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Delicious77 [7]
3 years ago
8

One problem when using rigid metal conduit in a residence is that the installation of the conduit may a weaken the structure b r

equire an equipment ground c be a fire hazard d be an electrical hazard
Physics
2 answers:
worty [1.4K]3 years ago
8 0
The correct answer is A. Installation of rigid metal conduit requires grounding and the grounding equipment used may weaken the structure.
astra-53 [7]3 years ago
6 0

Answer:

a). the installation of the conduit may a weaken the structure

Explanation:

RMC or Rigid metal conduit is a heavy duty steel tubing that is galvanized and is installed in the residence with threaded fittings. It provides structural support to the panels and electrical cables and provides protection from damage to the fittings. But this protection comes at the cost of weakening the structural member of the buildings.

Thus, the answer is a). the installation of the conduit may a weaken the structure.

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A bus start from rest and then moves with a constant acceleration of 2m/s square. calculate its time of traven if its final velo
pshichka [43]

Answer:

Its Answer is 10 s.

Explanation:

As acceleration is defined as time rate of change of velocity. So,

a =  ( vf - vi ) ÷ t

2 = (20 - 0 ) ÷ t

2 =  20 ÷  t

t = 20 ÷ 2

t  = 10 s

We have put initial velocity as zero because  body is starting from rest.

<em>Hope it helps.</em>

5 0
3 years ago
A box sits at the edge of a spinning disc. The radius of the disc is 0.5 m, and it is initially spinning at 5 revolutions per se
Furkat [3]

Answer:

a) α = 0.375 rad/s²

b) at = 0.1875 m/s²

c) ac =79 m/s²  

d) θ = 52 rad

Explanation:

The uniformly accelerated circular movemeis a circular path movement in which the angular acceleration is constant.

Tangential acceleration is calculated as follows:

at = α*R     Formula (1)

Centripetal acceleration is calculated as follows:

ac =ω² *R   Formula (2)

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t  Formula (3)

θ=  ω₀*t + (1/2)*α*t² Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed  ( rad/s)

R : radius of the circular path (m)

at:  tangential acceleration, (m/s²)

ac: centripetal acceleration, (m/s²)

Data:

R= 0.5 m  : radius of the disk

t₀=0 , ω₀ = 5 rev/s  

1 revolution = 2π rad

ω₀ = 5*(2π)rad/s  =10π rad/s  = 31.42 rad/s

ωf = 2*(2π)rad/s  =4π rad/s  = 12.57 rad/s

t = 8 s

(a) angular acceleration of the box

We replace data in the formula (3)

ωf= ω₀ + α*t

2 = 5 + α*(8)

2 -5 = α*(8)

-3 = (8)α

α=3 /8

α = 0.375 rad/s²

(b) Tangential acceleration of the box

We replace data in the formula (1)z

at =(α)*R

at = (0.375)*(0.5)

at = 0.1875 m/s²

c) Centripetal acceleration of the box at  t = 8 s

We replace data in the formula (2)

ac =ω² *R

ac =(12.57)² *(0.5)

ac = 79 m/s²  

d) Radians that the box has rotated over after t = 8 s

We replace data in the formula (4)

θ = ω₀*t + (1/2)*α*t²

θ = (5)*(8)+ (1/2)*( 0.375)*(8)²

θ = 52 rad

9 0
3 years ago
What name is given to the rate of flow of<br> electric charge?
RSB [31]

Answer:

<h2>Electric charge</h2>

Explanation:

The rate of the flow of electric charge is known as electric current. <u>By convention, the direction of electric current is always the direction of net flow of positive charge.</u>

3 0
3 years ago
To win the game, a place kicker must kick a
masya89 [10]

Answer:

0.57 m

Explanation:

First of all, we need to calculate the time it takes for the ball to cover the horizontal distance between the starting position and the crossbar. This can be done by analzying the horizontal motion only. In fact, the horizontal velocity is constant and it is

v_x = u cos \theta = (15)(cos 51.7^{\circ})=9.30 m/s

And the distance to cover is

d = 19 m

So the time taken is

t=\frac{d}{v_x}=\frac{19}{9.30}=2.04 s

Now we want to find how high the ball is at that time. The initial vertical velocity is

u_y = u sin \theta = (15)(sin 51.7^{\circ})=11.77 m/s

So the vertical position of the ball at time t is

y(t) = u_y t - \frac{1}{2}gt^2

where g = 9.8 m/s^2 is the acceleration of gravity. Substituting t = 2.04 s, we find

y=(11.77)(2.04)-\frac{1}{2}(9.8)(2.04)^2=3.62 m

The crossbar height is 3.05 m, so the difference is

\Delta h = 3.62 - 3.05 =0.57 m

So the ball passes 0.57 m above the crossbar.

8 0
3 years ago
a net force of 1,800 N is applied to a boat causing it to accelerate at 1.5 m/s^2. what is the mass of the boat?
Xelga [282]

Answer:

<h3>The answer is 1200 kg</h3>

Explanation:

The mass of an object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{1800}{1.5}  \\

We have the final answer as

<h3>1200 kg</h3>

Hope this helps you

7 0
4 years ago
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