Eddie and Val observed the picture of an athlete running in a race.

A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught

meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (Eq. 1)

Where:

$y_{o}$ - Initial height of the softball, measured in meters.

$y$ - Final height of the softball, measured in meters.

$v_{o}$ - Initial velocity of the softball, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $y = y_{o}$ , $t = 3.56\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the initial velocity of the softball is:

$v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0$

$3\cdot v_{o} -44.132\,m= 0$

$v_{o} = 14.711\,\frac{m}{s}$

The initial velocity of the softball is 14.711 meters per second.

8 0

3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0

3 years ago

Please help, my homework is due in a couple of days and I'm stuck. A plane is travelling 10.0 meters per second (this means the

ss7ja [257]

Answer:

< 25 m/s
triangle inequality
between north and east
45° < angle < 60°

Explanation:

(a) Just as one-dimensional numbers add on a number line by putting them end-to-end, so two-dimensional numbers add on a coordinate plane the same way.

Here, we choose to let the positive y-axis represent North, and the positive x-axis, East. This is the way a map is conventionally oriented. The velocity of the plane is represented by a vector pointing north (up). Its length represents the magnitude of the velocity. Likewise, the wind is represented by a vector of length 15 pointing east (right). The sum of these is the hypotenuse of the triangle they form.

The magnitude of the sum can be found here using the Pythagorean theorem, but for the purpose of this question, you're not asked to find that.

Instead, you're asked to estimate whether it is more or less than 25 (m/s).

Your knowledge of the triangle inequality will tell you that the hypotenuse (resultant) must be shorter than the sum of the lengths of the sides of the triangle, hence must be less than 10+15 = 25.

__

(b) The triangle inequality says the resultant is less than the sum of the other two sides of the triangle.

__

(c) Since the wind is blowing the plane toward the east, but the plane is traveling toward the north, the resulting direction is somewhere between north and east.

__

(d) "Somewhere between north and east" can be expressed as the inequality ...

0° < angle < 90°

4 0

3 years ago

Why is temperature a good criterion for searching for Earthlike exoplanets?

grandymaker [24]

Because it gives us the abilitity to find planets that have decent temperatures relative to earths temp so we can determine if the planet even has a possiblilty to sustain life hope this helps

4 0

3 years ago

Read 2 more answers

Need help I'll give you 30 points please.

Ksivusya [100]

Cathode Ray Tube i think?

7 0

3 years ago