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Hoochie [10]
3 years ago
6

Does anyone understand this stuff?

Physics
2 answers:
Airida [17]3 years ago
7 0
<span>This question can be solved as follows:

A discount offered = (100-18)(100-12)/(100*100)
B discount offered = (100-17)(100-13)/(100*100)</span>
denis23 [38]3 years ago
5 0

Answer:

Manufacturer A: 0.2784 or 27.84%

Manufacturer B: 0.2779 or 27.79%

Therefore, there is not much of a relevant difference but still manufacturer A has a better offer.

Explanation:

calculating the equivalent final discount rate using the formula:

FD= [1- (1-D1(%)) x (1-D2(%))]

D1: first discount to retail price (%)

D2: second discount to the first net price (%)

FD: Final equivalent discount rate (%)

observation: the expression can always be expanded to add more chain discounts using the same principles.

*Manufacturer A 18/12

Chain Discount:  D1= 18 D2= 12

FD= [1- (1- 0.18) x (1-0.12)] ≅ 0.2784

* Manufacturer B 17/13

Chain Discount: D1= 17 D2= 13

FD= [1- (1- 0.17) x (1-0.13)] ≅ 0.2779

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A force of 1 N will cause a mass of 1 kg to have an acceleration of 1 m/s2. Therefore, a force of 7 N applied to a mass of 7 kg
Lana71 [14]

1 m/s^2

Using F=ma,

7=7a

a=1

3 0
3 years ago
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A spherical shell with a net charge of 3Q surrounds a point charge of -q at the center of the shell. The charges on the inner an
aleksley [76]

Answer:

1) The charge on the outer shell is +4·Q

2) The charge on the inner shell is +Q

Explanation:

1) The given parameters of the spherical shell are;

The net charge on the spherical shell = 3·Q

The point charge surrounded by the spherical shell = -Q

Let 'x' represent the charge on the outer shell, and let 'y', represent the charge on the inner shell, we have;

The net charge, 3·Q = -q + x

∴ x = 3·Q + Q = 4·Q

The charge on the outer shell, x = 4·Q

2) The net charge in the shell is zero, therefore, the charge on the inner shell, 'y', is given as follows;

-Q + y = 0

∴ y = +Q

The charge on the inner shell, y = +Q

5 0
3 years ago
Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}}), where imax=\frac{E}{R}

Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

⇒\frac{L}{R}=\frac{2}{log2}

Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

5 0
3 years ago
Imagine an isolated positive point charge Q (many times larger than the charge on a single proton). There is a charged particle
egoroff_w [7]

Answer:

Explanation:

The magnitude of the electric force on this charged particle A depends upon the following

5. the distance between the point charge Q and the charged particle A

8. the amount of the charge on the point charge Q

9. the magnitude of charge on the charged particle A

7 0
3 years ago
A 10 kg block slides on a frictionless surface at 10 m/s . It hits a rough patch 3 meters long with a coefficient of kinetic fri
Valentin [98]

12 MPH

I DIDNT do the math my brother did  hes in colledge so good lucks guys

8 0
3 years ago
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