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ella [17]
3 years ago
15

Calculate the total energy of 4.0 kg object moving horizontally at 20 m/s 50 meters above the surface.

Physics
1 answer:
Serhud [2]3 years ago
4 0

Answer:

Correct answer:  E total = 2,800 J

Explanation:

Given:

m = 4 kg   the mass of the object

V = 20 m/s  the speed (velocity) of the object

H = 50 m the height of the object above the surface

E total = ? J

The total energy of an object is equal to the sum of potential and kinetic energy

E total = Ep + Ek

Ep = m g H   we take g = 10 m/s²

Ep = 4 · 10 · 50 = 2,000 J

Ek = m V² / 2

Ek = 4 · 20² / 2 = 2 · 400 = 800 J

E total = 2,000 + 800 = 2,800 J

E total = 2,800 J

God is with you!!!

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Give an example of how the apparent motion of an object depends on the observers motion
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A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0
2 years ago
Identify two size dependent properties and two size independent properties of an iron nail
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2 size dependent = volume and mass. 2 size independent properties = conductivity, bowing point, state of matter, and melting point. analyze, name two size dependent properties and two size independent properties of an iron nail.


hope this helps

6 0
3 years ago
Read 2 more answers
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