Ask question

Ask question

All categories

3 years ago

13

You throw a baseball straight up into the air with a speed of 24.5 m/s. How long does it take the baseball to reach its highest

point?

2 answers:

zzz [600]3 years ago

7 0

Answer:

Like the other person answered, the correct choice is 2.5 s

Explanation:

galben [10]3 years ago

3 0

Time=speed/acceleration
Gravitaional Acceleration=9.8 m/s^2
Speed=24.5 m/s
Time=24.5/9.8=2.5 s

You might be interested in

Compare and contrast what happens to an object's motion when balanced or

mario62 [17]

Answer: The motions shifts.

5 0

2 years ago

If the frequency of a FM wave is 8.85 × 107 hertz, what is the period of the FM wave?

patriot [66]

Answer:

1.1299 x 10^-8 second

Explanation:

Period = 1 / f = 1 / (8.85 * 10^7) = 1.1299 x 01^-8 sec

8 0

2 years ago

A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement

mamaluj [8]

Answer:

sup qwertyasdfghjk

Explanation:

3 0

2 years ago

What can you infer about the density of two objects if they object remains suspended in the liquid?

ZanzabumX [31]

Answer:

Equal Densities

Explanation:

if the density of the object was greater than that of the liquid, it would sink to the bottom. if the density od the object was lesser than the liquid, it would float :)

5 0

3 years ago

Read 2 more answers

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0

3 years ago