SOMONE PLEASE HELP ASAP!!!!

vlabodo [156]

X men those mutants are amazing

6 0

2 years ago

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s (~ 18 mph). What is the total time it is i

evablogger [386]

Answer:

The total time it is in the air for the ball is 1.6326 s

Given:

Initial velocity = 8 $\frac{m}{s}$

To find:

the total time it is in the air = ?

Formula used:

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

Solution:

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.

The time taken by the ball to reach the maximum height is given by,

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

t = $\frac{0-8}{-9.8}$

t = 0.8163 s

Thus, time taken by the ball to reach the ground again = time taken to reach maximum height

So, Total time required for ball to reach ground = 2t = 2 × 0.8163

Total time required for ball to reach ground = 1.6326 s

The total time it is in the air for the ball is 1.6326 s

4 0

3 years ago

The brain mass of a human fetus during a particular trimester can be accurately estimated from the circumference of the head by

Paladinen [302]

Answer:

a. If c = 20 cm, then the mass of the brain is m = 5 g.

b. At c = 20 cm, the brain's mass is increasing at a rate of 15.75 g/cm.

Explanation:

From the equation

$m\left(c\right) = \frac{c^3}{100}-\frac{1500}{c}$

we have

a. for c = 20 cm

$m\left(20\right)=\frac{20^3}{100}-\frac{1500}{20}=5,$

then the mass is m(20) = 5 g.

b. In order to find the rate of change, first we derivate

$\frac{dm}{dc}=\frac{3c^2}{100}+\frac{1500}{c^2}.$

Evaluated at c = 20 cm, we have

$\frac{dm}{dc}|_{c=20}=\frac{3\times 20^2}{100}+\frac{1500}{20^2}=15.75.$

So, at c = 20 cm, the mass of the brain is increasing at a rate of 15.75 g/cm.

3 0

3 years ago

6. A 2-kg ball B is traveling around in a circle of radius r1 = 1 m with a speed (vB)1 = 2 m/s. If the attached cord is pulled d

kipiarov [429]

Answer:

Explanation:

Given that,

Mass of ball m = 2kg

Ball traveling a radius of r1= 1m.

Speed of ball is Vb = 2m/s

Attached cord pulled down at a speed of Vr = 0.5m/s

Final speed V = 4m/s

Let find the transverse component of the final speed using

V² = Vr²+ Vθ²

4² = 0.5² + Vθ²

Vθ² = 4²—0.5²

Vθ² = 15.75

Vθ =√15.75

Vθ = 3.97 m/s.

Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

2 × 1 = 3.97 × r2

r2 = 2/3.97

r2 = 0.504m

The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

Then,

t = ∆r/Vr

t = (r1—r2) / Vr

t = (1—0.504) / 0.5

t = 0.496/0.5

t = 0.992 second

7 0

3 years ago

Six friends are at a pizzeria. They want to order enough pizza so that each person can eat at least 2/5 . start fraction, 2, d

Lyrx [107]

Friend #1 gets at least 2/5 of a pizza.

Friend #2 gets at least 2/5 .

Friend #3 gets at least 2/5 .

Friend #4 gets at least 2/5 .

Friend #5 gets at least 2/5 .

Friend #6 gets at least 2/5 .

Sum . . . . . . . . . at least 12/5 of a pizza.

Simplify . . . . . . at least 2.4 pizzas.

-- If pizzas can be bought by the half, they should order at least <em>2-1/2 pizzas.</em>

-- If only whole pizzas have to be ordered, then they should order at least <em>3 pizzas.</em>

4 0

3 years ago