A soccer ball is kicked with an initial velocity 13 m/s at an angle of 35 degrees from the horizontal. What is the maximum height reached by the ball?

Answers

Answer 1

Answer:

The maximum height reached by the ball is 2.84 m

Explanation:

Given;

initial velocity of the soccer, u = 13 m/s

angle of projection, θ = 35°

The maximum height reached by the ball = ?

Apply the following kinematic equation, to determine the maximum height reached by the ball.

Maximum height (H) is given as;

[tex]H = \frac{U^2sin^2\theta}{2g} \\\\H = \frac{(13)^2(sin \ 35^0)^2}{2*9.8}\\\\H = 2.84 \ m[/tex]

Therefore, the maximum height reached by the ball is 2.84 m


Related Questions

Which of the following is a characteristic of a base?
A. pH greater than 7
B. pH between 4 and 13
C. pH between 4 and 9
D. pH less than 7

Answers

Answer:

A. pH greater than 7

Explanation:

Because a base has a pH scale of 8 to 14

An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of the volume is occupied by liquid water and the rest by vapor. Now an electric resistor placed in the tank is turned on, and the tank is observed to contain saturated water vapor after 20 min. Determine:


(a) the volume of the tank

(b) the final temperature

(c) the electric power rating of the resistor

Answers

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

[tex]$m=\frac{V}{v}$[/tex]

[tex]$m=m_f+m_g$[/tex]

[tex]$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$[/tex]

[tex]$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$[/tex]       (taking the value of [tex]$v_g$[/tex] and [tex]$v_g$[/tex] at 200°C  )

[tex]$V=6.304 \times 10^{-3}$[/tex]

Now quality of vapor

[tex]$x=\frac{m_g}{m}$[/tex]

  [tex]$=3.377 \times 10^{-3}$[/tex]

Internal energy at state 1 can be found out by

[tex]$u_1=u_f+xu_{fg}$[/tex]

    [tex]$=850.65+3.377\times10^{-3}\times 1744.65$[/tex]

    = 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor [tex]$v_2=v_g \text { and }\ x=1$[/tex]

Tank is rigid, so volume of tank is constant.

[tex]$v_g=v_2=\frac{V}{m}$[/tex]

[tex]$v_g=\frac{6.304\times 10^{-3}}{1.4}$[/tex]

[tex]$v_g=4.502 \times 10^{-3} \ m^3 /kg$[/tex]

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

[tex]$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$[/tex]

   = 369.11° C

Internal energy at state 2

[tex]$u_2=2154.9 \ kJ/kg$[/tex]

Now power rating of the resistor

[tex]$P=\frac{m(u_2-u_1)}{t}$[/tex]

[tex]$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$[/tex]

  = 1.51 kW

1. Some penguins are playing turtle hockey on the frictionless ice. Ramona the penguin hits the turtle

with a force of magnitude F, and the turtle experiences an acceleration of magnitude a (Fand a are of

unknown size). Francisco the penguin hits the same turtle with a force of magnitude three times

greater than Ramona's force (3F). What will the resulting acceleration of the turtle be? Answer in

terms of a. Support your answer with reasoning and reference to Newton's Second Law.

Answers

Answer:

a' = 3a

Explanation:

Newton's Second Law gives the magnitude of the unbalanced force applied to an object as follows:

F = ma

where,

F = Force Applied

m = Mass of Object

a = acceleration

FOR RAMONA'S PUSH:

F = F

m = m

a = a

Therefore,

F = ma  

a = F/m   ---------------- equation (1)

FOR FRANSICO'S PUSH:

F = 3F

m = m  (mass of turtle will be same)

a = a'

Therefore,

3F = ma'  

a' = 3F/m

using equation 1

a' = 3a

PLS HELP THIS IS DUE IN TWO HOURS

Answers

I think C orrrrrrr A

Answer:

C

Explanation:

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