D(3, 6), E(-4,-3), F(6, 1) and<br> D'(1, 6), E'(8, -3), F(-2, 1)

A ballplayer catches a ball 3.4s after throwing it vertically upward. with what speed did he throw it, and what height did it re

Daniel [21]

From the equation of motion, we know,

$s=ut+\frac{1}{2}at^{2}$

Where s= displacement

u= initial velocity

a= gravitational force

t= time

Displacement is 0 since the ball comes back to the same point from where it was thrown.

A = $-9.8m/s^{2}$ since the ball is thrown upwards.

Plug the known values into the equation.

=> $0=u\left ( 3.4 \right )+\frac{1}{2}\left ( -9.81 \right )(3.4^{2})$

Solving for u gives :

u= 16.67 m/ sec ....... equation (1)

At maximum height, final velocity i.e v is 0

Time take to reach the top = $\frac{3.4}{2} = 1.7 sec$

$v^{2}=u^{2}+2as$

=> $0=(16.67)^{2}+2(-9.81)(s)$

Solving for s we get

s= 14.16 m

4 0

2 years ago

Can soemone help meee ):

bogdanovich [222]

I don't know the answer but my tip would be to split the shape the way I did it, find the lengths of each side, and then find the area for each individual shape.

Then add all the areas together.

6 0

2 years ago

Someone please help meee

slamgirl [31]

Use the formula:
Y-yo=m(x-xo)

8 0

2 years ago

A friend asks to borrow $250 for 100 days anD aGREES tO pAY 36.5%

aksik [14]

95% percent that friend is not going to pay you back

7 0

2 years ago

A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and

nasty-shy [4]

Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

so

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

so

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

3 0

3 years ago

D(3, 6), E(-4,-3), F(6, 1) and D'(1, 6), E'(8, -3), F(-2, 1)