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2 years ago

13

Preston tossed a red ball upward and it reaches a maximum height of 3.0. What is the final velocity when it returns to prestons

hand?

1 answer:

Leokris [45]2 years ago

5 0

That will depend on the units of the 3.0. We need to know if it's 3 feet, 3 yards, 3 meters, or 3 miles. Each one will have a different answer.

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Caculate the total charge transferred through a wire carrying a current of 0.5 Ameter for a period of 20 seconds.

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1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa

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Answer:

$ans \: = \boxed{{4.8 \times 10}^{ - 4} Hz}$

Explanation:

$given \to \\ f_{r} = 350 \: \\ v_{r} = {3 \times 10}^{8} \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{r} = \frac{v_{r}}{f_{r}} = \frac{3 \times 10^{8} }{350} = \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\ f_{w} = water \: frequency = \: \boxed{ ?}\: \\ v_{w} = 14 50 \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{w} = \frac{v_{w}}{f_{w}} = \frac{1}{100} \times \gamma _{r} = \frac{1}{100} \times 857,142.85714 \\\gamma _{w} = \boxed{8,571.4285714 \: m} : hence \to \: \\ f_{w} = \frac{v_{w}}{ \gamma _{w}} = \frac{1450}{8,571.4285714} = \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) = \frac{f_{w}}{f_{r}} = \frac{0.1691666667}{350} = 0.0004833333 \\ hence \to \\ the \: frequency \: of \: the \: radio \: wave \: is \to \: \boxed{{4.8 \times 10}^{ - 4} }\: \\ that \: of \: the \: wave \: created \: in \: the \: water.$

♨Rage♨

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3 years ago

5. In 1947 Bob Feller, a pitcher for the Cleveland Indians, threw a baseball across the

Darina [25.2K]

The maximum height reached by the ball is 99.2 m

Explanation:

When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration ( $g=9.8 m/s^2$ towards the ground). Therefore, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

In this problem, we have:

u = 44.1 m/s is the initial vertical velocity of the ball

v = 0 is the final velocity when the ball reaches the maximum height

s is the maximum height

$a=-g=-9.8 m/s^2$ is the acceleration of gravity (downward, so negative)

Solving for s, we find the maximum height reached by the ball:

$s=-\frac{u^2}{2a}=-\frac{44.1^2}{2(-9.8)}=99.2 m$

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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3 years ago