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3 years ago

Gravity is a force of attraction that acts

Physics

1 answer:

bixtya [17]3 years ago

5 0

Answer:gravity is a force of attraction that acts between two masses separated by distance

Explanation:

Gravity is a force of attraction that acts between two masses separated by distance

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Find the expression for pressure exerted by fluid with proper description

fredd [130]

Explanation:

The pressure exerted by a column of liquid of height h and density ρ is given by the hydrostatic pressure equation p = ρgh, where g is the gravitational acceleration

3 0

2 years ago

A battery is connected to a light bulb that has 3 of resistance . If there is 0.5 A of current flowing what is the voltage on th

Alenkasestr [34]

Answer:

1.5 V

Explanation:

E = IR = 0.5(3) = 1.5 V

8 0

3 years ago

Three masses are located in the x- y plane as follows: a mass of 6 kg is at (0 m, 0 m), a mass of 4 kg is at (3 m, 0 m), and a m

enot [183]

Answer:

The center of mass of three mass in the x-y plane is located at (1,0.5).

Explanation:

It is given that, a mass of 6 kg is at (0,0), a mass of 4 kg is at (3,0), and a mass of 2 kg is at (0,3). We need to find the center of mass of the system. Center of mass in x direction is :

$C_x=\dfrac{6\times 0+4\times 3+2\times 0}{6+4+2}\\\\C_x=1$

The center of mass in y direction is :

$C_y=\dfrac{6\times 0+4\times 0+2\times 3}{6+4+2}\\\\C_y=0.5$

So, the center of mass of three mass in the x-y plane is located at (1,0.5).

3 0

2 years ago

determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. assume that the

Alenkasestr [34]

Answer:

92.81 psia.

Explanation:

The density of water by multiplying its specific gravity by the density of sea water.

SG = density of sea water/density of water

ρ = SG x ρw

1 kg/m3 = 62.4 lbm/ft^3

= 1.03 * 62.4

= 64.27lbm/ft^3.

The absolute pressure at 175 ft below sea level as this is the location of the submarine.

P = Patm +ρgh

= 14.7 + 64.27 * 32.2 * 175

Converting to pound force square inch,

= 14.7 + 64.27 * (32.2ft/s^2) * (175ft) * (1lbf/32.2lbm⋅ft/s^2) * (1ft^2/144in^2 )

= 14.7 + 78.11 psia

= 92.81 psia.

8 0

2 years ago

An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9∘ above the horizontal. part a determine x-v

ankoles [38]

Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal

Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s

The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s

The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.

Answer:

t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37

5 0

2 years ago

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