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3 years ago

If I fall for 18 s what would my final velocity be, if I started from rest.

Mathematics

1 answer:

ozzi3 years ago

8 0

Answer:

176.58 m/s

Step-by-step explanation:

assume that the acceleration due to gravity, g = 9.81 m/s²

I also assume we neglect air resistance (hence we assume no terminal velocity)

recall that one of the equations of motions can be expressed as

v = u + at, where

u = initial velocity = 0 (because it says you started from rest)

v = final velocity (we are asked to find this)

a = acceleration due to gravity = 9.81 m/s²

t = time = given as 18s

simply substitute the above into the equation.

v = 0 + (9.81)(18)

v = 176.58 m/s

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Suppose you are taking the bus home, and can take either Route 51 or Route 82. You know that the Route 82 will arrive at the bus

kykrilka [37]

Answer:

E[T] = 10

Step-by-step explanation:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distribution has two bounds, a and b, and the expected value of the uniform distribution is given by:

$E = \frac{a+b}{2}$

Uniformly distributed between 0 and 20 minutes.

This means that $a = 0, b = 20$

Let T be the number of minutes you wait until you board a bus. Find E[T].

This is

$E[T] = \frac{a + b}{2} = \frac{0 + 20}{2} = 10$

E[T] = 10

8 0

2 years ago

I need help ASAP?!!!!

Roman55 [17]

Answer:

B. 1 3/8

The second one

Step-by-step explanation:

4 1/8 ÷ 3 = 1.375 = 1 3/8

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6 0

2 years ago

Answer. it fast please

mojhsa [17]

Answer:

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{15}$

Step-by-step explanation:

Given

$(\frac{6}{10})^3 * (\frac{5}{9})^2$

Required

Solve:

$(\frac{6}{10})^3 * (\frac{5}{9})^2$

Simplify 6/10

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = (\frac{3}{5})^3 * (\frac{5}{9})^2$

Express 9 as 3^2

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = (\frac{3}{5})^3 * (\frac{5}{3^2})^2$

Apply law of indices

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3}{5^3}* \frac{5^2}{(3^2)^2}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3}{5^3}* \frac{5^2}{3^4}$

Express as a sing;e fraction

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{3^3*5^2}{5^3*3^4}$

Apply law of indices:

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5^{3-2}*3^{4-3}}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5^1*3^1}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{5*3}$

$(\frac{6}{10})^3 * (\frac{5}{9})^2 = \frac{1}{15}$

5 0

3 years ago

Help would be much appreciated pic of question

Westkost [7]

Answer:

see attached

Step-by-step explanation:

The graph is a series of horizontal lines. A solid dot goes wherever there is an "or equal to" symbol in the inequality. An open dot goes where the point is not included in the function definition (but nearby points are).

7 0

3 years ago

Which of the following fractions is equivalent to 10/12? 40/48 20/30 4/5 12/14

Sergeeva-Olga [200]

Answer:

40/48

Step-by-step explanation:

10*4=40

12*4=48

6 0

3 years ago

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