Help me with this Weather Station Model by answering 5 questions. It is due tomorrow.

An object's weight is dependent upon its location in the universe. Why is this true? A. This is true because weight is the amoun

Westkost [7]

I think its either C or D.

5 0

3 years ago

A spring with a pointer attached is hanging next to a scale marked in millimeters. Three different packages are hung from the sp

SOVA2 [1]

solution;

$the expression for force applied on the spring due to the load is\\f=k\Delta x\\here,\Delta x is the extension in the spring due to appling force\\given three case as following\\110N=k(40-x_{o})----------1\\240N=k(60-x_{o})----------2\\w=k(30-x_{o})-------------3\\$ $To calculate the accrual length of the spring,solve th equation 1 and 2\\\frac{110N}{240N}=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458(60mm-x_{o})=(40mm-x_{o})\\x_{o}(1-0.458)=(40-60(0.458))mm\\x_{o}\frac{12.52}{0.542}\\=23.1mm\\to calculate the force on the spring in case,\\solve the equation 1 and 2\\\frac{110}{w}=\frac{k(40-x_{o})}{k(60-x_{o})}\\\frac{110}{w}=\frac{(40mm-23.1mm)}{30mm-23.1mm}\\w=\frac{110}{2.45}=44.9N$

8 0

3 years ago

A fish is hooked on a line that is wound around the spool of a fishing reel. The reel has a mass of 652 g and a diameter of 10 c

iren2701 [21]

The magnitude of the force exerted by the fish is 3,052.5 N.

The given parameters;

mass of the reel, m = 652 g = 0.652 kg

diameter of the reel, d = 10 cm

radius of the reel, r= 5 cm = 0.05 m

time of motion, t = 5 s

angular speed of the reel, ω = 306 rad/s

According to Newton's third law, the force exerted by the reel is equal in magnitude to the force fish exerted on the reel.

The centripetal force of the reel = force exerted by the fish

$F_c = m\omega^2 r\\\\F_c = 0.652 \times (306)^2 \times 0.05\\\\F_c = 3,052.5 \ N$

Thus, the magnitude of the force exerted by the fish is 3,052.5 N.

Learn more here: brainly.com/question/14905888

3 0

3 years ago

The Sankey diagrams below show the energy transfers in two light bulbs. What is the efficiency of light bulb 1? 1: 80j Input ene

tatyana61 [14]

Answer:

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7 0

3 years ago

To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 130 m. The car

marysya [2.9K]

Answer:

0.739

Explanation:

If we treat the four tire as single body then

W ( weight of the tyre ) = mass × acceleration due to gravity (g)

the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r

where v is speed 25.6 m/s and r is the radius of the circle

centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²

net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²

coefficient of static friction between the tires and the road = frictional force / force of normal

frictional force = m × net acceleration / m×g

where force of normal = weight of the body in opposite direction

coefficient of static friction = (7.2567 × m) / (9.81 × m)

coefficient of static friction = 0.739

4 0

3 years ago