Answer:
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g
U is the speed = 300m/s
H is the maximum height = 78.4m
g is the acceleration due to gravity = 9.8m/s²
Substitute into the fromula;
R = 300√2(78.4)/9.8
R = 300 √(16)
R = 300*4
R = 1200m
Hence the projectile travelled 1200m before hitting the ground
In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, the note at 110 Hz has two strings at this frequency. If one string slips from its normal tension of 600 N to 540 N, what beat frequency is heard when the hammer strikes the two strings simultaneously?
Answer:
f = 5.65 Hz
Explanation:
The fundamental frequency of a string is given by following formula:
f = v/2L
where,
f = fundamental frequency
v = speed of wave = √(TL/m)
L = Length of String
m = Mass of String
T = Tension in String
Therefore,
f = √(TL/m)/2L
2f = √(T/Lm)
For initial condition:
T₁ = 600 N
f₁ = 110 Hz
2(110 Hz) = √(600 N/Lm)
√(600 N)/220 Hz = √Lm
Lm = 0.01239 N/s²
Now, for changed tension:
2f₂ = √(540 N/0.01239 N/s²)
f₂ = 208.7 Hz/2
f₂ = 104.35 Hz
So, the beat frequency will be:
f = f₁ - f₂
f = 110 Hz - 104.35 Hz
f = 5.65 Hz
lucia raced her car on a raceway
Answer:
Good question to ask in physics, sir maam
Answer:
That's good
Explanation:
A ball is attached to one end of a string such that the ball travels in a vertical circular path near Earth's surface. The force diagram
of the ball at its lowest point in the circular path is shown above. What is the net centripetal force exerted on the ball?
A) 10N
B) 15 N
C
25 N
D
35 N
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is [tex]F_{centripetal} = 15 \ N[/tex]
Explanation:
From the diagram we see that
The tension force on the ball is [tex]F_{Tension} = 25 \ N[/tex]
The gravitational force on the ball is [tex]F_{Gravity } = 10 \ N[/tex]
Generally the net centripetal force exerted on the ball is mathematically represented as
[tex]F_{centripetal} = F_{Tension} - F_{Gravity}[/tex]
=> [tex]F_{centripetal} = 25 - 10[/tex]
=> [tex]F_{centripetal} = 15 \ N[/tex]
The net centripetal force exerted on the ball is 15 Newton.
Hence, Option B) 15N is the correct answer.
Given the data in the question;
Tension force on the ball; [tex]F_{Tension} = 25N[/tex]Gravitational force on the ball; [tex]F_{Gravity} = 10N[/tex]Net centripetal force exerted on the ball; [tex]F_{centripetal} = \ ?[/tex]
From the diagram below, net force acting on the ball gives rise to centripetal force. Hence
[tex]F{net} = F_1 + f_2 = F_{centripetal}[/tex]
Now, from the diagram, force acting towards the center of the circular track is is positive while force directly pointing away from center is negative.
Hence
[tex]F{net} = F_1 + f_2 = F_{centripetal} = F_{Tension} - F_{Gravity}[/tex]
We substitute in our given values
[tex]F_{centripetal} = 25N - 10N\\\\ F_{centripetal} = 15N[/tex]
The net centripetal force exerted on the ball is 15 Newton.
Hence, Option B) 15N is the correct answer.
Learn more: https://brainly.com/question/8050835
The upper arm muscle is _______________ to the skin.
Answer:
The pectoralis major, latissimus dorsi, deltoid, and rotator cuff muscles connect to the humerus and move the arm. The muscles that move the forearm are located along the humerus, which include the triceps brachii, biceps brachii, brachialis, and brachioradialis.
Which of the following is a characteristic of active galaxies? A. They generally exhibit no signs of explosive activity. B. Their energy emission cannot be explained as the accumulated emission of their stars. C. They are generally less luminous than normal galaxies. D. Their energy output is steady in time.
Answer:
The correct option is B
Explanation:
Active galaxies are characterized by a small core of emission at the center which is usually brighter than the rest of the galaxy. The energy emitted from active galaxy is way more than that emitted from normal galaxy. As compared to normal galaxy where it's energy emission is thought to be the sum of the emission from each star found in the galaxy, this is not true or the same in active galaxy. There (active galaxy) huge energy emission cannot be explained as the sum of the emission of their stars.
A spring-loaded gun has a spring for which
k =180N /m
and is initially compressed by
14cm
. It fires
a
0.024kg
projectile vertically. Find the maximum height above the initial position?
Answer:
7.5 m
Explanation:
k = Spring constant = 180 N/m
x = Displacement of spring = 14 cm
m = Mass of projectile = 0.024 kg
a = g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
s = Displacement of projectile
v = Final velocity = 0
u = Initial velocity
The potential energy of the spring will be equal to the kinetic energy of the object
[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mu^2\\\Rightarrow u=\sqrt{\dfrac{kx^2}{m}}\\\Rightarrow u=\sqrt{\dfrac{180\times 0.14^2}{0.024}}\\\Rightarrow u=12.12\ \text{m/s}[/tex]
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-12.12^2}{2\times -9.81}\\\Rightarrow s=7.5\ \text{m}[/tex]
The maximum height reached above the initial position is 7.5 m
Can someone let me know if I got these questions right please
A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the opposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.
Answer:
42.67NExplanation:
Step one:
Given
mass m= 0.32kg
intital velocity, u= 14m/s
final velocity v= 22m/s
time= 0.06s
Step two:
Required
Force F
the expression for the force is
F=mΔv/t
F=0.32*(22-14)/0.06
F=(0.32*8)/0.06
F=2.56/0.06
F=42.67N
The average force exerted on the bat 42.67N
. State any three applications of conduction
-Two pickup trucks each have a mass of 2,000 kg. The gravitational force between the
trucks is 3.00 x 10-5 N. One pickup truck is then loaded with 1,000 kg of bricks. Which
of the following would be the new gravitational force between the two trucks?
Answer:
[tex]F'=4.5\times 10^{-5}\ N[/tex]
Explanation:
The gravitational force between two pickups is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
It means,
[tex]F\propto m_1m_2[/tex]
So,
[tex]\dfrac{F}{F'}=\dfrac{m_1m_2}{m_1'm_2'}[/tex]
We have,
[tex]m_1=m_2=2000\ kg\\\\F=3\times 10^{-5}\ N\\\\m_1'=2000+1000 =3000\ kg\\\\m_2'\ \text{remains the same i.e. 1000 kg}[/tex]
F' is the new force
So, putting all the values,
[tex]\dfrac{3\times 10^{-5}}{F'}=\dfrac{2000\times 2000}{3000\times 2000}\\\\\dfrac{3\times 10^{-5}}{F'}=\dfrac{2}{3}\\\\F'=\dfrac{9\times 10^{-5}}{2}\\\\F'=4.5\times 10^{-5}\ N[/tex]
So, the new force between the two trucks is [tex]4.5\times 10^{-5}\ N[/tex].
Explain how you would determine how much error there is between a vector addition and the displacement of the actual trip taken. Why aren’t vectors true straight - line paths in the real world? Would you expect the vector addition to be larger or smaller than the actual trip taken? Why?
Answer:
find the difference"vector addition" is likely to be larger than displacementExplanation:
No doubt your curriculum materials have an adequate discussion of error and how it is determined.
a) The error between two values is the difference between an approximation and the true value. You determine it by subtracting the true value from the approximation.
In the case of a trip, you need to decide what is the "true value" and what is the "approximation." The wording here suggests that the "displacement of the actual trip taken" is to be considered the "true value." Then, you determine the error by subtracting that displacement from the "vector addition."
__
b) In Euclidean geometry, a vector is a straight-line path. In the real world, we use the term vector to refer to the direction taken by an object, generally along a curved path at some relative distance from the Earth's surface. The direction reference may change along the path, so that following a given "vector" will usually result in a curved path (not a great circle) on the Earth's surface.
__
c) The curved path from a point A to a point B on the Earth's surface will always be longer than the straight-line path between the same two points. In the case of points on opposite sides of the world, the straight-line path is through the center of the Earth, whereas the "vector addition" will be some path along the surface of the Earth.
You need to decide the meaning of "actual trip taken," as any actual trip will generally involve all the changes in direction and ups and downs along the way. If you consider "the actual trip taken" to be the difference between starting and ending coordinates, then the "vector addition" will always be longer.
An element with 5 protons, 6 neutrons, and 8 electrons has an atomic number of
(2 Points)
Answer:
I've attached a screenshot of a document from K12 that should give you the answer to your question, and many more question you might have. If this is not helpful please say so in the replies to this answer and I'll try my best to find some more information.
Explanation:
May I have brainliest please? :)
Which information can be determined using half-life?
In,radioactive decay,the half-life is the length of time after which there is a 50 percent chance that an atom will have undergone nuclear decay.It varies depending on the atom type and isotope,and is usually determined experimentally.
A stone is thrown horizontally. In
0.5s
after the stone began to move, the numerical value of its
velocity was 1.5 times its initial velocity. Find the initial velocity of the stone. Disregard the
resistance of the air.
Answer:
Approximately [tex]4.39 \; \rm m \cdot s^{-1}[/tex] (assuming that [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].)
Explanation:
Let the initial velocity of this stone be [tex]v\; \rm m \cdot s^{-1}[/tex] ([tex]v \ge 0[/tex] because speed should be non-negative.) That should be numerically equal to the initial horizontal speed of this stone. Because the stone was thrown horizontally, its vertical speed would initially be zero ([tex]0\; \rm m \cdot s^{-1}[/tex].)
The weight of this stone will accelerate the stone downwards. Because the weight of the stone is in the vertical direction, it will have no effect on the horizontal speed of the stone.
Therefore, if air resistance is indeed negligible, the horizontal speed of this stone will stay the same. The horizontal speed of the stone after [tex]0.5\; \rm s[/tex] should still be [tex]v\; \rm m \cdot s^{-1}[/tex]- same as its initial value.
On the other hand, if [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex], the (downward) vertical speed of this stone will increase by [tex]9.81\; \rm m \cdot s^{-2}[/tex] every second. After [tex]0.5\; \rm s[/tex], the vertical speed of this stone would have become:
[tex]0.5\; \rm s \times 9.81\; \rm m \cdot s^{-2} = 4.905\; \rm m \cdot s^{-1}[/tex].
[tex]t = 0\; \rm s[/tex]:
Horizontal speed: [tex]v\; \rm m \cdot s^{-1}[/tex].Vertical speed: [tex]0\; \rm m \cdot s^{-1}[/tex].[tex]t = 0.5\; \rm s[/tex]:
Horizontal speed: [tex]v\; \rm m \cdot s^{-1}[/tex].Vertical speed: [tex]4.905\; \rm m \cdot s^{-1}[/tex].Refer to the diagram attached. Consider the vertical speed and the horizontal speed of this rock as lengths of the two legs of a right triangle. The numerical value of the velocity of this stone would correspond to the length of the hypotenuse of that right triangle. Apply the Pythagorean Theorem to find that numerical value:
[tex]\begin{aligned} &\text{numerical value of velocity} \\ &= \sqrt{{\left(\text{horizontal speed}\right)}^2 + {\left(\text{vertical speed}\right)}^2}\end{aligned}[/tex].
At [tex]t = 0\; \rm s[/tex], the magnitude of the velocity of this stone would be [tex]\sqrt{v^{2} + 0^2} = v\; \rm m \cdot s^{-1}[/tex].
At [tex]t = 0.5\; \rm s[/tex], the magnitude of the velocity of this stone would be [tex]\left(\sqrt{v^2 + 4.905^2}\right)\; \rm m \cdot s^{-1}[/tex].
The question requires that the magnitude of the velocity of the stone at [tex]t = 0.5\; \rm s[/tex] should be [tex]1.5[/tex] times the value at [tex]t = 0\; \rm s[/tex]. In other words:
[tex]\sqrt{v^2 + 4.905^2} = 1.5 \, v[/tex].
Square both sides of this equation and solve for [tex]v[/tex]:
[tex]v^2 + 4.905^2 = 2.25\, v^2[/tex].
[tex]v \approx 4.39[/tex] ([tex]v \ge 0[/tex] given that speed should be non-negative.)
Therefore, the initial velocity of this stone should be approximately [tex]4.39\; \rm m \cdot s^{-1}[/tex].
Calculate the drag force acting on a 105 kg (including the bicycle) cyclist moving at 10 m/s, 20 m/s, an at 30 m/s. Use a drag coefficient of 0.8, a frontal surface area of 1.0 m2 , and a density of 1.0 kg/m3.
Answer:
40N
160N
360N
Explanation:
The drag force can be calculated using the formula below
FD = 0.5CPAV²
The drag coefficient = c = 0.8
The frontal density = p = 1.0kg
The frontal surface area = A = 1.0
The velocity v = 10m/s, 20m/s, 30m/s
When V = 10m/s
Fd = 0.5(0.8)(1)(1)(10)²
Fd = 40N
When V = 20m/s
FD = 0.5(0.8)(1)(1)(20)²
FD = 160N
When V = 30m/s
FD = 0.5(0.8)(1)(1)(30)²
FD = 360N
Thank you
Why is people to come together and combine their efforts?
Answer:
people who had a hard time to help other people because they feel that if they help
that the work of the person is much harder than his own work
Individuals who find it difficult to help others because they believe that if they do, the person's job will be much tougher than their own.
A group of people working together to achieve a shared purpose.
Collaboration, cooperation, and coordination are all words that come to mind when thinking of teaming.
Learn more:
https://brainly.com/question/10552710?referrer=searchResults
There are_
periods included in the periodic table.
Answer here
SUBMIT
s
Answer:
7
Explanation
There are currently seven complete periods in the periodic table, comprising the 118 known elements. Any new elements will be placed into an eighth period; see extended periodic table.
A 52.0-kg sandbag falls off a rooftop that is 22.0 m above the ground. The collision between the sandbag and the ground lasts for a total of 17.0 ms. What is the magnitude of the average force exerted on the sandbag by the ground during the collision?
Answer:
F = 7,916,955.0N
Explanation:
According to newtons second law
Force = mass * acceleration
Given
mass = 52.0kg
distance S = 22.0m
time t = 17.0 ms = 0.017s
We need to get the acceleration first using the formula;
S = ut+ 1/2at²
22 = 0 + 1/2 a(0.017²)
22 = 0.0001445a
a = 22/0.0001445
a = 152,249.13m/s²
The magnitude of the average force exerted will be;
F = ma
F = 52 * 152,249.13
F = 7,916,955.0N
Which major regions had the Romans controlled?
A car is traveling at 15m/s on a horizontal road. the brakes are applied and the car skids to a stop in 4.0s . the coefficient of Kinetic friction between the tires and road is:_________.
A) .38
B) .69
C) .76
D) .92
E) 1.11
Answer:
the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
Explanation:
Given that;
final velocity v = 0
initial velocity u = 15m/s
time taken t = 4 s
acceleration a = ?
from the equation of motion
v = u + at
we substitute
0 = 15 + a × 4
acceleration a = -15/4 = - 3.75 m/s²
the negative sign tells us that its a deacceleration so the sign can be ignored.
Deacceleration due to friction a = μ × g
we substitute
3.75 = μ × 9.8
μ = 3.75 / 9.8 = 0.3826 ≈ 0.38
Therefore the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
Another name for a pivot point is the:
a.
output
b.
torque
c.
fulcrum
Answer:
I think it's c .... fulcrum
How much work is done when 425 N of force is applied for 60.0m
Answer:
Workdone = 25500Nm
Explanation:
Given the following data
Force = 425N
Distance = 60m
To find the workdone
Workdone = force *distance
Substituting into the equation, we have
Workdone = 425*60
Workdone = 25500Nm
A merry-go-round rotates from rest with an angular acceleration of 1.45 rad/s2. How long does it take to rotate through (a) the first 3.70 rev and (b) the next 3.70 rev?
Answer:
(a) t = 5.66 s
(b) t = 8 s
Explanation:
(a)
Here we will use 2nd equation of motion for angular motion:
θ = ωi t + (1/2)∝t²
where,
θ = Angular Displacement = (3.7 rev)(2π rad/1 rev) = 23.25 rad
ωi = initial angular speed = 0 rad/s
t = time = ?
∝ = angular acceleration = 1.45 rad/s²
Therefore,
23.25 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²
t² = (23.25 rad)(2)/(1.45 rad/s²)
t = √(32.06 s²)
t = 5.66 s
(b)For next 3.7 rev
θ = ωi t + (1/2)∝t²
where,
θ = Angular Displacement = (3.7 rev + 3.7 rev)(2π rad/1 rev) = 46.5 rad
ωi = initial angular speed = 0 rad/s
t = time = ?
∝ = angular acceleration = 1.45 rad/s²
Therefore,
46.5 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²
t² = (46.5 rad)(2)/(1.45 rad/s²)
t = √(64.13 s²)
t = 8 s
The time taken in each case is 2.55 s
Let us recall that the equations of circular motion are almost like those of linear motion;
α = ω2 - ω1/t
α = angular acceleration
ω2 = final angular velocity
ω1 = initial angular velocity
t = time taken
a) 1.45 rad/s2 =3.70 rev/s - 0 rev/s/t
t = 3.70 rev/s/1.45 rad/s2
t = 2.55 s
b) For, ) the next 3.70 rev:
1.45 rad/s2 =3.70 rev/s - 0 rev/s/t
t = 3.70 rev/s/1.45 rad/s2
t = 2.55 s
Learn more about angular velocity; https://brainly.com/question/1301963
a race car with a mass of 500 kg on a bridge 45 m above a river what is the potential energy of the car?
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of mass at 40% of the arm length from the shoulder.
a. What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight out to his side, parallel to the floor?
b. What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight, but 45∘ below horizontal?
Answer:
(a) τ = 26.58 Nm
(b) τ = 18.79 Nm
Explanation:
(a)
First we find the torque due to the ball in hand:
τ₁ = F₁d₁
where,
τ₁ = Torque due to ball in hand = ?
F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N
d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m
τ₁ = (29.4 N)(0.6 m)
τ₁ = 17.64 Nm
Now, we calculate the torque due to the his arm:
τ₁ = F₁d₁
where,
τ₂ = Torque due to arm = ?
F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N
d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m
τ₂ = (37.24 N)(0.24 m)
τ₂ = 8.94 Nm
Since, both torques have same direction. Therefore, total torque will be:
τ = τ₁ + τ₂
τ = 17.64 Nm + 8.94 Nm
τ = 26.58 Nm
(b)
Now, the arm is at 45° below horizontal line.
First we find the torque due to the ball in hand:
τ₁ = F₁d₁
where,
τ₁ = Torque due to ball in hand = ?
F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N
42.42 cm = 0.4242 m
τ₁ = (29.4 N)(0.4242 m)
τ₁ = 12.47 Nm
Now, we calculate the torque due to the his arm:
τ₁ = F₁d₁
where,
τ₂ = Torque due to arm = ?
F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N
d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m
τ₂ = (37.24 N)(0.1696 m)
τ₂ = 6.32 Nm
Since, both torques have same direction. Therefore, total torque will be:
τ = τ₁ + τ₂
τ = 12.47 Nm + 6.32 Nm
τ = 18.79 Nm
A)The magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor will be 26.58 Nm.
What is torque?Torque is the force's twisting action about the axis of rotation. Torque is the term used to describe the instant of force. It is the rotational equivalent of force. Torque is a force that acts in a turn or twist.
The amount of torque is equal to force multiplied by the perpendicular distance between the point of application of force and the axis of rotation.
m is the mass of steel ball = 3.0 kg in his hand.
L is the length of the arm is 60 cm long
M is the mass of arm=3.8 kg,
The torque is given as;
The magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor is found as;
[tex]\rm \tau_1 = F \times d \\\\ \rm \tau_1 = 29.4 \times 0.63 \\\\ \rm \tau_1 =17.64 \ Nm[/tex]
The torque due to the arm;
[tex]\rm \tau_2= F \times d \\\\ \rm \tau_2 = mg \times d \\\\ \rm \tau_2 =3.8 \times 9.81 \times 0.24 \ Nm \\\\ \rm \tau_2=8.94 \ Nm[/tex]
The net torque for case 1 will be;
[tex]\tau = \tau_1 + \tau_2\\\\ \rm \tau = 17.64 Nm + 8.94 Nm\\\\ \tau = 26.58 Nm[/tex]
Hence the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor will be 26.58 Nm.
B) the magnitude of the torque about his shoulder if he holds his arm straight, but 45∘ below horizontal will be 18.79 Nm.
The torque is due to the weight of the ball;
[tex]\rm \tau_1= F \times d \\\\ \rm \tau_1 = mg \times d \\\\ \rm \tau_1 =3.8 \times 9.81 \times 0.4242 \ Nm \\\\ \rm \tau_1= 12.47 \ Nm[/tex]
The torque due to the arm will be;
[tex]\rm \tau_2= F \times d \\\\ \rm \tau_2 = mg \times d \\\\ \rm \tau_2 =3.8 \times 9.81 \times 0.1696 \ Nm \\\\ \rm \tau_2=6.32 \ Nm[/tex]
The net torque in case 2 will be
[tex]\tau = \tau_1 + \tau_2\\\\ \rm \tau = 12.47 Nm + 6.32 Nm\\\\ \tau = 18.79 Nm[/tex]
Hence the magnitude of the torque about his shoulder if he holds his arm straight, but 45° below horizontal will be 18.79 Nm
To learn more about the torque refer to the link;
brainly.com/question/6855614
A jet plane lands with a velocity of 100 m/s and can accelerate at a maximum of -9.0 m/s2 as it comes to rest. The minimum time needed before it can come to rest is seconds.
Explanation:
That must be the right ans.
1. Find the magnitude of the gravitational force (in N) between a planet with mass 8.00 ✕ 1024 kg and its moon, with mass 2.75 ✕ 1022 kg, if the average distance between their centers is 2.40 ✕ 108 m.2. What is the moon's acceleration (in m/s2) toward the planet? 3. What is the planet's acceleration (in m/s2) toward the moon?
Answer:
1.358 10e8 have good day please mark brainliest
The solid shaft with a 20 mm radius is used to transmit the torques applied to the gears. Determine the maximum torsional shear stress developed in in the shaft.
Answer:
τ = (7.96 x 10⁴ m⁻³)T
This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.
Explanation:
The maximum allowable shear stress on the solid shaft can be given by the torsional formula as follows:
τ = Tc/J
where,
τ = Maximum Allowable Shear Stress = ?
T = Maximum Torque Applied to the Shaft
c = maximum distance from center to edge = radius in this case = 20 mm = 0.02 m
J = Polar Moment of inertia = πr⁴/2 = π(0.02 m)⁴/2 = 2.51 x 10⁻⁷ m⁴
Therefore,
τ = T(0.02 m)/(2.51 x 10⁻⁷ m⁴)
τ = (7.96 x 10⁴ m⁻³)T
This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.
A car turns from a road into a parking lot and into an available parking space. The car’s initial velocity is 4 m/s [E 45° N]. The car’s velocity just before the driver decreases speed is 4 m/s [E 10° N]. The turn takes 3s. What's the average acceleration of the car during the turn? The answer should have directions with an angle.
Write the velocity vectors in component form.
• initial velocity:
v₁ = 4 m/s at 45º N of E
v₁ = (4 m/s) (cos(45º) i + sin(45º) j)
v₁ ≈ (2.83 m/s) i + (2.83 m/s) j
• final velocity:
v₂ = 4 m/s at 10º N of E
v₂ = (4 m/s) (cos(10º) i + sin(10º) j)
v₂ ≈ (3.94 m/s) i + (0.695 m/s) j
The average acceleration over this 3-second interval is then
a = (v₂ - v₁) / (3 s)
a ≈ (0.370 m/s²) + (-0.711 m/s²)
with magnitude
||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²
and direction θ such that
tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92
→ θ ≈ -62.5º
which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.
Which statement best explains why running on a track with constant speed at 3m/s is not work but climbing a mountain at 1m/s is work?
1. At constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy
2. At constant speed, change in the potential energy is zero, but climbing a mountain produces change in the kinetic energy.
3. At constant speed, change in the kinetic energy is finite, but climbing a mountain produces no change in the potential energy.
4. At constant speed, change in the potential energy is finite, but climbing a mountain produces no change in the kinetic energy.
Answer:
1. At constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy
Explanation:
According to work-energy theorem, work done in moving an object from one point to another is equal to change in mechanical energy ( kinetic energy or potential energy ) of the object.
Kinetic energy is given by;
K.E = ¹/₂m(Δv)²
where;
Δv is change in speed
at constant speed, Δv = 0
Potential energy is given by;
P.E = mgΔh
where;
Δh is change in height,
there is change in height in climbing a mountain
Therefore, the best explanation in the given options is "1".
"At constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy"
