The father of scientific management is- (

1. A race car travels one lap around a track with a radius of 80 m and a speed

gulaghasi [49]

Answer:

15

Explanation:

6 0

3 years ago

In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate th

Radda [10]

Answer:

a) Option D is correct.

The center of mass between the Eartg and the moon is inside the Earth.

Explanation:

Given,

Mass of the moon = (7.35×10²²) kg

Mass of the Earth = (6.00×10²⁴) kg

Mass of the Sun = (2.00×10³⁰) kg

Distance between the Earth and the moon = (3.80×10⁵) km

Distance between the Earth and the Sun = (1.50×10⁸) km

With the assumption that all.of the bodies being considered are on the same straight line on the x-axis,

Note that Centre of mass is given as

C.M = (Σmx)/(Σm)

For the Earth-moon system, let the earth be x=0, then the moon is at x = (3.80 × 10 5) km away.

C.M = (Σmx)/(Σm)

Σmx = (6.00×10²⁴)) × (0) + (7.35×10²²) × (3.80×10⁵) = (2.793 × 10²⁸) kg.km

Σm = (6.00×10²⁴) + (7.35×10²²) = (6.0735 × 10²⁴) kg

CM = (2.793 × 10²⁸) ÷ (6.0735 × 10²⁴)

CM = (4.60 × 10³) km = 4600 km

This means the centre of mass is 4600 km from the Earth.

The Earth's radius = 6378 km

Hence, the centre if mass is inside the Earth.

Hope this Helps!!!

8 0

3 years ago

Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surf

Vladimir [108]

Answer:

1. 0.574 kJ/kg

2. 315.7 MW

Explanation:

1. The mechanical energy per unit mass of the river is given by:

$E_{m} = E_{k} + E_{p}$

$E_{m} = \frac{1}{2}v^{2} + gh$

Where:

Ek is the kinetic energy

Ep is the potential energy

v is the speed of the river = 3 m/s

g is the gravity = 9.81 m/s²

h is the height = 58 m

$E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg$

Hence, the total mechanical energy of the river is 0.574 kJ/kg.

2. The power generation potential on the river is:

$P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW$

Therefore, the power generation potential of the entire river is 315.7 MW.

I hope it helps you!

4 0

3 years ago

At a distance of 0.208 cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485 n/c . wh

Anna35 [415]

We have the equation for electric field E = kQ/ $d^{2}$

Where k is a constant, Q is the charge of source and d is the distance from center.

In this case E is inversely proportional to $d^{2}$

So, $\frac{E_{1} }{E_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}$

$E_{1}$ = 485 N/C

$d_{1}$ = 0.208 cm

$d_{2}$ = 0.620 cm

$E_{2}$ = ?

$\frac{485 }{E_{2}} = \frac{0.620^{2}}{0.208^{2}}$

$E_{2}$ = $\frac{485*0.208^{2}}{0.628^{2}}$

$E_{2}$ = 53.20 N/C

7 0

3 years ago

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Suppose a rocket is traveling through space and moving at a speed close to the speed of light. Three minutes pass on the rocket

Sedbober [7]

Answer:

B

Explanation:

on edge2020

4 0

3 years ago

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