Answer:
Increasing the tension on a string increases the speed of a wave, which increases the frequency (for a given length). Pressing the finger at different places changes the length of string, which changes the wavelength of standing wave, affecting the frequency.
Explanation:
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.
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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
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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.
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
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
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].
lucia raced her car on a raceway
Answer:
Good question to ask in physics, sir maam
Answer:
That's good
Explanation:
Which of the following is an example of a force that acts at a distance? *
a bear walking through the woods
a jet ski skimming across the water
a leaf falling off of a tree to the ground
a tennis ball hit by a tennis player's racket
Which two components must a vector quantity
have?
a. Magnitude and velocity
b. Acceleration and direction
c. Force and speed
d. Direction and magnitude
Answer:
d. Direction and magnitude
Explanation:
The two components of a vector are its magnitude and direction.
Magnitude is the quantity of the substance
Direction is the path.
Other quantities are called scalar quantities. Scalar quantities have only magnitude but no direction.Examples of vector quantities are velocity, displacement, acceleration.
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.
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.
-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].
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
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? :)
A 0.0138-m3 container is initially evacuated. Then, 4.73 g of water is placed in the container, and, after some time, all of the water evaporates. If the temperature of the water vapor is 358 K, what is its pressure?
Answer:
56.7kpa
Explanation:
V = 0.0138-m³
M = 4.73g
T = 358k
R = ideal gas constant = 8.314
Number of moles of water n = 4.73/18 = 0.2628 (Remember that the molecular mass of water is 18).
Pv = nRT
P = nRT/V
P = (0.2628x8.314x358)/0.0138
P = 56681.23
P = 56.7Kpa
56.7Kpa is therefore the pressure of the water vapor.
What is a concern about recovered memories?
A. Some recovered memories reveal child abuse.
B. Some recovered memories are false.
C. Recovered memories help people grow emotionally stronger.
D .choices A and B
Answer:
D because those are both concerning.
An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and 0.1250 T respectively. The particle passes out the electric field but the magnetic field continues, and the particle makes a semicircle of diameter 25.05 cm. Part A. What is the particles charge to mass ratio?Part B. Can you identify the particle?
a. can't identify
b. proton
c. electron
d. neutron
Answer:
(A) q/m = 9.58 x 10⁷ C/kg
(B) b. Proton
Explanation:
(A)
In order to pass un-deflected between magnetic and electric field both electric and magnetic forces must be equal:
Felectric = Fmagnetic
Eq = Bqv
v = E/B ------------ equation (1)
Now, when the particle forms semi-circle under influence of magnetic field. At this, point magnetic field is equal to centripetal force:
Centripetal Force = Magnetic Force
mv²/r = qvB
mv = qrB
v = qrB/m ------------ equation (2)
comparing equation (1) and equation (2), we get:
E/B = qrB/m
q/m = E/B²r
where,
q/m = charge to mass ratio = ?
E = Electric Field = 187500 V/m
B = Magnetic Field = 0.125 T
r = radius of circular path = 25.05 cm/2 = 12.525 cm = 0.12525 m
Therefore,
q/m = (187500 V/m)/(0.125 T)²(0.12525 m)
q/m = 9.58 x 10⁷ C/kg
(B)
This is the charge to mass ration of a proton:
q/m = 1.6 x 10⁻¹⁹ C/1.67 x 10⁻²⁷ kg
q/m = 9.58 x 10⁷ C/kg
Therefore, correct option is:
b. proton
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 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.
Suppose a boat is moving 6m/s relative to the pier, but you are walking on the boat -1 m/s relative to the boat' What would be your relative velocity to the pier? What direction would the boat be moving? Is this a one or two-dimensional problem?
Answer:
V₁ = 5 m/s
This is the problem of one dimension
Explanation:
Before using the concepts of relative velocity, we must first define the variables for each velocity here:
V₁ = Person's Velocity Relative to Pier = ?
V₂ = Boat's Velocity Relative to Pier = 6 m/s
V₁₂ = Person's Velocity Relative to Boat = - 1 m/s
Now, from the relative velocity, we know that:
V₁₂ = V₁ - V₂
- 1 m/s = V₁ - 6 m/s
V₁ = 6 m/s - 1 m/s
V₁ = 5 m/s
This is the problem of one dimension
because, the velocity is along single dimension with opposite directions. No, second dimension is involved.
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
Another name for a pivot point is the:
a.
output
b.
torque
c.
fulcrum
Answer:
I think it's c .... fulcrum
Begin any simulation, and turn on Gravity Force in the central menu on the right. The gravity force arrow shows the direction and strength of the gravitational force each body feels. How do the gravity force arrows change throughout the orbit?
Answer:
The gravity arrow for each body rotates, always pointing toward the other body. Both arrows grow longer when the bodies come closer to one another and shorter when they move farther apart. This change shows that the gravitational force is stronger the closer together the bodies are.
Explanation:
A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies. Thus, a gravitational field—measured in newtons per kilogram (N/kg)—is employed to describe gravitational processes.
What strength of the gravitational force each body feels?Each body's gravity arrow spins while constantly pointing in the direction of the other body. When the bodies move apart, the arrows go shorter and get longer as they get closer together. This modification demonstrates that the gravitational force increases with increasing body proximity.
Therefore, All objects with mass are attracted to one another by the gravitational attraction. Which has a magnitude that is directly proportional to the masses of the two objects and inversely proportional to the square of their distance from one another.
Learn more about gravitational force here:
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. State any three applications of conduction
Which major regions had the Romans controlled?
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
Which TWO statements describe the ocean floor giving brainliest please help
Answer:
A and B.
Explanation:
Answer C is wrong - there are no tall trees underwater
Answer D is wrong, there are mountain chains underwater due to plate tectonics.
What is a safe following distance between your automobile and the vehicle in front of you?
Answer:
Many drivers follow the “three-second rule.” In other words, you should keep three seconds worth of space between your car and the car in front of you in order to maintain a safe following distance.
Thank you and please rate me as brainliest as it will help me to level up
hello, I need help with a question
what force noncontact force:
a. buoyancy
b. friction
c. air reistanceses
d. Gravity
(please if youre not sure whats the answer dont answer it
Answer:
The Answer is (A) . . .
I Think . . .
Answer:
a. buoyancy
....................
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?
