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docker41 [41]
3 years ago
15

8. A receptacle, plug, or any other electrical device whose design limits the ability of an electrician to come in contact with

any of its electrical
connections is identified as
safe.
A. touch
B. removal
C. recessed
n feel
Physics
1 answer:
zloy xaker [14]3 years ago
5 0
Recessed is the correct answer
You might be interested in
There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co
mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

q_1=q_2=2.06\mu C=2.06\times 10^{-6} C

q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force,F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2

F_1=F_3=F

F_{net}=\sqrt{F^2+F^2+0}-F_2

F_{net}=\sqrt 2F-F_2

F=\frac{kq^2}{d^2}

F_2=\frac{Kq^2}{2d^2}

F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

Where k=9\times 10^9

F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})

F_{net}=0.208 N

3 0
3 years ago
Q¹=0,07Mc Q²=2C r=1,08 cm F=.......?
Nostrana [21]
The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = <span>(k * q1 * q2) / (r^2)  where:
</span>q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9<span> N. m</span>2/ C<span>2
</span>r is the distance between the two charges.

Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9<span> x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n </span> 
5 0
4 years ago
4. A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exer
Rudik [331]

Answer:

0.61°

Explanation:

Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.

Pulling force= resistance force

From the formula for pulling force,

F(x)= Fcos(θ)

= 425×cos(35.2)

=347N

The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)

=425×0.567=245N

Resistance force= (325N+ 245N) (α)= 570N(α)

We can now equates the pulling force to resistance force

570 (α)= 347N

(α)= 347/570

= 0.61

3 0
3 years ago
Need help with my science quiz
MaRussiya [10]

Explanation:

1.sugar

................

5 0
3 years ago
How much heat is evolved when 1234 g of water condenses to a liquid at 100.°c?
lbvjy [14]
During the phase transition vapour --> liquid water, the temperature of the water does not change; the molecules of water release heat and the amounf of heat released is equal to
Q=mL_e
where
m is the mass of the water
L_e is the latent heat of evaporation.

For water, the latent heat of evaporation is L=2257 kJ/kg, while the mass of the water is
m=1234 g=1.234 kg

so, the amount of heat released in the process is
Q=mL_e = (1.234 kg)(2257 kJ/kg)=2785 kJ
6 0
3 years ago
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