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Dennis_Churaev [7]
2 years ago
7

A spring with a spring constant of 4 N/m is stretched by Sarah's dog until extended by 2 m. How much elastic potential energy is

stored by the spring?
Physics
1 answer:
azamat2 years ago
8 0

Hi there!

Recall the equation for spring potential energy:

PE = \frac{1}{2}kx^2

k = Spring constant (N/m)

x = extension of spring from equilibrium (m)

PE = Potential Energy (J)

Plug in the given values:

PE = \frac{1}{2}(4)(2^2) = \boxed{8 J}

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A pendulum is swinging next to a wall. The distance from the bob of the pendulum to the wall varies in a periodic way that can b
nexus9112 [7]

The formula of the trigonometric function that models the distance HHH from the pendulum's bob to the wall after t seconds is

H(t) = 15 -6sin(2.5π(t -0.5))

Detailed explanation:

The function can be expressed as the following for the midline M, amplitude A, period T, and time t0 at which the function deviates from the midline:

 H(t) = M -Asin(2π/T(t -t0))

The equation is based on the parameters M=15, A=6, T=0.8, and t0 = 0.5.

H(t) is equal to 15 -6sin(2.5π(t -0.5))

<h3>What is function?</h3>

The trigonometric functions in mathematics are real functions that connect the right-angled triangle's angle to the ratios of its two side lengths .In all areas of study that involve geometry, such as geodesy, solid mechanics, celestial mechanics, and many others, they are widely used.

To learn more about functions visit:

brainly.com/question/15607563

#SPJ4

The correct question is:

A pendulum is swinging next to a wall. The distance from the bob of the swinging pendulum to the wall varies in a periodic way that can be modeled by a trigonometric function.

The function has period 0.80.80, point, 8 seconds, amplitude 6 \text{ cm}6 cm6, start text, space, c, m, end text, and midline H = 15 \text{ cm}H=15 cmH, equals, 15, start text, space, c, m, end text. At time t = 0.5t=0.5t, equals, 0, point, 5 seconds, the bob is at its midline, moving towards the wall.

Find the formula of the trigonometric function that models the distance HHH from the pendulum's bob to the wall after t seconds. Define the function using radians.

5 0
2 years ago
A 65.0 kg ice skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. What is the vel
spin [16.1K]

Answer:

(d) 0.07 m/s

Explanation:

Given Data

Snowball mass m₁=0.15 kg

Ice skater mass m₂=65.0 kg

Snowball velocity v₁=32.0 m/s

To find

Velocity of Skater v₂=?

Solution

From law of conservation of momentum

m_{1}v_{1}=m_{2}v_{2}\\  v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s

So Option d is correct one

5 0
3 years ago
Determine the thrust that a boat with a volume of 1.2m³ receives when it is stranded at sea. The density of seawater is 1020kg /
puteri [66]

Answer:

The maximum possible up-thrust on the boat is 11,995.2 N

Explanation:

According to Archimedes' principle, the thrust received by an object immersed a fluid is equal to the weight of the fluid displaced;

The given parameter of the boat in sea water are;

The volume of the boat = 1.2 m³

The density of seawater = 1020 kg/m³

Density = Mass/Volume

Therefore, Mass = Density × Volume

The maximum volume of water that the boat displaces = 1.2 m³

The mass of the water displaced by the boat = (Density of seawater) × (Volume of seawater displaced)

∴ The maximum possible mass of the water displaced by the boat = 1.2 m³ × 1020 kg/m³ = 1224 kg

The maximum possible mass of the water displaced by the boat, m = 1224 kg

Weight = Mass, m × g

Where;

g = The acceleration due to gravity = 9.8 m/s²

The up-thrust on the boat = The weight of the seawater displaced

∴ The maximum possible up-thrust on the boat = m × g = 1224 kg × 9.8 m/s² = 11,995.2 N

The maximum possible up-thrust on the boat = 11,995.2 N.

3 0
3 years ago
The 480 g bar is rotating as shown what is the angular momentum of the bar about the axle?
Greeley [361]
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:

Angular momentum L = Iω, where 
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>

<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>

<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>

<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>

<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
4 0
3 years ago
How to convert 1gcm⁻³ in to kg⁻³??
dimulka [17.4K]

The answer is 0.001 , 1000 grams = 1 Kilograms

3 0
4 years ago
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