 * 0.18² / (2 * 0.13 * 9.81)
h = 12.32 m
Height above the equilibrium point = 12.32 - 0.18
Height above the equilibrium point = 12.14 m
Velocity is calculated1 as follows:
Half of the maximum height relative to the equilibrium point = 12.14/2 + (0.18) = 6.25 m
¹/₂ * m * v² = ¹/₂ * K * d²
m * v² = K * d²
v = √(K * d² / m)
v = √(970 * 6.25² / 0.13)
v = 458.8 m/s
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If an airplane is running low on fuel, the pilot may decide to dump unneededweight. As the airplane gets lighter, the engines need less fuel to generate thesame amount of acceleration for flight. The pilot has taken advantage ofNewton's law of motion.A. fourthB. thirdO C. firstD. second
Answers
According to Newton's second law we have that the force is equivalent to the product of the mass and the acceleration:
[tex]F=ma[/tex]This means that if we decrease the mass "m" of the object the required force is decreased also. Since the force is proportional to the required energy and the energy comes from the fuel consumption this means
Which label goes through the horizontal axis?
A. Distance
B. Acceleration
C. Force
D. Mass
Answers
Answer:
A: Distance
Explanation:
You can not change gravity with acceleration or force in a weird way. Only mass and distance really affect gravity in normal physics.
Gravity is stronger when two objects are closer. Therefore gravity would decrease as distance increased.
Mass increases gravity.
The image below shows that gravity (depth) increases as the distance from the black hole decreases.
A 50 kW tractor moves at a speed of 2.5 m/s. What is the traction force of the tractor?
Answers
In order to calculate the force, we can use the formula below:
[tex]P=F\cdot v[/tex]Where P is the power (in W), F is the force (in N) and v is the velocity (in m/s).
So we have:
[tex]\begin{gathered} 50000=F\cdot2.5\\ \\ F=\frac{50000}{2.5}\\ \\ F=20000\text{ N} \end{gathered}[/tex]Therefore the traction force is 20 kN.
Two plastic blocks of the same plastic material have the following characteristics:
Block A: Mass 10 g, Volume 25 mL, and
Block B: Mass 20 g, Volume 50 mL.
Which of the statements below is true? (Density of water is 1g/mL at room temperature.)
Group of answer choices
Both blocks will sink equally under water.
Block A will sink deeper under water.
Both blocks will float equally over water.
Block B will float more than Block A over water.
Answers
The true statement, given the data from the question is: Both blocks will float equally over water.
How to determine the true statementTo know which statement is true, we shall obtain the density of each blocks. Details below
For block A:
Mass of block A = 10 gramsVolume of block A = 25 mL Density of block A = ?Density = mass / volume
Density of block A = 10 / 25
Density of block A = 0.4 g/mL
For block B:
Mass of block B = 20 gramsVolume of block B = 50 mL Density of block B = ?Density = mass / volume
Density of block B = 20 / 50
Density of block B = 0.4 g/mL
From the above calculation, we can see that the two blocks have the same density.
Thus, we can conclude that the true statement is both blocks will float equally over water.
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Two identical point charges exert a repulsive force of 0.500 N on one another when separated by 1.5 m. What is the magnitude of the net charge of either point charge?
Answers
Given,
The repulsive force exerted by the charges, F=0.500 N
The distance between the charges, d=1.5 m
From Coulomb's law,
[tex]F=\frac{\text{kqq}}{r^2}[/tex]Where q is the magnitude of the charge of each point charge and k is the coulomb's constant.
On rearranging the above equation,
[tex]\begin{gathered} F=\frac{kq^2}{r^2} \\ \Rightarrow q=\sqrt[]{\frac{F}{k}}r \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} q=\sqrt[]{\frac{0.5}{9\times10^9}}\times1.5 \\ =1.1\times10^{-5}\text{ C} \end{gathered}[/tex]Thus the magnitude of the charge of each point charge is 1.1×10⁻⁵ C
Therefore the correct answer is option B.
cassy can get more force on the bricks she breaks with a blow of her bare hand when _______.
Answers
Answer:Her hand is made to bounce from the bricks
Explanation:
Cassy can get more force on the bricks that she breaks with a blow of her bare hand when her hand is made to bounce from the bricks,
What is Force?A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's speed can vary, or accelerate, as a result of a force. Intuitively, a pull or a push can also be used to describe force. Being a vector quantity, a force also has magnitude and direction. The SI unit of newton is used to measure it (N). The letter F stands for force.
According to Newton's second law's original formulation, an object's net force is equal to the speed at which momentum is changing over time.
Objects' velocities can be altered by the concepts of push, drag, and torque. Thrust causes an object's velocities to increase, while torque causes an object's velocities to decrease. Each part of an extended body typically exerts pressures on its neighboring sections, and the internal mechanical force is determined by how these stresses are distributed.
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Writing Simple ExpressionsChoose all of the TRUE statement(s).Add 6 and 5, then multiply by 4 is the same as 4(6 + 5).4 times greater than 80 + 25 is the same as 4 x (80 + 25).Subtract 15 from 42 is the same as 15 − 42.9 times greater than 11 + 12 is the same as 9 + 11 + 12.8 times greater than 21 + 15 is the same as 8(21 + 15).
Answers
Add 6 and 5, then multiply by 4 is the same as 4(6 + 5). TRUE
4 times greater than 80 + 25 is the same as 4 x (80 + 25). TRUE
Subtract 15 from 42 is the same as 15 − 42. TRUE
9 times greater than 11 + 12 is the same as 9 + 11 + 12. FALSE
9 (11+12)
8 times greater than 21 + 15 is the same as 8(21 + 15). TRUE
You and some friends are duck pin bowling. You’re up and you roll your bowling ball (m = 1.6kg) down the lane. It collided with one pin (0.68kg) on the end in a perfectly elastic one dimensional collision. If the ball was moving at a velocity of 6 m/s just before it hits the pin, what is the velocity of the bowling pin after the collision?
Answers
Given:
The mass of the bowling ball is m1 = 1.6 kg
The mass of the pin is m2 = 0.68 kg
The initial velocity of the ball is
[tex]v_i=\text{ 6 m/s}[/tex]Required: Velocity of the bowling pin after the collision.
Explanation:
According to the conservation of momentum, the velocity after the collision will be
[tex]\begin{gathered} m1v_i+m2\times0\text{ =\lparen m1+m2\rparen v}_f \\ v_f=\frac{m1v_i}{m1+m2} \\ =\frac{1.6\times6}{1.6+0.68} \\ =4.21\text{ m/s} \end{gathered}[/tex]Final Answer: The velocity of the bowling pin after the collision is 4.21 m/s
When a rubber band with a force constant of 58.7 N/m is stretched a certain distance,
there is 1.94 J of elastic potential NRG stored in it. How far has the band been stretched
Answers
According to the given statement the band has been stretched far is 0.25 m.
What is the meaning of elastic potential?The energy that is stored when a force is used to bend an elastic object is known as elastic potential energy. Until the force is released as well as the object springs returns to its original form, doing work in the process, the energy is retained. The object may be squeezed, stretched, or bent during the deformation.
What's an example of elastic potential energy?For example, if you pull a spring, it will return to its initial form when you release it (energy input equals energy output.) Elastic potential energy is rendered possible by this.
Briefing:U= 1.94 J
k = 58.7 N/m
U = 1/2 kΔ x²
1.94 = 1/2 kΔ x²
3.88/58.7 = Δ x²
x = √(3.88 /58.7)
x = 0.25 m
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What is the gravitational force between two trucks, each with a mass of 2.0 x 10^4 kg, that are 2.0m apart? (G=6.673 x 10^-11 N•m^2/kg^2)
Answers
Firs we will use the next formula
[tex]F=G\text{ }\frac{m_1\cdot m^{}_2}{r^2}[/tex]Where G is the gravitational force, m nd m2 are the masses and r is the distance between the masses
In our case
m1=m2= 2.0 x 10^4 kg
r= 2m
G=6.673 x 10^-11 N•m^2/kg^2
We substitute
[tex]F=6.673\times10^{-11}\cdot\frac{2.0x10^4\cdot2.0x10^4}{2^2}=0.0066743N=6.7\times10^{-3}[/tex]ANSWER
The gravitational force is 0.00667=6.7x10^-3N
In a system of 2 large round objects, R1 and R2 (R1 is larger), what properties will affect the force of gravity between them? (select all that apply)
Answers
According to Newton's Law of Universal Gravitation, if two particles with masses M and m are located a distance r from each other, an attractive force between them appears, and its magnitude is given by:
[tex]F=G\frac{Mm}{r^2}[/tex]Where G is the gravitational constant:
[tex]G=6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2}[/tex]From the equation, we can see that the properties that affect the force between them are their masses and the distance between them.
Then, the correct choices are:
- Mass of R1
- Mass of R2
- Distance from the center of R1 to the center of R2.
When you apply a force F on an object of mass of m, it would produce acceleration a. If you apply the same force on another object of mass 2m, how would be the acceleration of the second object?
Answers
ANSWER
[tex]\frac{a}{2}[/tex]EXPLANATION
When a force is applied on an object of mass m, it produces an acceleration of a.
We can represent this relationship using Newton's second law of motion:
[tex]\begin{gathered} F=ma \\ \Rightarrow a=\frac{F}{m} \end{gathered}[/tex]Now, the same force is applied on an object with a mass of 2m.
Let the acceleration experienced by the object be a1. This implies that:
[tex]\begin{gathered} F=\left(2m\right)\left(a_1\right) \\ \Rightarrow a_1=\frac{F}{2m} \end{gathered}[/tex]We can write this new acceleration in terms of a as follows:
[tex]\begin{gathered} a_1=\frac{1}{2}\left(a\right) \\ a_1=\frac{a}{2} \end{gathered}[/tex]That would be the acceleration of the second object.
The answer is the third option.
A flywheel with a moment of inertia of 3.45 kg·m2is initially rotating. In order to stopits rotation, a braking torque of -9.40 N·m is applied to the flywheel. Calculate the initialangular speed of the flywheel if it makes 1 complete revolution from the time the brake isapplied until it comes to rest
Answers
Given data
*The given moment of inertia is I = 3.45 kg.m^2
*The given braking torque is T = -9.40 N.m
*The angular distance traveled is
[tex]\theta=(1\times2\pi)rad_{}[/tex]*The final angular speed is
[tex]\omega=0\text{ rad/s}[/tex]The angular acceleration of the flywheel is calculated by using the torque and moment of inertia relation as
[tex]\begin{gathered} T=I\alpha \\ \alpha=\frac{T}{I} \\ =\frac{-9.4}{3.45} \\ =-2.72rad/s^2 \end{gathered}[/tex]The formula for the initial angular speed of the flywheel is given by the rotational equation of motion as
[tex]\omega^2-\omega^2_0=2a\theta[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} (0)^2-\omega^2_0=2\times(-2.72)(2\pi) \\ \omega_0=\sqrt[]{2\times2.72\times2\pi} \\ =5.88\text{ rad/s} \end{gathered}[/tex]Hence, the initial angular speed of the flywheel is 5.88 rad/s
If a space ship traveling at a 1000 miles per hour enters and area free of Gravitational forces,it’s engine must run at some minimum level in order to maintain the ships velocity. is this statement true or false
Answers
The given statement is false.
If a spaceship traveling at 1,000 miles per hour enters an area free of gravitational forces, its engine must run at some maximum level in order to maintain the ship’s velocity
On a standard day the speed of sound is 345 meters per second. A whistle whose frequency is 725 Hz is movingtoward an observer at a speed of 25.2 meters per second. What is the wavelength of the sound at the observer?(a) 0.367 m(b) 0.441 m(c) 0.511 m(d) 0.623 m
Answers
Take into account that this is a situation where the source moves presenting the Doppler effect.
In order to determine the wavelength of the sound generatd by the whistle at the observer, first calculate the frequency at the observer by using the following formula:
[tex]f=\frac{v}{v-v_s}f_s[/tex]where:
f: frequency at the observer = ?
fo: source frequency = 725 Hz
vs: source speed = 25.2 m/s
v: speed of sound = 345 m/s
replace the previous values of the parameters into the fomrula for f:
[tex]\begin{gathered} f=\frac{345m/s}{345m/s-25.2m/s}725Hz \\ f=782.1Hz \end{gathered}[/tex]Next, use the following formula to determine the wavelength of the sound at observer, by using the previous result:
[tex]\lambda=\frac{v}{f_s}=\frac{345m/s}{782.1Hz}=0.441m[/tex]Hence, the wavelength of the sound at the observer is 0.441 m