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1 year ago

Energy released by the sun results from the process where atomic nuclei

Physics

1 answer:

Ray Of Light [21]1 year ago

6 0

Answer:

b) combine

Explanation:

The energy from the sun is due to Nuclear Fusion where lighter nuclei combine to form heavier and stable nuclei this releasing a lot of energy in this process.

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A 0.25-kg ball sits on the roof of a building that is 10 meters tall. Find the GPE. (Gravity = 9.8 on Earth)

Tasya [4]

Gravitational potential energy = mgh or mass times acceleration due to gravity times the height

Here the mass is 0.25kg, the height is 10m, and gravity is 9.8m/s^2 so...

GPE = (0.25)(10)(9.8)

GPE = 24.5 J

7 0

3 years ago

Which example possesses mechanical potential energy?. A. a taut guitar string. B. an oscillating pendulum. C. a roller coaster r

Andrej [43]

<span>
The taut guitar string haspotencial energy which we can see in action.</span> <span>· so option a is correct.</span>

6 0

2 years ago

Read 2 more answers

A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of

lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

8 0

2 years ago

Read 2 more answers

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At thi

tatiyna

Answer:

50 revolutions

Explanation:

Data provided:

case I: From rest to top spin

The initial angular speed of the washer, ωi = 0 rev /s

Final angular speed of the washer ωf = 5 rev /s

Time taken, t₁ = 8 s

now,

The angular displacement or the number of revolutions taken (θ₁) is calculated as:

θ₁ = ωi t₁ + (1/2)α₁t₁²

where,

α is the angular acceleration

The angular acceleration can be calculated as:

ωf - ωi = α₁t₁

on substituting the values, we get

8α₁ = 5 - 0

α₁ = 0.625 rev/s²

substituting the values in the equation for the number of revolutions, we get

θ₁ = 0 + (1/2) (0.625)(8)²

θ₁ = 20 revolutions

also,

For the case II: From top spin to rest

we have

The initial angular speed, ωi = 5 rev /s

and the final angular speed, ωf = 0 rev /s

Total time taken, t₂ = 12 s

Now, angular acceleration for this case

ωf - ωi = α₂t₂

on substituting the values, we have

12α₂ = 0 - 5

α₂ = - 0.4166 rev/s²

Therefore, the number of revolutions ( i.e angular displacement )

θ₂ = ωit₂ + (1/2)α₂t₂²

on substituting the values, we have

θ₂ = 5 × 12 + (1/2)(-0.4166)(12)²

θ₂ = 30 rev

Hence,

the total number of revolutions made by the washer during the 20s is

θ = θ₁ + θ₂

θ = 20 rev + 30 rev

θ = 50 revolutions

7 0

2 years ago

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0

3 years ago