A 1.60-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. you may want to review ( pages 204 - 205) . part a if you ignore the weight of the bar itself, how far from the left end of the barbell is the center of gravity

The elevation difference is 55.3 mm when a pressure of 98.0 Pa (gage) is applied to the right tube of the manometer, displacing water downward and kerosene upward.

When a pressure of 98.0 Pa (gage) is applied to the right tube of the manometer, the water in the right tube is displaced downward by the same distance that the kerosene in the left tube is displaced upward, maintaining equilibrium.

To find the elevation difference, we can use the hydrostatic pressure formula:

P = ρgh

Where:

P = pressure difference (98.0 Pa in this case)

ρ = density of the fluid (for water, ρ_water ≈ 1000 kg/m³, for kerosene, ρ_kerosene ≈ 820 kg/m³)

g = acceleration due to gravity (approximately 9.81 m/s²)

h = elevation difference

Since the water and kerosene columns have opposite elevation changes, we can write:

98.0 Pa = (ρ_water * g * h) - (ρ_kerosene * g * h)

Now, solve for h:

98.0 Pa = (1000 kg/m³ * 9.81 m/s² * h) - (820 kg/m³ * 9.81 m/s² * h)

98.0 Pa = (9810 h) - (8034 h)

98.0 Pa = 1776 h

h = 98.0 Pa / 1776 = 0.0553 meters = 55.3 mm

So, when a pressure of 98.0 Pa (gage) is applied to the right tube, the elevation difference between the two free surfaces in the manometer is 55.3 mm.

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First let us calculate for the angle of inclination using the sin function,

sin θ = 1 m / 4 m

θ = 14.48°

Then we calculate the work done by the movers using the formula:

W = Fnet * d

So we must calculate for the value of Fnet first. Fnet is force due to weight minus the frictional force.

Fnet = m g sinθ – μ m g cosθ

Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48

Fnet = 84.526 N

So the work exerted is equal to:

W = 84.526 N * 4 m

W = 338.10 J

Answer:

Occulation

Explanation:

The percent uncertainty of a 5 lb bag with an uncertainty of ±0.4 lb is calculated using the formula (SA / A) x 100%, resulting in a percent uncertainty of 8%.

To calculate the percent uncertainty for a 5 lb bag with an uncertainty of ±0.4 lb, you would use the following formula:

% uncertainty = (SA / A) × 100%

Where A is the expected value (5 lb in this case) and SA is the uncertainty in the value (0.4 lb). Plugging in the numbers:

% uncertainty = (0.4 lb / 5 lb) × 100% = 8%

This means the percent uncertainty of the bag's weight is 8%.

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Answer:

Electric charge, q = 323910 C

Explanation:

It is given that,

Net electric flux, [tex]\phi=3.66\times 10^{16}\ Nm^2/C[/tex]

We have to find the total electric charge on the planet. We can find it using Gauss's law. It is as follows :

[tex]\phi=\dfrac{q}{\epsilon_0}[/tex]

where

q is the net electric charge

[tex]\epsilon_0[/tex] is the electric permeability

So, net electric charge is given by :

[tex]q=\phi\times \epsilon_0[/tex]

[tex]q=3.66\times 10^{16}\ Nm^2/C\times 8.85\times 10^{-12}\ F/m[/tex]

[tex]q=323910\ C[/tex]

Hence, this is the required solution.

In physics, the definition of work is simply the product of force exerted and the displacement. Therefore in this case, work done by the ant is:

Work = 0.00805 N * 9.80 cm * (1 m / 100 cm)

Work = 7.889 x 10^-4 J

The sedimentary layers at Capitol Reef are tilted due to the Waterpocket Fold, a monocline formed by the Laramide Orogeny, which caused an ancient buried fault to reactivate and the west side to shift upwards, tilting the layers.

The sedimentary layers at Capitol Reef are tilted due to geological activity associated with the Waterpocket Fold, a classic monocline formed during the Laramide Orogeny. This major mountain-building event reactivated an ancient buried fault, causing movement along the fault which shifted the west side upwards relative to the east side. This process lifted the sedimentary rock layers and created a distinctive tilt. Other regions like the Grand Canyon, Zion National Park, and the Colorado Plateau were also affected by uplift, but they often display a more uniform 'layer cake' appearance because their uplift was relatively even, unlike the dramatic tilting observed at Capitol Reef.

Answer:

B: Fnety = (FT)(sin 32°) – Fg

Explanation:

The total length of the spring when a 17.20 kg mass is suspended from it is 34.85 cm, calculated using Hooke's Law and the original unstretched length of the spring.

To find the total length of the spring when a 17.20 kg mass is suspended from it, we use Hooke's Law, which is F = kx, where F is the force applied to the spring, k is the spring constant, and x is the extension from the spring's original unstretched length. First, we need to calculate the extension caused by the 8.40 kg mass: F = (8.40 kg)(9.8 m/s2) = 82.32 N. With k = 875 N/m, the extension (x) is F/k = 82.32 N / 875 N/m = 0.0941 m or 9.41 cm. This extension is in addition to the unstretched length of the spring (L0), given by the total length minus the extension for the 8.40 kg mass: L0 = 0.250 m - 0.0941 m = 0.1559 m.

Next, for the 17.20 kg mass, the force is F = (17.20 kg)(9.8 m/s2) = 168.56 N. The additional extension caused by this mass is x = F/k = 168.56 N / 875 N/m = 0.1926 m or 19.26 cm. Therefore, the total length of the spring with the 17.20 kg mass is L = L0 + x = 0.1559 m + 0.1926 m = 0.3485 m or 34.85 cm.

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Answer: Viscous property

Explanation: The asthenosphere is the layer that lies below the crust. This layer is generally viscous, where the rocks floats due to deformation. It is the top part of the mantle and generation of convection current pushes the materials to move, thereby breaking the crust, causing it to move in convergent or divergent direction or slide past each other. The composition is same for both the lithosphere and the asthenosphere but it has its viscous property.

Using the principles of Physics and the equation for vertical displacement in projectile motion, it is calculated that the mountain lion, felis concolor, leaves the ground at a speed of approximately 8.3 m/s in order to reach a height of 11.7ft when jumping at an angle of 41.1 degrees.

The question is about the kinematics of projectile motion. To answer it, we can use the equation for the vertical displacement in projectile motion, which is h = v*sin(θ)*t - 0.5*g*t^2. Given that the mountain lion, felis concolor, jumps to a height of 11.7ft, we can substitute this value for h and convert it to SI units, which gives us 3.56m. The angle, θ, is given to be 41.1 degrees. We can assume that at the highest point of the jump, the velocity is zero, so we can set t as the time it takes for the lion to reach the peak of the jump and rearrange our equation to v*sin(θ) = 0.5*g*t. By substituting the known values of g and θ into this equation, we find that the mountain lion leaves the ground at a speed of about 8.3 m/s, calculated to one decimal place.

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The puma leaves the ground at a speed of approximately 12.79 m/s to make an 11.7 ft jump at an angle of 41.1°.

To determine the speed at which a puma, cougar, or mountain lion (Felis concolor) leaves the ground to achieve its impressive 11.7 ft (3.57 meters) jump at an angle of 41.1°, we apply projectile motion principles.

We are given:

The formula to find the initial vertical velocity component (V_oy) for a projective motion is:

H = (V_oy)² / (2g)

Here, 'g' is the acceleration due to gravity, which is approximately 9.81 m/s².

First, we solve for Voy:

V_oy = √(2gH)

Substitute the values:

V_oy = √(2 × 9.81 × 3.57) ≈ 8.35 m/s

Next, we use V_oy to find the initial speed (Vo) with the formula:

V_oy = Vo × sin(θ)

Rearranging gives:

Vo = V_oy / sin(θ)

Substitute the values:

Vo = 8.35 / sin(41.1°) ≈ 12.79 m/s

Therefore, the speed at which the puma leaves the ground to make this leap is approximately 12.79 m/s.

A person is not only dehydrated during summer time but also in cold dry weather.

During cold dry weather we loss more fluids due to loss of water in respiratory parts.At that time our blood vessels constrict which makes it difficult for blood to flow freely to extremities.

During winter time one feels less thirsty.It is so because sweat evaporates very quickly in cold weather which decreases our thirst response. At that time more urine is also produced.

Hence the statement given in the question is true(T).

Answer:

6.6°C

Explanation:

The question involves using the free fall equation to calculate the positions of an object falling from rest every second, from the top of a tower. The positions calculated are 4.9m, 19.6m, 44.1m, 78.4m and 122.5m for each of the 5 seconds. On the position-time graph, the points would be plotted and connected by a curved line, indicating acceleration.

The subject of this question involves the concept of free fall in Physics. In this scenario, an object falls from a height under the influence of gravity alone, with an acceleration of 9.8 m/s². We will utilize the displacement equation to calculate the object's position from the top of the tower every 1.0 seconds.

The equation used is: d = 0.5 * g * t² where g = acceleration due to gravity (9.8 m/s²), t = time, and d = distance fallen.

Please note that these are positions from the top of the tower, so the object is falling downwards.

Unfortunately, a pictorial representation of a position-time graph cannot be shown here. However, your graph should have time (in s) on the x-axis, and position (in m) on the y-axis, with data points plotted for each second up to 5 seconds, connected by a curved line—an implication of the accelerating object.

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When Dana accelerates onto the highway, the sports medal leans away from the windshield due to inertia. Dana's acceleration is approximately 2.51 m/s^2 away from the wall.

When Dana accelerates onto the highway, the sports medal suspended from her rearview mirror leans away from the windshield. This can be attributed to the inertia of the medal. As the car accelerates, the medal tends to remain at rest due to inertia while the car moves forward. Thus, the medal appears to lean away from the windshield.

To find Dana's acceleration, we can use the given information that the medal hangs at an angle of 15 degrees from the vertical. The acceleration can be calculated using the equation:

acceleration = g * tan(angle)

Substituting the given values, we get:

acceleration = 9.8 m/s^2 * tan(15 degrees)

Therefore, Dana's acceleration is approximately 2.51 m/s^2 away from the wall.

The scale read, in grams, as the hamster slides down along the wedge-shaped block is 931.5 g.

Given that:

Mass of the hamster, m = 190 g

Mass of the block, M = 820 g

The angle of the wedge block = 40°

Since the block does not have friction, the only force acting on the hamster is the normal force.

The normal component of the force is:

[tex]F_n=mgcos(\theta)[/tex]

This normal force has a downward component equal to:

[tex]F_d=F_ncos(\theta)[/tex]

[tex]F_d=mgcos^2(\theta)[/tex]

The net force acting on the hamster is:

F = Mg + mgcos²(θ)

  = (0.82 × 9.8) + (0.19 × 9.8 × cos²(40°))

  = 9.128 N

In grams, this can be found as:

[tex]\text{F}=\frac{9.128}{9.8} \times1000[/tex]

  [tex]=931.5\text{ g}[/tex]

Hence, the force is 931.5 g.

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The complete question with the figure is given below.

The figure shows a 190 g hamster sitting on an 820 g wedge-shaped block. The block, in turn, rests on a spring scale. An extra-fine lubricating oil having μs=μk=0 is sprayed on the top surface of the block, causing the hamster to slide down. The friction between the block and the scale is large enough that the block does not slip on the scale. What does the scale read, in grams, as the hamster slides down?

The question explores a physics scenario where a hamster slides off a lubricated block due to negligible friction. The block doesn't move due to sufficient friction with the scale. Concepts of forces, friction, and mechanical equilibrium are integral to understanding this.

The subject in question revolves around a scenario in Physics involving a hamster on a wedge-shaped block sprayed with lubricating oil, creating a scenario with negligible friction. In this case, the block is on a spring scale and the friction between the block and scale is enough to prevent slipping. This situation explores concepts of mass, force, friction, and mechanical equilibrium.

When the block is sprayed with lubricant, the friction between the hamster and the block decreases to nearly zero (μs=μk=0). This causes the hamster to slide off. The block, however, doesn't move because the friction between the block and the scale is high enough.

This scenario can be analyzed in a few different ways. One way could be through the use of Newton's laws of motion, considering the forces applied to the hamster and block separately. Another way could be using principles of energy conservation, though this might be more complex due to the dynamic nature of the situation.

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The magnitude of the net of force is about 470 Newton

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

period of the circular motion = T = 60 s

mass of the the bicycle = m = 76 kg

radius of the circuit = R = 140 m

Unknown:

magnitude of the net force = ΣF = ?

Solution:

We will use this following formula to find the tangential acceleration of the cyclist:

[tex]s = ut + \frac{1}{2}at^2[/tex]

[tex]2\pi R = 0(T) + \frac{1}{2}a(T)^2[/tex]

[tex]2\pi (140) = 0 + \frac{1}{2}a(60)^2[/tex]

[tex]280\pi = 1800a[/tex]

[tex]a = 280 \pi \div 1800[/tex]

[tex]a = \frac{7}{45} \pi \texttt{ m/s}^2[/tex]

[tex]\texttt{ }[/tex]

Next we will find the centripetal acceleration of the cyclist as it crosses the finish line:

[tex]a_c = v^2 \div R[/tex]

[tex]a_c = ( u + aT )^2 \div R[/tex]

[tex]a_c = ( 0 + \frac{7}{45} \pi(60))^2 \div 140[/tex]

[tex]a_c = \frac{28}{45} \pi^2 \texttt{ m/s}^2[/tex]

[tex]\texttt{ }[/tex]

Finally we could calculate the magnitude of the net force by using Newton's 2nd Law Of Motion as follows:

[tex]\Sigma F = m\sqrt{a^2 + a_c^2}[/tex]

[tex]\Sigma F = 76 \sqrt{(\frac{7}{45}\pi)^2+(\frac{28}{45}\pi^2)^2}[/tex]

[tex]\Sigma F \approx 470 \texttt{ Newton}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

IT IS 0.7 THANK YOU. Also if your not sure if your answer is right or not don't put it down cause other people look at it and enter it and get a bad grade cause of your selfish ahh. Thank you

The problem is about finding the reaction components at a ball-and-socket joint and the tension in wires supporting a sign. You can do this by breaking the forces and moments into their x,y,z components, setting up force balance and moment balance equations, and solving for the unknowns.

This problem pertains to static equilibrium, where both the sum of forces and the sum of torques are zero. To begin, let's establish a coordinate system where the ball-and-socket joint 'a' is considered the origin.

Assuming the tension in the wires is T and the angles of the wires to the x, y, z axes are known, you can break the tensions into their respective components using trigonometric principles. Let's consider Tbc and Tbd to be the tensions in the strings.

For the x-component we would sum up the forces in the x-direction and set it equal to zero. This can be represented as ∑Fx = Tbc,x + Tbd,x=0

The same goes for the y and z components with the z component taking into account the force due to gravity on the sign. Hence, ∑Fy = Tbc,y + Tbd,y=0 and ∑Fz = Tbc,z + Tbd,z - weight of the sign = 0. Here, the weight of the sign is 100kg*9.81m/s2

By balancing torques about each axis, you can find the tensions in the strings. The tension force should be the same in each wire if they make the same angles with the axes and are of equal length.

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This problem in Physics requires setting up equilibrium equations based on the given mass and calculating forces at the ball-and-socket joint and cable pressures.

The subject of this question falls under the branch of

Physics

referred to as statics, specifically the study of systems in equilibrium. Equilibrium implies that an object is neither accelerating nor rotating, meaning the sum of all forces and the sum of all torques acting on the object must both be zero. The 'x', 'y' and 'z' components referred to are likely part of a coordinate system used to describe the forces at work on the sign.

To determine these, one would need to set up equilibrium equations based on the known details about the system, which could include the known weight (provided by the mass of the sign and gravity), any relevant distances for calculating torques, and any other forces present such as tension in wires. The process of solving this system of equations would then yield the desired reaction components at the ball-and-socket joint and the tensions in the wires.

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Answer:

9800 W

Explanation:

The work done with the force of 75 N and displacement of 15 meters and angle of 45 ° is 795.4 J.

Work done in physics is the dot product of force and displacement. Work done is a vector quantity thus, it is characterised with a magnitude and direction.

When a force applied on an object is resulted in a displacement for the object, it is said to be work is done on the object. The work done by a force acting in angle θ in the horizontal direction is:

W = F d cos θ.

Given that, the force applied by the dog = 75 N

angle = 45 °

displacement = 15 m

Then, the work done W = 75 × cos 45 = 795.4 J.

Therefore, the work done by the dog on the sled is 795.4 J.

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Argon and krypton gases are typically used for colorful lighting when an electric current is applied.

The electromagnetic radiation spectrum produced when an electron changes from a high energy state to a lower energy state is known as the emission spectrum of a chemical element or chemical compound.

The energy difference between the two states is equal to the photon energy of the emitted photon. Each atom contains a large number of potential electron transitions, and each transition has a distinct energy difference.

An emission spectrum is made up of a variety of transitions that result in various radiated wavelengths. The emission spectra of each element is distinct. Therefore, components in matter with an unknown composition can be identified via spectroscopy.

Due to emission spectrum of Argon and krypton gases, colorful lighting happens when an electric current is applied.

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When both cars A and car B leave school at the same time, traveling in the same direction. car A travels at a constant speed of 75 km/h, while car B travels at a constant speed of 91 km/h, then car a would be 127.5 kilometers away from the school.

The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second.

As given in the problem when both cars A and car B leave school at the same time, traveling in the same direction. car A travels at a constant speed of 75 km/h, while car B travels at a constant speed of 91 km/h,

The distance covered by car A after 1.7 hours = 75km/h × 1.7 hours

                                                                                  =127.5 kilometers

Thus, car A would be 127.5 kilometers away from the school after 1.7 hours.

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