Answer:
The value is [tex]P = 14.7 \ N[/tex]
Explanation:
From the question we are told that
The weight of the hollow cylinder is [tex]W = 40 \ N[/tex]
The radius of the hollow cylinder is [tex]r = 0.25 \ m[/tex]
The distance which it is pulled is [tex]d = 5 \ m[/tex]
The velocity of the end of the rope is [tex]v = 6 \ m/s[/tex]
Gnerally the mass of the hollow cylinder is
[tex]m = \frac{W}{g }[/tex]
=> [tex]m = \frac{ 40 }{ 9.8 }[/tex]
=> [tex]m = 4.081 \ kg[/tex]
Generally angular displacement for the distance covered is mathematically represented as
[tex]\theta = 2 \pi * \frac{ d } {2\pi r }[/tex]
=> [tex]\theta = 2 \pi * \frac{ 5 } {2\pi r }[/tex]
=> [tex]\theta = \frac{ 5 } { 0.25}[/tex]
=> [tex]\theta =20[/tex]
Generally the torque experienced by the hollow cylinder is mathematically represented as
[tex]P * r = I * \alpha[/tex]
Here I is the moment of inertia
=> [tex]P * r = m r^2 * \alpha[/tex]
=> [tex]\alpha = \frac{P }{ mr }[/tex]
Generally from kinematic equation
[tex]w_f ^2 = w_i ^2 + 2\alpha \theta[/tex]
=> [tex]w_f ^2 = w_i ^2 + 2\alpha \theta[/tex]
Generally the final angular velocity is mathematically
[tex]w_f = \frac{v}{r}[/tex]
=> [tex]w_f = \frac{ 6 }{ 0.25 }[/tex]
=> [tex]w_f = 24 \ m/s[/tex]
Generally the initial angular velocity is Zero given that the hollow cylinder was at rest before rolling
[tex]24^2 = 0^2 + 2* \frac{P}{4.081 *0.25 } * 20[/tex]
=> [tex]24^2 = 0^2 + 2* \frac{P}{mr} * 20[/tex]
=> [tex]P = 14.7 \ N[/tex]
In a place covered by shadow of cloud sun cannot be seen . Explain with reasons .
Answer:
Because even though our eyes have a huge dynamic range (ability to pick out details in sharply lit and lesser lit areas simultaneously) than any camera, there's a limit.
When there's strong sunlight, your pupils contract and let less light in, which makes the shadows look darker.
When it's cloudy, your pupils widen and let more light in, which makes the shadows look less dark.
Do some experiments with a camera and you'll soon get the hang of it.
NOTE: Also test HDR (high dynamic range) photography, where the camera takes three or more pictures in quick succession, with different exposure settings, and combines them to get the most detail of both bright and dark areas. The result is more or less what we percieve.
A golf ball is sitting on a tee. The ball is struck with a golf club and flies
through the air. How does the force on the club compare with the force on the
ball when momentum is transferred between the club and ball?
Answer:
c i kn now it is
Explanation:
Describe what happens to the moving boat when the oars are out of the water and the forward thrust is zero
Answer:
The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.
A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the spring
The forces constant (spring constant) of the spring will be 41.09 N/m.
What is spring force?The force required to extend or compress a spring by some distance scales linearly concerning that distance is known as the spring force. Its formula is;
F = kx
The given data in the problem is;
F is the spring force = 30 N
K is the spring constant= ?
x is the displacement of spring = 0.73 m
The spring constant is;
K =F/x
K=30/0.73
K=41.09 N/m
Hence the force constant (spring constant) of the spring will be 41.09 N/m.
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When two ocean plates come together, one ocean plate __________________
under the other, causing a chain of ________________ __________________
to form.
Answer:
A subduction zone is also generated when two oceanic plates collide — the older plate is forced under the younger one — and it leads to the formation of chains of volcanic islands known as island arcs.
Explanation:
an object has an mass of 15 kg and is falling at a rate of 2.0 m/s what is the momentum?
Answer:
30 kg.m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 15 × 2
We have the final answer as
30 kg.m/sHope this helps you
help please asap due 20 minutes please help me
A man speeding at 40m/s decides to outrun the cops and starts to
accelerate at a rate of 2.5m/s2 for 12 seconds. What is the criminal's new
speed?
Answer:
70 m/s.
Explanation:
Given that,
Initial speed, u = 40 m/s
Acceleration = 2.5 m/s²
Time, t = 12 s
We need to find criminal's new speed. Let it is v. Using equation of motion to find it as follows :
v = u +at
Substitute all the values
v = 40 + 2.5(12)
v = 70 m/s
So, the new speed is 70 m/s.
What travels by vibrating particles? Mechincal Waves or ElecrtoMagnetic Waves.
Answer:mechanical waves.
Explanation:
Mechanical waves require the particles of the medium to vibrate in order for energy to be transferred. For example, water waves, earthquake/seismic waves, sound waves, and the waves that travel down a rope or spring are also mechanical waves.
show your work. john uses a 25N force to push a boulder off a cliff that is 312m tall. What is the work done on the boulder?
Answer:7800
work=force x distance
Force in Newtons
Distance in Meters
Work in Joules
a metallic cube whose each side is 10 cm is subjected to a shearing force of 100 kg. The top force is displaced through 0.25 cm with respect to the bottom. calculate the shearing stress strain and modulus
Answer:
9.8×104Nm−2,0.025,3.92×106Nm−2
Solution :
Here, L=10cm=10×10−2m
F=100kgf=100×9.8N
ΔL=0.25cm=0.25×10−2m,
Shearing stress =FL2=100×9.8(10×10−2) Sheraing strain =ΔLL=0.25×10−210×10−2 = 0.025 Shear Modulus of elasticity, G=Shearing stressShearing strain=9.8×1040.025
=3.92×106Nm−2
Explanation:
A car traveling with 500,000 J of kinetic energy is brought to a kinetic energy of
100,000 J in 12 seconds. What is the force acting on the car to accomplish this?
A:41,666.67
B:3x10^-5N
C:33,333.33N
D:8,333.33N
Answer:
33,333.33 N
Explanation:
Given that :
Initial kinetic energy = 500,000 J
Final kinetic energy = 100,000 J
Using the relation :
Force * time = change in momentum (Newton's law)
Force (F) * 0.12 = (500,000 - 100,000)
0.12F = 400,000 J
Force = (400,000 J) / 0.12s
Force = 33333.333
Force = 33,333.33 N
(Blank) is caused by plate motion.
It’s please help.
Answer:
heat
Explanation:
please answer asignment due today
Answer:
Give me some time okkkk
A 0.15 kg ball is moving with a velocity of
35 m/s. Find the momentum of the ball.
Answer:
5.25 kg.m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 0.15 × 35
We have the final answer as
5.25 kg.m/sHope this helps you
A fish swimming at a rate of .6 m/s notices a huge shark. Three seconds later, the fish is swimming at a speed of 3 m/s. What is the fish's acceleration?
0.8 m/s/s
-0.8 m/s/s
12.5 m/s/s
-12.5 m/s/s
Answer:
C
Explanation:
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i think
9. A student notices that wearing darker colors in sunlight makes him feel warmer, so he decides to conduct an experiment. He takes five pieces of different
colored cloth and wraps
each one around a water bottle. He then places all five bottles in direct sunlight and measures the temperature of the water in each bottle an hour later
What is the dependent variable in this experiment?
O the time he leaves it in the sunlight
O the amount of water in each bottle
O the color of the cloth
O the temperature of the water
Answer: 4
Explanation:
The dependent variable is the temperature of the water.
A(n) ____________ stretch is one done where antagonist muscles are used to stretch the muscles. But the _____________ stretch is one that is done that the muscle needs help from something or someone to stretch the muscle.
Answer: Dynamic - Static Flexibility
Explanation:
Can u anser 5,6 on the picture
Answer: Number 6 is Periods
Explanation:
a squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy
What is the squirrels mass
Answer:
yeet yeet yeet yeet
Explanation:
Kinetic energy (K.E):-
So, the Mass of the Squirrel is 0.51 Kg (or) 510 grams.
A squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy.
What is the squirrel’s mass?
Answer: 0.51 kg
A 75kg bicyclist (including the bicycle), initially at rest at the top of a hill, coasts down the hill, reaching a speed of 14.6m/s at the bottom of the hill. The distance and height of the hill are shown. Neglect any friction impeding the motion and the rotational energy of the wheels. List the energy types at the initial and final time and whether work and loss (due to non-conservative forces) occur as well as the corresponding amounts of energy.
The energy type at the initial time is potential energy and the energy at the final time or position is kinetic energy.
What is the law of conservation of energy?The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.
Based on the law of conservation of mechanical energy, the formula for the change in the kinetic energy and the potential energy of the bicyclist is given as;
K.Ei + P.Ei = K.Ef + P.Ef
where;
K.Ei is the initial kinetic energy of the bicyclistK.Ef is the final kinetic energy of the bicyclistP.Ei is the initial potential energy of the bicyclistP.Ef is the final potential energy of the bicyclistThe kinetic energy of the bicyclist increases with increase in the velocity of the bicyclist while the potential energy increases with increase in the height of the bicyclist.
At the initial position when the bicyclist is at rest, the kinetic energy is zero, so the only energy at the initial position is potential energy because the height is maximum.
In addition, at the final position, the velocity of the bicyclist is maximum and the height is zero, so the only energy at the final position is kinetic energy.
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draw position time graph when speed is increasing
Explanation:
We need to draw position-time graph when the speed is increasing.
The slope of position-time graph gives the speed of an object.
Position means distance covered.
When the speed of an object is increasing with time. It means it is moving with increasing speed.
The attached figure shows the position -time graph when speed is increasing.
An object that falls and accelerates solely as a result of gravity is said to be in
(2 points)
A. terminal velocity
B. free fall
C. air resistance
D. terminal acceleration
The magnitude of vector vector A is 84.9 m and it points in the +y axis direction. The magnitude of vector vector B is 195.0 m and it points at an angle of 41.0° counterclockwise from +x axis. The magnitude of vector vector C is 126.2 m and it points in the +x axis direction.
Solution:
The magnitude of A vector is 84.9 m in the positive y-axis direction.
So the X component of A =0
the Y component of A = 84.9 m
Now the magnitude of B vector is 195 m and it makes an angle of 41° in the direction from the positive x-axis direction.
So the X component of B = B cos 41°
= 195 x cos 41°
= 195 x 0.75 = 146.25 m
the Y component of B = B sin 41°
= 195 x sin 41°
= 195 x 0.65 = 126.75 m
Now it is given that vector C has a magnitude of 126.2 m and it makes a direction towards the positive x-axis.
So the X component of C =126.2 m
the Y component of C = 0
Comparing all these, we get
1. B vector has the largest X component
2. B vector has the largest Y component
How much would a pair of 0.5 kg shoes weigh on Earth? (Include units in
your answer) *
Answer:
1.1 lbs
Explanation:
To convert kg to lbs you multiply kilograms by 2.2. So 0.5kg × 2.2 equals to 1.1 lbs
Es muy común que cuando se viaja hacia un río o lago se juegue "ranita", el cual consiste en lanzar una piedra horizontalmente hacia adelante para que cuando ésta toque la superficie del agua haga varios "saltos" sobre el agua. Durante un juego de estos, un desocupado nota que una de las piedras que arroja se demora 0,4 s en tocar la superficie del agua y la toca a 2,5 m de la orilla del lago, desde donde fue lanzada. Encuentre: a) La altura de la que fue lanzada la piedra. b) La velocidad con la que fue lanzada.
Answer:
a) La piedra es lanzada desde una altura de 0,785 metros.
b) La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.
Explanation:
a) Dado que la piedra es lanzada horizontalmente, tenemos que la piedra experimenta un movimiento horizontal a velocidad constante y uno vertical uniformemente acelerado debido a la gravedad. La altura de la que fue lanzada la piedra se puede determinar mediante la siguiente ecuación cinemática:
[tex]y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (1)
Donde:
[tex]y[/tex] - Altura final, medida en metros.
[tex]y_{o}[/tex] - Altura inicial, medida en metros.
[tex]v_{o,y}[/tex] - Componente vertical de la velocidad inicial, medida en metros por segundo.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo cuadrado.
Si sabemos que [tex]y = 0\,m[/tex], [tex]v_{o,y} = 0\,\frac{m}{s}[/tex], [tex]t = 0,4\,s[/tex] y [tex]g = -9,807\,\frac{m}{s^{2}}[/tex], entonces la altura inicial de la piedra es:
[tex]y_{o} = y-v_{o,y}\cdot t -\frac{1}{2}\cdot g\cdot t^{2}[/tex]
[tex]y_{o} = 0\,m-\left(0\,\frac{m}{s} \right)\cdot (0,4\,s)-\frac{1}{2}\cdot \left(-9,807\,\frac{m}{s^{2}} \right) \cdot (0,4\,s)^{2}[/tex]
[tex]y_{o} = 0,785\,m[/tex]
La piedra es lanzada desde una altura de 0,785 metros.
b) Ahora, obtenemos el componente horizontal de la velocidad inicial a partir de la siguiente ecuación cinemática:
[tex]v_{o,x} = \frac{x-x_{o}}{t}[/tex] (2)
Donde:
[tex]x_{o}[/tex], [tex]x[/tex] - Posiciones horizontales iniciales y finales, medidas en metros.
[tex]t[/tex] - Tiempo, medido en segundos.
Si tenemos que [tex]x_{o} = 0\,m[/tex], [tex]x = 2,5\,m[/tex] y [tex]t = 0,4\,s[/tex], entonces el componente horizontal de la velocidad inicial es:
[tex]v_{o,x} = \frac{2,5\,m-0\,m}{0,4\,s}[/tex]
[tex]v_{o,x} = 6,25\,\frac{m}{s}[/tex]
La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.
WRITE A PARAGRAPH ABOUT PRESSURE
Answer:
pressure is a horrible thing to go through, it can lead to many bad side effects, like burn out and it can also lead to high blood pressure, head aches, heart problems, depression, anxiety, and many more damaging effects. Pressure can destroy someones mental health if not dealed worth properly
I hope this is okay! I'm not sure what pressure you meant
5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley to a 0.0150 kg mass.
Draw a free body diagram showing all the forces acting on each object
Calculate the acceleration of the masses
Calculate the tension force in the cord
How long will it take the block to get to go 1.2 m to the edge of the table.
What will the velocity be as soon as it gets to the edge?
Answer:
1) Please find attached, created with Microsoft Visio
2) The acceleration of the masses connected by the light string is 0.00735 m/s²
3) The tension in the cord is 0.147 N
4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s
5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s
Explanation:
1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio
2) The acceleration of the masses connected by the light string is given as follows;
F = Mass, m × Acceleration, a
The mass of the truck, M = 20.0 kg
The mass attached to the string, hanging rom the pulley, m = 0.0150 kg
The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N
The pulling force acting on the cart, F = M × a
∴ F = 0.147 N = 20.0 kg × a
a = 0.147 N/(20.0 kg) = 0.00735 m/s²
The acceleration of the truck = a = 0.00735 m/s²
3) The tension in the cord = F = 0.147 N
4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²
Where;
s = The distance to the edge of the table = 1.2 m
u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)
a = The acceleration of the cart = 0.00735 m/s²
t = The time taken
Substituting the known values, gives;
s = u·t + 1/2·a·t²
1.2 = 0 × t + 1/2 ×0.00735 × t²
1.2 = 1/2 ×0.00735 × t²
t² = 1.2/(1/2 ×0.00735) ≈ 326.5306
t = √(1.2/(1/2 ×0.00735)) ≈ 18.07
The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s
5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;
v² = u² + 2·a·s
u = 0 m/s
v² = 0² + 2 × 0.00735 × 1.2 = 0.001764
v = √(0.001764) = 0.042
The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.
Answer:
There's no answer I'm just taking points like you did me, so thank you for your points I'll put them to good use ;)
13. Austin rode his bike 10 m/s for two minutes. How far did he travel? A. 200 meters B. 1200 meters C. 1000 meters D. 20 meters
Answer:
B. 1200
Explanation:
60 sec in one min in 2 min there will be 120 sec. 10x120=1200
It takes a crane 59s to lift a flagstone using 342 W of power. How much work is done on the flagstone?
Answer: The work done on the flagstone is 20178 J
Explanation:
Power is the rate at which work is done . It is equal to the amount of work done divided by the time it takes to do the work.
[tex]Work=Power\times time[/tex]
Given : work = ?
Power = 342 W = 342J/s
Time = 59 s
[tex]Work=342J/s\times 59s=20178J[/tex]
Thus the work done on the flagstone is 20178 J