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Rufina [12.5K]
3 years ago
9

What forces are not present in space

Engineering
1 answer:
ss7ja [257]3 years ago
7 0

Answer:

C.) Weight and distance I believe

Explanation:

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You are in charge of ordering the concrete for a basement wall concrete pour. The wall forms are all set up and ready. The wall
choli [55]

Answer:

189.15cy

Explanation:

To understand this problem we need to understand as well the form.

It is clear that there is four wall, two short and two long.

The two long are \rightarrow 120ft5in+2(10ft)

The two long are \rightarrow 122ft1in=122.08ft

The two shors are \rightarrow 86ft4.5in = 86.375ft

The height and the thickness are 14ft and 0.83ft respectively.

So we only calculate the Quantity of concrete,

Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3

That in cubic yards is equal to 180.15 (1cy=27ft^3)

Hence, we need order 5% plus that represent with the quantity

Q_{ordered}=1.05*180.15=189.15cy

8 0
3 years ago
A system consists of N very weakly interacting particles at a temperature T sufficiently high so that classical statistical mech
algol [13]

Answer:

the restoring force is = 3/4NKT

Explanation:

check the attached files for answer.

7 0
4 years ago
The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc
Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy  

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.  

Given Data  

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist  

∅=4°

∅=4*\pi/180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist  

∅=TL/GJ

∅=TL/G*\pi/2*c^4

∅=2TL/G*\pi*c^4

c=\sqrt[4]{2TL/\pi G }∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J*\pi/2*c^4)]*c

г =[2T/(J*\pi*c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required  

We will use larger of the two values  

c= 18.0569 x 10^-3 m  

c = 18.0569 mm  

3 0
3 years ago
Q10: Technician A says that nearly all brands of scan tools will pull DTCs from the ABS
Llana [10]

Hey! How are you? My name is Maria, 19 years old. Yesterday broke up with a guy, looking for casual sex.

Write me here and I will give you my phone number - *pofsex.com*

My nickname - Lovely

7 0
3 years ago
g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,
irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0
3 years ago
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