The velocity of the object is 4.1 m/s. Recall that the centripetal acceleration is the acceleration of a body moving in a circular path.
Given that the centripetal acceleration is obtained using the formula;
ac = v^2/r
ac= centripetal acceleration
v = velocity
r = radius
v^2 = rac
v= √ rac
ac = 5.6m/s^2
r = 3 meters
Substituting values;
v = √3 × 5.6m/s^2
v = √16.8
v = 4.1 m/s
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Most gasoline engines in today's automobiles are belt driven. This means that the crankshaft, a rod which rotates and drives the
pistons, is timed to the camshaft, the mechanism which actuates the valves, by means of a belt. Starting from rest, assume it
takes t = 0.0320 s for a crankshaft with a radius of r = 3.75 cm to reach 1250 rpm. If the belt does not stretch or slip, calculate
the angular acceleration ay of the larger camshaft, which has a radius of r2 = 7.50 cm, during this time period.
The angular acceleration of the larger camshaft is 995.72 rad/s².
The given parameters;
initial angular velocity, [tex]\omega _i[/tex] = 0time of motion, t = 0.032 sradius of the crankshaft, r = 3.75 cm final angular speed, [tex]\omega _f[/tex] = 1250 rpmThe angular acceleration of the 3.75 cm camshaft is calculated as follows;
[tex]\omega _f = \omega _i + \alpha t\\\\\omega _f =0 + \alpha t\\\\\omega _f = \alpha t\\\\(1250 \ \frac{rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1\min}{60 \ s} ) = 0.032 \alpha \\\\130.92 = 0.032\alpha \\\\\alpha = \frac{130.92}{0.032} = 4091.25 \ rad/s^2[/tex]
The angular momentum of the camshaft is calculated as follows;
[tex]I_1 \alpha _1 = I_2 \alpha_2 \\\\\frac{1}{2} mr_1^2 \alpha _1 = \frac{1}{2}m R^2 \alpha_2\\\\r_1^2 \alpha _1 = R^2 \alpha_2\\\\\alpha_2 = \frac{r_1^2 \alpha _1 }{R^2} \\\\\alpha_2 =\frac{(0.037)^2 \times (4091.25)}{(0.075)^2} \\\\\alpha _2 = 995.72 \ rad/s^2[/tex]
Thus, the angular acceleration of the larger camshaft is 995.72 rad/s².
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--[50 POINTS]--
1)A block of mass 25 kg is placed on flat ground. The coefficient of static friction and kinetic friction are 0.73 and 0.16
a.If a person pushes the block and the block is moving, what will be the acceleration of the block?
2) A block has a mass of 79 kg. The coefficient of static and kinetic friction between the sled and the ground is 0.87 and 0.37. Person A tries to pull the block with 210N, but fails.
a) Person B successfully pulls the sled with 909N. What is the acceleration of the sled?
Newton's second law allows us to find the results for the acceleration of the blocks are:
1) The acceleration is a = 559 m / s²
2) The acceleration is a = 7.88 m / s²
Newton's second law states that the net force is equal to the product of the mass and the acceleration of the body.
∑ F = m a
Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body.
The reference system is a coordinate system with respect to which the decomposition of the forces is carried out, in the attached we have a free body diagram of the system.
1) They indicate that the body mass is 25 kg.
y-axis
N - W = 0
N = W = m g
x-axis
F -fr = ma
The friction force is a macroscopic force that results from the sum of all the microscopic interactions between the two surfaces, it has the formula
fr = μ N
Where fr is the friction force, N the normal and very the friction coefficient.
This friction coefficient has two values:
Static. For when with there is not relative motion between the two surfaces. Dynamic. When there is relative motion between the two surfaces.
We substitute.
F - μ m g = m a
a) The system moves which is the acceleration.
Suppose that the force that star to move the system keeps constant, just before the system begins to move the coefficient of friction is static, let's find the applied force.
F = μ m g
F = 0.73 25 9.8
F = 178.85 N
The block begins to move and the friction coefficient decreases to the dynamic value, we look for the acceleration.
a = [tex]\frac{F - \mu \ m g}{m}[/tex]
a = [tex]\frac{178.85 - 0.16 \ 25 \ 9.8 }{25}[/tex]
a = 5.59 m / s²
2) In this case the mass of the block is 79 kg and the applied force is
F = 909 N
We look for acceleration.
a = [tex]\frac{909 - 0.37 \ 79 \ 9.8 }{79}[/tex]
a = 7.88 m / S²
In conclusion using Newton's second law we can find the results for the acceleration of the blocks are:
1) The acceleration is a = 559 m / s²
2) The acceleration is a = 7.88 m / s²
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A block slides down an incline plane that makes a 30 degree angle with the
horizontal. If the coefficient of kinetic friction is 0.3. Calculate the acceleration of the block.
Hi there!
We know that:
Force due to gravity = Mgsinθ
Force due to friction = μMgcosθ
Let the positive direction be directed in the direction of the block's acceleration, which is downward.
Thus:
ΣF = Mgsinθ - μMgcosθ
Solving for acceleration requires diving all terms by the mass, so:
a = gsinθ - μgcosθ
Substitute in given values. (g = 9.8 m/s²)
a = 9.8sin(30) - 0.3(9.8)cos(30) = 2.354 m/s²
In hockey activities, a warm hockey puck and a frozen hockey puck has a different coefficient of restitution: 0.5 for a warm hockey puck, and 0.35 for a frozen one. NHL requires the frozen pucks to be used in games. To make sure the puck can be used in the game, the referee drops the puck on its side from a height of 2.5 m. How high should the puck bounce if it is a frozen puck
If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.
Given the data in the question;
Since the hockey puck was initially in the referee's hands
Initial velocity; [tex]u = 0m/s[/tex]Distance or height from which it was dropped; [tex]h = 2.5m[/tex]Acceleration due to gravity; [tex]g = 9.8 m/s^2[/tex]Coefficient of restitution a frozen puck; [tex]0.35[/tex]First we will find the velocity of the Puck when it hits the ground
From the Third Equation of Motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.
Since the pluck is under gravity, we will have:
[tex]v^2 = u^2 + 2gh[/tex]
We substitute in our value and find "v"
[tex]v^2 = 0 + (2 \ *\ 9.8m/s^2\ *\ 2.5m )\\\\v^2 = 47.04m^2/s^2\\\\v= \sqrt{47.04m^2/s^2}\\\\v = 6.85857m/s[/tex]
Now, Velocity of the hock puck after it hits the ground and bounce back;
We know that; Coefficient of restitution [tex]= \frac{Relative\ velocity\ after\ collision}{Relative\ velocity\ before\ collision}[/tex]
Hence, Relative Velocity after collision = Coefficient of restitution × Relative Velocity before collision
we substitute in our values;
Relative Velocity after collision [tex]= 0.35 \ *\ 6.85857m/s[/tex]
Relative Velocity after collision [tex]= 2.4 m/s[/tex]
Now, to determine how high should the puck bounced back
We use the Third Equation of Motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.
Since the pluck is under gravity, we will have:
[tex]v^2 = u^2 + 2gh[/tex]
Now, since the hockey puck bounces back, it is experiencing a negative acceleration
Hence, the equation becomes
[tex]v^2 = u^2 - 2gh[/tex]
We substitute our values into the equation and find "h"
[tex](0m/s)^2 = (2.4m/s)^2 - ( 2*9.8m/s^2*h)\\\\0 = 5.76m^2/s^2 - (19.6m/s^2*h)\\\\(19.6m/s*h) = 5.76m^2/s^2 \\\\h= \frac{ 5.76m^2/s^2 }{19.6m/s^2}\\\\h = 0.3m[/tex]
Therefore, If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.
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Once you start pulling your object with less force than friction, what should you expect your object to do? What about when your object is pulled with more force than friction?
NO LINK S
#Case -1
If Pulling force is less than frictional force the object won't move .
#Case-2
If Pulling force is greater than frictional force then object will be .
In order to calculate friction force you need Limiting friction first .
[tex]\\ \sf\longmapsto F_L=\mu sN[/tex]
u s is coefficient of static friction and N is normal reaction
Or
[tex]\\ \sf\longmapsto F_L=\mu smg[/tex]
As N=mgWhich of the following is the current best hypothesis for the formation of the solar system?
A. Formed by an exploding super nova star which then collapsed and coalesced into a spinning
disk forming Sun and planets
B. Our solar system has always been here and has never changed
C. Formed from the Sun’s explosion releasing particles into space forming planets and other
objects
D. Our solar system was formed by a great collision of other stars with one another
Answer:
A
Explanation:
all galaxies exploded in order to create the sun/stars
*Sorry for the bad quality picture!*
A frictionless pendulum with a mass of 0.4 kg and a length of 2.1 m starts at point A, at an angle 0 of 60°. As it swings downward, it passes through point B, which is 30 degrees from equilibrium. What is the kinetic energy of the pendulum at point B?
A) 3.9 J
B) 3.0 J
C) 1.1 J
D) 4.1 J
The conservation of mechanical energy allows finding the result for the speed of the pendulum when it is at 30º is:
The speed is: 3.88 m / s
The conservation of mechanical energy is a theorem of greater importance in physics and ordinary life, it states that if there is no friction force the total mechanistic energy remains constant at all points.
Mechanical energy is the sum of kinetic energy plus all potential energies. In the attachment we see a diagram of the pendulum's movement at the two points of interest.
They indicate that the pendulum is released from an initial angle of θ₁ = 60º, let's find the mechanical energy at that point.
Em₀ = U = m g h
Where the height is measured from the lowest point of the movement.
h = L - L cos tea1 = L (1 cos tea1)
The second point of interest occurs for θ₂ = 30º.
At this point part of the energy is indica and part gravitational potential.
[tex]Em_f[/tex] = K + U₂
[tex]Em_f[/tex] = ½ m v² + m g h ’
There is no friction in the system, therefore mechanical energy is conserved.
Em₀ = Em₀_f
mg L (1 - cos θ₁) = ½ m v² + m g L (1 - cos θ₂)
v² = 2g L (cos θ₂ - cos θ₁)
Let's calculate.
v² = 2 9.8 2.1 (cos 30 - cos 60)
v² = 41.16 0.366
v = 3.88 m / s
In conclusion using the conservation of mechanical energy we can find the result for the speed of the pendulum when it is at 30º is:
The speed is: 3.88 m / s
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Answer:
3.0 J
Explanation:
Just took the test
Un péndulo de 4 m de longitud tiene una frecuencia de 5Hz. Calcular la longitud de otro péndulo que en el mismo lugar tiene una frecuencia de 4Hz
Conociendo la longitud y frecuencia de un péndulo, queremos encontrar la longitud de otro pendulo de tal forma que tenga otra frecuencia.
Veremos que la longitud del nuevo péndulo debe ser 6.25m
Sabemos que un péndulo de 4m de longitud tiene una frecuencia de 5Hz.
La frecuencia de un péndulo está dada por:
[tex]f = \frac{1}{2*\pi} *\frac{g}{l}[/tex]
Donde g es la aceleración gravitatoria y l es la longitud del péndulo, remplazando los datos que tenemos en esa ecuación obtenemos:
[tex]5 Hz = \frac{1}{2*3.14} *\sqrt{\frac{g}{4m} } \\\\(5Hz*2*3.14)^2*4m = g = 3,943.8 m/s^2[/tex]
Ahora debemos encontra la longitud de tal forma que la frecuencia sea 4Hz, entonces debemos resolver:
[tex]4Hz = \frac{1}{2*3.14} *\sqrt{ \frac{3943.8m/s^2}{l} }\\\\4hz*2*3.14 = \sqrt{ \frac{3943.8m/s^2}{l} }\\\\l =\frac{3943.8m/s^2}{ (4hz*2*3.14)^2} = 6.25m[/tex]
La longitud del nuevo péndulo deve ser 6.25m
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A football is kicked with a velocity of 5 m/s at an angle of 53° above the horizontal. What is its speed at the
maximum height?
A) 3 m/s
B) 6 m/s
C) 9 m/s
D) 12 m/s
E) 15 m/s
How can stretching affect the range of motion of the neck? Hypothesis
Answer:
reduce passive stiffness and increase range of movement during exercise.
Explanation:
stretching performed as part of a warm up prior to exercise is thought to reduce passive stiffness and increase range of movement during exercise. in general it appears that is static stretching is most beneficial for athletes requiring flexibility for their sports.
Select the correct answer.
The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's motion?
OA
It's not moving
ОВ.
It's moving at a constant speed.
OC. It's moving at a constant velocity.
OD. It's speeding up.
Reset
Next
Answer:
A.It’s not moving.
Explanation:
Position-Time graphs display the motion of a object by showing the changes of velocity with respect to time.
The motion of a car on a position-time graph that is represented with a horizontal line indicates that the car has stopped moving.
A straight line with a positive slope indicates that the car is moving at a constant velocity, and thus the slope is constant. On the other hand, a curve with a changing slope, shows that the velocity is changing.
Which of the following formulas describes the change in momentum of an
object?
A. change in momentum = force x time over which force is applied
B. change in momentum = acceleration distance over which
acceleration is applied
C. change in momentum = force x distance over which force is
applied
O D. change in momentum = acceleration time over which
acceleration is applied
A player kicks a soccer ball.
What is the reaction force?
A. The ball pushes back on the player's
foot.
B. The ground pushes back on the player.
C. The ground pushes back on the ball.
Answer:
the ground pushes back on the player
Answer:
c
Explanation:
force is every where so anything that goes up must come down
Which of the following is the least important factor of a personal fitness program? A. the individual's personal conditions B. the availability of resources C. the level of motivation D. the time of day physical activity will be performed Please select the best answer from the choices provided. A B C D Mark this and return
Answer:
I think it's B
Explanation:
I think its trying to tell you that no matter who you are you could still do regular fitness but I don't know♀️
Help !!!
I solved Anthor one but didn’t understand this one ??
Answer:
Explanation:
It's a velocity•time chart. As distance = vt, the area under the curve between limits is the distance traveled.
Pic is fuzzy so I will ASSUME the horizontal axis is seconds and vertical is meters per second.
0 s ≤ t ≤ 5 s = ½(5 - 0)(30 - 0) = 75 m
5 s ≤ t ≤ 10 s = (10 - 5)(30 - 0) = 150 m
10 s ≤ t ≤ 15 s = ½(30 + 20)(15 - 10) = 75 m
0 s ≤ t ≤ 25 s = 75 + 150 + 75 + 20(20 - 15) + ½(20)(25 - 20) = 450 m
Ex 2) A cannon ball is shot straight up into the air with an initial velocity of 25 m/s[Up).
What is the maximum height of the cannonball?
Explanation:
S=(V^2-U^2)/2a a=g (gravity) a=10
=(0^2-25^2/2*(-10)
=625/20
=31.25m
a person on a bike (m=90 kg) is traveling 4 m/s at the top of a 2 m hill. what is the bikers total energy?
A) 1,944 J
B) 2,484 J
C) 1,764 J
D) 720 J
Answer:
720 J
Explanation:
Kinetic Energy = 1/2 * m *(v*v)
M= 90 kg
V=4 m/s
KE=1/2 * 90 * (4*4)
KE= 45*16= 720 J
[tex]\frac{m}{s}[/tex]Answer: D) 720 J
Explanation:
Given : m=90kg
Total Energy = K. E. + P. E. =[tex]\frac{1}{2}[/tex][tex]mv^{2}[/tex]+mgh
Since bike is present over a surface of Earth (top of hill) hence h=0m.
Therefore,
Total Energy= [tex]\frac{1}{2}[/tex]*90kg*[tex](4m/s)^{2}[/tex]+(90kg)*(9.81[tex]\frac{m}{s}^2[/tex])x(0)
= 720 J
Hence total energy of bike is equal to 720 J
Give two examples of contact and non-contact forces and explain why they are contact and non-contact forces respectively.
The Two examples of contact forces are:
frictional force Contact force.The two examples of non contact forces are:
Gravitational forcemagnetic force.Contact forces happens due to the contact between two objects
Non Contact forces happens because there is no contact between two objects. There is no attraction.
The displacement of an object in SHM is described by the equation
[tex] x = cos\binom{2\pi}{3}t[/tex]
where x is in meters and t in seconds. Determine the velocity of the object at t = 0.6 s.
Answer:
[tex]-1.99\:\mathrm{m/s}[/tex]
Explanation:
Assuming that the equation is intended to be [tex]\displaystyle x=\cos\left(\frac{2\pi}{3}t\right)[/tex], we can find the velocity vs. time equation by taking the first derivative with respect to [tex]t[/tex]:
[tex]\displaystyle \frac{dx}{dt}=\frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)[/tex]
Recall the chain rule:
[tex]\displaystyle f(g(x))'=f'(g(x))\cdot g'(x)[/tex]
Therefore,
[tex]\displaystyle \frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex]
Therefore, the velocity vs. time equation of the object is [tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex].
Substitute [tex]t=0.6\text{ s}[/tex] into this equation to find the velocity at that given time:
[tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}(0.6)\right)\cdot \frac{2\pi}{3}\approx \boxed{-1.99\text{ m/s}}[/tex]
The model represents a fluorine (F) atom. What is the mass of the atom?
Answer:
19
Explanation:
The mass of an atom is found in the nucleus: number of protons + number of neutrons; 9 + 10 = 19
The mass number of fluorine is 19
The volume of an ideal gas is increased from 0.6 m3 to 2.4 m3 while maintaining a constant pressure of 1000 Pa (1000 N/m2). Determine, in J, the amount of work done by the gas in this expansion.
The amount of work done by the gas in the given expansion is 1800 J.
The given parameters;
initial volume of the ideal gas, V₁ = 0.6 m³final volume of the ideal gas, V₂ = 2.4 m³constant pressure of the gas, P = 1000 PaThe amount of work done by the gas in the given expansion is calculated as follows;
W = PΔV
where;
ΔV is the change in volume of the gasSubstitute the given parameters and solve for the work done;
W = 1000(2.4 - 0.6
W = 1800 J
Thus, the amount of work done by the gas in the given expansion is 1800 J.
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A car accelerates at 4 m/s/s from rest. What is the car's velocity after it travels 20 m?
The first motor abilities a new born exhibits are
Answer:
here your answer
i am sorry if wrong
What is the intensity level for a sound of Intensity 10-⁶w/m²
There are 6 foundation of sports and which one you think is the most important?
I just need the points
Explanation:
But just pick any and say something like "it stans out to me most/more" or "it sounds/looks more intristing to me"
A 10 kg box initially at a speed of 10 m/s accelerates uniformly to a speed of 20 m/s in 2
seconds. Determine the
A. energy gained by the block
Answer:
What Um Sorry Where's The Answer?....
If you traveled 50m/s for 60 seconds, how far did you travel? Remember speed=distance/time
Question options:
300 m/s
500 m/s
3,000 m/s
300 km/h
for some reason my question got removed-_-
A man with a mass of 60 kg rides a bike with a mass of 13 kg. What is the force needed by the man to accelerate the bike at 0.90 m/s2?
An object starts from rest and uniformly accelerates at a rate of 1.25 m/s2 for 7.0 seconds.
(a) What is the object's displacement during this 7-second time period?
(b) What is the object's final velocity?
(c) How many seconds does it take the object to have a displacement of 22 meters?
Explanation:
Since its accelerating, the velocity vs time graph is linear
For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula
A= (vf - vi)/t
a= vf-0/t
1.25=vf / 7
1.25*7=vf
8.75 = vf
Now for displacement plug all the values in
X = 1/2(vf-vi)/t formula
The displacement (x) is 30.625 m
For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above
The answer is t = 5.02
Im pretty sure all the answers are correct
Fossils show that some animals _____.
are extinct
had not seen rain
liked the cold
made noise
Answer:I think Fossils show that some animals are extinct
Explanation:
Please mark as brainliest for me.Thanks