If the force acting on a body of mass 40 k.g is doubled by how much will the acceleration change

A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

centripetal force acting on satellite = $\dfrac{mv^2}{r}$

gravitational force = $\dfrac{GMm}{r^2}$

equating both the above equation

$\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$

$v = \sqrt{\dfrac{GM}{r}}$

$v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}$

v = 5578.5 m/s

b) $T= \dfrac{2\pi\ r}{v}$

$T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}$

T = 14416.92 s

$T = \dfrac{14416.92}{3600}\ hr$

T = 4 hr

c) gravitational force acting

$F = \dfrac{GMm}{r^2}$

$F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}$

F = 5202 N

4 0

3 years ago

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

bixtya [17]

Answer:

1340.2MW

Explanation:

Hi!

To solve this problem follow the steps below!

1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid

W=αhQ

α=specific weight for water =9.81KN/m^3

h=height=220m

Q=flow=690m^3/s

W=(690)(220)(9.81)=1489158Kw=1489.16MW

2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator

Wr=(0.9)(1489.16MW)=1340.2MW

the maximum possible electric power output is 1340.2MW

3 0

2 years ago

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

vredina [299]

Answer:

The angular speed after 6s is $\omega = 1466.67s^{-1}$ .

Explanation:

The equation

$I\alpha = Fd$

relates the moment of inertia $I$ of a rigid body, and its angular acceleration $\alpha$ , with the force applied $F$ at a distance $d$ from the axis of rotation.

In our case, the force applied is $F = 22N$ , at a distance $d = 6cm =0.06m$ , to a ring with the moment of inertia of $I =mr^2$ ; therefore, the angular acceleration is