20 pts

there are 5 times as many fiction books in the class library as there are nonfiction books. the number of information books is 7

algol [13]

Answer:

um

Step-by-step explanation:

7 0

3 years ago

Find the volume of the cone. Use 3.14 and round to the nearest cubic meter

Vaselesa [24]

First, you need to πr2h3

5 0

3 years ago

4. What is the range of the graph below

nata0808 [166]

Answer:

R: (Y| E R {-4<=y<=-1})

Step-by-step explanation:

Just look at where the graph is limited to vertically, in this case -1 and -4.

4 0

3 years ago

I need help aklllllllllllllllllfdp[;lgjfg

Katyanochek1 [597]

Answer:

i think its 17

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Q1 A ball is thrown upwards with some initial speed. It goes up to a height of 19.6m and then returns. Find (a) The initial spee

lubasha [3.4K]

Answer:

(a) 19.6 ms⁻¹

(b) 2 s

(c) 9.8 ms⁻¹

(d) 4 s

Step-by-step explanation:

<u>Constant Acceleration Equations (SUVAT)</u>

$\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}$

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Acceleration due to gravity = 9.8 ms⁻².

When the ball reaches its maximum height, its velocity will momentarily be zero.

<u>Given values</u> (taking up as positive):

$s=19.6 \quad v=0 \quad a=-9.8$

$\begin{aligned}\textsf{Using} \quad v^2&=u^2+2as\\\\\textsf{Substitute the given values:}\\0^2&=u^2+2(-9.8)(19.6)\\0&=u^2-384.16\\u^2&=384.16\\u&=\sqrt{384.16}\\\implies u&=19.6\; \sf ms^{-1}\end{aligned}$

Therefore, the initial speed is 19.6 ms⁻¹.

Using the same values as for part (a):

$\begin{aligned}\textsf{Using} \quad s&=vt-\dfrac{1}{2}at^2\\\\\textsf{Substitute the given values:}\\19.6&=0(t)-\dfrac{1}{2}(-9.8)t^2\\19.6&=4.9t^2\\t^2&=\dfrac{19.6}{4.9}\\t^2&=4\\t&=\sqrt{4}\\\implies t&=2\; \sf s\end{aligned}$

Therefore, the time taken to reach the highest point is 2 seconds.

As the ball reaches its maximum height at 2 seconds, one second before this time is 1 s.

<u>Given values</u> (taking up as positive):

$u=19.6 \quad a=-9.8 \quad t=1$

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=19.6+(-9.8)(1)\\v&=19.6-9.8\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

The velocity of the ball one second before it reaches its maximum height is the <u>same</u> as the velocity one second after.

<u>Proof</u>

When the ball reaches its maximum height, its velocity is zero.

Therefore, the values for the downwards journey (from when it reaches its maximum height):

$u=0 \quad a=9.8 \quad t=1$

(acceleration is now positive as we are taking ↓ as positive).

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=0+9.8(1)\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

Therefore, the velocity of the ball one second before <u>and</u> one second after it reaches the maximum height is 9.8 ms⁻¹.

From part (a) we know that the time taken to reach the highest point is 2 seconds. Therefore, the time taken by the ball to travel from the highest point to its original position will also be 2 seconds.

Therefore, the total time taken by the ball to return to its original position after it is thrown upwards is 4 seconds.

4 0

1 year ago