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1 year ago

What is the net torque on the bar shown in (Figure 1) about the axis indicated by the dot? Suppose that F=6.0N

Physics

1 answer:

AysviL [449]1 year ago

6 0

Answer:

net torque on the system is 2.5 Nm

Explanation:

As we know that net torque on the system is given as

$\tau = r_1F_1 - r_2F_2$

now we have

$r_1 = 75 cm$

$F_1 = 6 N$

$r_2 = 25 cm$

$F_2 = 8 N$

now we have

$\tau = 0.75 (6) - 0.25(8)$

$\tau = 2.5 Nm$

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A manometer is used to measure the air pressure in a tank. the fluid used has a specific gravity of 1.25, and the differential h

BartSMP [9]

Specific Gravity of the fluid = 1.25
Height h = 28 in
Atmospheric Pressure = 12.7 psia
Density of water = 62.4 lbm/ft^3 at 32F
Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
Density of the Fluid p = 78 lbm/ft^3
Difference in pressure as we got the differential height, dP = p x g x h dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
Difference in pressure = 1.26 psia
(a) Pressure in the arm that is at Higher
P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
(b) Pressure in the tank that is at Lower
P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia

4 0

2 years ago

An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion,

Ne4ueva [31]

A) f = 1.8 rev/s = 2 Hz
T = 1 / f = 0.55s

B) not really sure..srry

C) T = 2 pi √ ( L / g ) 
0.57 = 2 x 3.14 x √ ( 0.2 / g )

g = 25.5 m/s²

Hope this helps a little at least.. :)

5 0

1 year ago

How high above the earth's surface is g reduced to 8.80m/^2?

Sladkaya [172]

Gravity changes as the altitude change. The gravitational force is proportional to 1/R2, where R is the distance from the center of the Earth the radius of earth where gravity is 9.8 m/s^2 is 6400 km this will serve as the zero mark.

g1/(g2) = R2^2/(R1)^2

so we set the constant values to R1 and the unknown distance as x

(9.8)/(8.80) = (6400-x)2/(6400)^2

solving for x we will get

x = 345.85 km above the earths surface

Hope my answer would be a great help for you.
If you have more questions feel free to ask here at Brainly.

6 0

2 years ago

Read 2 more answers

Temperature difference in the body. The surface temperature of the body is normally about 7.00 ∘C lower than the internal temper

egoroff_w [7]

Answer:

7 K.

12. 6 °F

Explanation:

Convert the individual temperatures to Kelvin (Surface temperature and internal temperature) before calculating the temperature difference of the body,

Let The Surface temperature Be = X °C

And the internal Temperature will be = (X + 7) °C

Converting the surface and the internal temperature to temperature in Kelvin

Surface Temperature of the body (K) = (X + 273) K

Internal Temperature of the body (K) = (X + 7) + 273 = (X + 280) K.

∴ Temperature difference of the body (K) = Internal temperature(K) - surface temperature(K) = (X + 279) - (X + 280)

= X - X + 280 - 273 = 7 K.

∴Temperature difference of the body (K) = 7 K

Also for Fahrenheit, Convert the individual temperatures (Surface temperature and internal temperature) to Fahrenheit before calculating the temperature difference of the body.

We use , F = 1.8C + 32

Where C = temperature in Celsius.

also,

Let The Surface temperature Be = X °C

And the internal Temperature of the body will be = (X + 7) °C

Converting to Fahrenheit

Surface Temperature of the body = 1.8X + 32 °F

Internal Temperature of the body = 1.8(X+7) + 32 = 1.8X + 12.6 + 32

Internal Temperature of the body = 1.8X + 44.6 °F

∴ The temperature difference of the body (°F) = Internal temperature(°F) - surface temperature(°F) = (1.8X + 44.6) - (1.8X + 32)

surface temperature(°F) = 1.8X - 1.8X + 44.6 - 32

surface temperature(°F) = 12. 6 °F.

3 0

2 years ago

"The two equations below express conservation of energy and conservation of mass for water flowing from a circular hole of radiu

Leno4ka [110]

Answer:

The two equations below express conservation of energy and conservation of mass for water flowing from a circular hole of radius 3 centimeters at the bottom of a cylindrical tank of radius 10 centimeters. In these equations, delta m is the mass that leaves the tank in time delta t, v is the velocity of the water flowing through the hole, and h is the height of the water in the tank at time t. g is the acceleration of gravity, which you should approximate as 1000 cm/s2.

shdh

5 0

1 year ago