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padilas [110]
2 years ago
8

A wheelbarrow full of bricks is lifted to the top of a wall. If the mass of the loaded wheelbarrow is 3000.0 g and the height of

the wall is 0.45 meters, what is its GPE?
Physics
1 answer:
Rina8888 [55]2 years ago
6 0

Answer:

the Gravitational potential energy is 13.23 J

Explanation:

The computation of the GPE is shown below:

GPE stands for Gravitational potential energy

The following formula should be used for the same

= mass × gravity × height

= 3000 g × 9.8m/sec^2 × 0.45 m

= 13.23 J

Hence, the Gravitational potential energy is 13.23 J

We simply applied the above formula so that we can easily determine the GPE

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A cylindrical tank has a radius of 53 cm and a height of 1200 mm. It weighs 9.6 kN
xxTIMURxx [149]

Answer:

1985kg

Explanation:

assuming that

pi =3.14

oil density = 950kg/ cubic meter

g= 9.8m/s

M=\frac{9600}{9.8 } + 950 * 3.14*.53^{2} *1.2=1985kg

4 0
3 years ago
Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example
iragen [17]

Given Information:

Magnetic field = B = 1×10⁻³ T

Frequency = f = 72.5 Hz

Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m

Required Information:

Maximum Emf = ?

Answer:

Maximum Emf = 20.66×10⁻¹² volts

Explanation:

The maximum emf generated around the perimeter of a cell in a field is given by

Emf = BAωcos(ωt)

Where A is the area, B is the magnetic field and ω is frequency in rad/sec

For maximum emf cos(ωt) = 1

Emf = BAω

Area is given by

A = πr²

A = π(d/2)²

A = π(7.60×10⁻⁶/2)²

A = 45.36×10⁻¹² m²

We know that,

ω = 2πf

ω = 2π(72.5)

ω = 455.53 rad/sec

Finally, the emf is,

Emf = BAω

Emf = 1×10⁻³*45.36×10⁻¹²*455.53

Emf = 20.66×10⁻¹² volts

Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts

8 0
3 years ago
With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz t
VladimirAG [237]

Answer:

the tension of the rope is 34.95 N

Explanation:

Given;

length of the rope, L = 3 m

mass of the rope, m = 0.105 kg

frequency of the wave, f = 40 Hz

wavelength of the wave, λ = 0.79 m

Let the tension of the rope = T

The speed of the wave is given as;

v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu  = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v =  \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N

Therefore, the tension of the rope is 34.95 N

4 0
3 years ago
A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters
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Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

I'm pretty sure its wrong

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