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3 years ago

14

Choose the true statement from those shown below: A Merchant Account allows you to use SSL on your web site. Disadvantages of us

ing an instant online storefront for your e-commerce site include limited templates and potential limits on the number of products that can be sold. Instant storefronts are what most large-scale E-Commerce sites use. None of the above is true

1 answer:

Slav-nsk [51]3 years ago

7 0

Answer:

The answer is A. that is, a merchant account allows you to use SSL on your website.

Explanation:

SSL means Secure Sockets Layer and this is an encryption-based Internet security protocol.

For an e-commerce merchant website or account, it is advised that an SSL package be installed to prevent any potential loss of private information.

For this reason, a merchant account allows use of SSL on your website because this also boost the confidence of client and customers visiting the website to purchase products.

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Which - type of service shop is least likely to provide service to all

MatroZZZ [7]

E. None of the above

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2 years ago

The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a ve

stellarik [79]

Answer:

Explanation:

Given

velocity at A is $v=4\ ft/s$

For $r=500\ ft$

velocity is increasing at $\dot{v}=0.004t\ ft/s^2$

Tangential acceleration is given by

$a_t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int 0.004tdt=\int dv$

$\int dv=\int 0.004tdt$

$v=0.002t^2+c$

at $t=0\ v=4\ ft/s$

$4=0.002\cdot 0+c$

$c=4\ ft/s$

thus $v=0.002t^2+4$

Velocity in terms of Displacement is given by

$v=\frac{\mathrm{d} s}{\mathrm{d} t}$

$\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt$

$\Rightarrow s=\frac{0.002t^3}{3}+4t$

When car has traveled $\frac{3}{4}$ th of distance i.e.

$s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}$

$s=750\pi$

$750\pi =\frac{0.002t^3}{3}+4t$

$\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0$

on solving we get $t=139.23\ s$

Thus velocity at $t=139.23\ s$

$v=42.76\ s$

(b)Acceleration when car has traveled three-fourth the way of track

normal acceleration $a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}$

$a_n=3.658\ m/s^2$

Tangential acceleration $a_t at t=139.23\ s$

$a_t=0.556\ m/s^2$

Net acceleration $a_t=\sqrt{(a_n)^2+(a_t)^2}$

$a_n=\sqrt{(3.658)^2+(0.556)^2}$

$a_n=3.7\ m/s^2$

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3 years ago

Why do you suppose a value of 5 is used? Do you think other values might work?

Shtirlitz [24]

Answer:

YES

Explanation:

values other than five will work

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2 years ago

A small pad subjected to a shearing force is deformed at the top of the pad 0.12 in. The heigfit of the pad is 1.15 in. What is

ser-zykov [4K]

Answer:

Shearing strain will be 0.1039 radian

Explanation:

We have given change in length $\Delta L=0.12inch$

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Shearing strain is always in radian so we have to change angle in radian

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Kipish [7]

Answer:

tHE answer is b

Explanation:

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3 years ago

Read 2 more answers