Please help ASAP guys !!! Thanks

What are the conditions under which the resultant of three coplanar forces is zero?

serg [7]

When three or more coplanar forces are acting at a point and the vector diagram closes, there is no resultant. The forces acting at the point are in equilibrium.

4 0

2 years ago

What is a blue moon

Irina18 [472]

.
a phenomenon whereby the moon appears bluish owing to smoke or dust particles in the atmosphere

8 0

3 years ago

The downsprue leading into the runner of a certain mold has a length of 175 mm. The cross-sectional area at the base of the spru

adoni [48]

Answer:

(a) Velocity at bottom is 1.85 m/s

(b) Volume flow rate is 7.4 x 10⁻⁴ m³/s.

(c) The time required to fill the mold is 1.35 s.

Explanation:

(a)

Applying Bernoulli's Equation on both ends of the down sprue, with the assumptions that every point is at atmospheric pressure and the liquid metal at the pouring basin is at zero velocity. The equation then becomes:

V = √2gh

where,

V = velocity at bottom of down sprue

h = height of down sprue = 175 mm = 0.175 m

V = √2(9.8 m/s²)(0.175 m)

V = 1.85 m/s

(b)

The volume flow rate is given as:

Volume Flow Rate = (V)(A)

where,

V = velocity at bottom = 1.85 m/s

A = Area of bottom = 400 mm² = 0.0004 m²

Therefore,

Volume Flow Rate = (1.85 m/s)(0.0004 m²)

Volume Flow Rate = 7.4 x 10⁻⁴ m³/s = 740 cm³/s

(c)

The time required to fill the cavity is given as:

Volume Flow Rate = V/t

where,

V = Volume of mold Cavity = 0.001 m³

t = time required to fill the cavity = ?

Therefore,

t = V/Volume Flow Rate

t = 0.001 m³/7.4 x 10⁻⁴ m³/s

t = 1.35 s

5 0

3 years ago

Chegg Given that the mean radius of the Moon’s orbit is 3.84 x 108 m and its period is 2.36 x 106 sec, at what altitude above th

alukav5142 [94]

Answer:

The altitude of geostationary satellite is $3.58\times10^{7}\ m$

Explanation:

Given that,

Radius of moon's orbit $r=3.84\times10^{8}\ m$

Time period $T=2.36\times10^{6}\ sec$

We need to calculate the orbital radius of geostationary satellite is

Using formula of time period

$T=\sqrt{\dfrac{4\pi^2}{GM}a^3}$

$a=((\dfrac{GM}{4\pi^2})T^2)^{\dfrac{1}{3}}$

Where, G = gravitational constant

M = Mass of earth

T = time period of geostationary satellite orbit

Put the value in to the formula

$a=((\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{4\times\pi^2})\times(86160)^2)^{\dfrac{1}{3}}$

$a=4.217\times10^{7}\ m$

We need to calculate the altitude of geostationary satellite

Using formula of altitude

$h = a-R_{e}$

Where, R = radius of earth

a = radius of geostationary satellite

Put the value into the formula

$h =4.217\times10^{7}-6.38\times10^{6}$

$h =35790000\ m$

$h=3.58\times10^{7}\ m$

Hence, The altitude of geostationary satellite is $3.58\times10^{7}\ m$

4 0

3 years ago

How much force acts on one surface of a rectangular bedroom wall that is 2.20 m by 3.20 m when atmosspheric pressure is?

irakobra [83]

We know that pressure is the force applied into a surface, in our case the wall of the room, so then first we will calculate the surface of this wall: S = 2.2 * 3.2 = 7.04 m2 Then we also know the atmospheric pressure in normal conditions is 1 atm. That is the same 1 atm = 101325 Pascals or 101325 N/m2 Now we need to use the formula : P = F/S where P is pressure, F is force and S is surface to calculate the force: F = P * S = 101325 * 7.04 = 713,328 Newtons Conclusion: the force acts on the wall due the air inside the room is 713,328 N

3 0

3 years ago