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3 years ago

Select the correct answer from each drop down menu.

1 answer:

3 0

Hooke's Law is a principle of physics that states that the that the force needed to extend or compress a spring by some distance is proportional to that distance.

volume, plastic behavior .

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Which of the following describes beta testing?

Sergio039 [100]

Answer:

What is Beta Testing? Beta testing is an opportunity for real users to use a product in a production environment to uncover any bugs or issues before a general release. Beta testing is the final round of testing before releasing a product to a wide audience.

3 0

3 years ago

!!!!!!!!!!!!!!PLEASE ANSWER THIS!!!!!!!!!!!!!!!!

dmitriy555 [2]

Answer:

age = int(input())

ticket = 100

total = 0

passengers = 1

while passengers <= 5:

if age < 3:

continue

else:

total += ticket

passengers += 1

print(total)

Explanation:

UuU

6 0

2 years ago

Given : f(x) = x³- 7x²+ 36 Draw the graph of f neatly on F graph paper. Clearly indicate an Intercepts and coordinates of turnin

vichka [17]

Answer:

Explanation:

xx=33 vvalues

7 0

3 years ago

A train starts from rest at station A and accelerates at 0.4 m/s^2 for 60 s. Afterwards it travels with a constant velocity for

mash [69]

<h3><u>The distance between the two stations is</u><u> </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

$\\$

Explanation:

<h2>Given:</h2>

$a_1 \:=\:0.4\:m/s²$

$t_1 \:=\:60\:s$

$v_{i1} \:=\:0\:m/s$

$a_2 \:=\:0\:m/s²$

$t_2 \:=\:25\:min\:=\:1500\:s$

$a_3 \:=\:-0.8\:m/s²$

$v_{f3} \:=\:0\:m/s$

$\\$

<h2>Required:</h2>

Distance from Station A to Station B

$\\$

<h2>Equation:</h2>

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v\:=\:\frac{d}{t}$

$\\$

<h2>Solution:</h2><h3>Distance when a = 0.4 m/s²</h3>

Solve for $v_{f1}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$0.4\:m/s²\:=\:\frac{v_f\:-\:0\:m/s}{60\:s}$

$24\:m/s\:=\:v_f\:-\:0\:m/s$

$v_f\:=\:24\:m/s$

$\\$

Solve for $v_{ave1}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{0\:m/s\:+\:24\:m/s}{2}$

$v_{ave}\:=\:12\:m/s$

$\\$

Solve for $d_1$

$v\:=\:\frac{d}{t}$

$12\:m/s\:=\:\frac{d}{60\:s}$

$720\:m\:=\:d$

$d_1\:=\:720\:m$

$\\$

<h3>Distance when a = 0 m/s²</h3>

$v_{f1}\:=\:v_{i2}$

$v_{i2}\:=\:24\:m/s$

$\\$

Solve for $v_{f2}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$0\:m/s²\:=\:\frac{v_f\:-\:24\:m/s}{1500\:s}$

$0\:=\:v_f\:-\:24\:m/s$

$v_f\:=\:24\:m/s$

$\\$

Solve for $v_{ave2}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{24\:m/s\:+\:24\:m/s}{2}$

$v_{ave}\:=\:24\:m/s$

$\\$

Solve for $d_2$

$v\:=\:\frac{d}{t}$

$24\:m/s\:=\:\frac{d}{1500\:s}$

$36,000\:m\:=\:d$

$d_2\:=\:36,000\:m$

$\\$

<h3>Distance when a = -0.8 m/s²</h3>

$v_{f2}\:=\:v_{i3}$

$v_{i3}\:=\:24\:m/s$

$\\$

Solve for $v_{f3}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$-0.8\:m/s²\:=\:\frac{0\:-\:24\:m/s}{t}$

$(t)(-0.8\:m/s²)\:=\:-24\:m/s$

$t\:=\:\frac{-24\:m/s}{-0.8\:m/s²}$

$t\:=\:30\:s$

$\\$

Solve for $v_{ave3}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{24\:m/s\:+\:0\:m/s}{2}$

$v_{ave}\:=\:12\:m/s$

$\\$

Solve for $d_3$

$v\:=\:\frac{d}{t}$

$12\:m/s\:=\:\frac{d}{30\:s}$

$360\:m\:=\:d$

$d_3\:=\:360\:m$

$\\$

<h3>Total Distance from Station A to Station B</h3>

$d\:= \:d_1\:+\:d_2\:+\:d_3$

$d\:= \:720\:m\:+\:36,000\:m\:+\:360\:m$

$d\:= \:37,080\:m$

$d\:= \:37.08\:km$

$\\$

<h2>Final Answer:</h2><h3><u>The distance between the two stations is </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

7 0

3 years ago

an object of mass 2kg is released from a top of inclined plane 30° and height 6m. The coefficient of kinetic friction of the sur

mel-nik [20]

Explanation:

1) Work done = force x distance x cos(θ)

= 0.15 x 6 x cos(30)

= 0.779

2) Ek = ½mv²

v = acceleration due to gravity so 9.81

Ek = ½(2)(9.81)²

Ek = 96.2361

3) v = (√(2em)) / m

= (√(2(96.2361)(2)) / 2

= 9.81 so especially with no time given, I can only assume the acceleration due to gravity but take it with a pinch of salt.

5 0

2 years ago