Exercise and health

An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one

aalyn [17]

Answer:

20.78 m/s that we can approximate to option d (21 m/s)

Explanation:

The solution involves a lot of algebra and to be familiar with different convenient formulas for launching an object vertically under the action of gravity.

First you need to recall (or derive) the formula for the maximum height reached by an object with launches with initial velocity $v_0$ :

Maximum height = $\frac{{v_0}^2}{2g}$

Therefore one fourth of such height would be: $\frac{{v_0}^2}{8g}$

Second, find what would be the time needed to reach that height by solving for the time in the equation for the vertical position:

$y(t)=v_0*t-\frac{g}{2} t^2 \\\frac{{v_0}^2}{8g} = v_0*t-\frac{g}{2} t^2\\\frac{g}{2} t^2-v_0*t+\frac{{v_0}^2}{8g}=0$

And now, solve for t in the last equation using the quadratic formula to find the time needed for the object to reach that height (one fourth of the max height):

$t=\frac{v_0+/-\sqrt{{v_o}^2-4*\frac{g}{2}*\frac{{v_o}^2}{8g} } }{g} = \frac{v_0+/-\sqrt{{v_o}^2-\frac{{v_o}^2}{4} } }{g} =\frac{v_0+/-\sqrt{\frac{3{v_o}^2}{4} } }{g} = \\=\frac{v_0+/-v_0\sqrt{\frac{3}{4} } }{g} =\frac{v_0+/-v_0\frac{\sqrt{3}}{2} } {g}$

Next, use this expression for t in the equation for the velocity at any time t in the object's trajectory that comes from the definition of acceleration;

$v(t)=v_0-g*t$

Then for the time we just found, this new equation becomes:

$v=v_0-g(\frac{v_0+/-v_0\frac{\sqrt{3} }{2}}{g}) =v_0-v_0+/- v_0\frac{\sqrt{3} }{2} = +/- v_0 \frac{\sqrt{3} }{2}$

Now, using that the velocity at this height is 18 m/s, and solving for the unknown velocity $v_0$ , we get:

$v_0=\frac{18*2}{\sqrt{3} } =\frac{36}{\sqrt{3} }= 20.78\frac{m}{s}$

7 0

3 years ago

Using what you know about the periodic table, choose all of the correct answers,

Klio2033 [76]

Answer:

What are the choices?

7 0

3 years ago

A baseball is hit almost straight up into the air with a speed of 26 m/s . Estimate how high it goes.

Natalka [10]

Answer:

The maximum height of the ball is 34.5 m.

The ball is 5.31 s in the air.

Explanation:

Hi there!

The equations for the height and velocity of the baseball that is hit straight up are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the baseball at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t.

If we place the origin of the frame of reference at the place where the baseball is hit, then, y0 = 0.

To calculate how high it goes, we have to obtain the time at which the ball is at maximum height. At that point, the velocity is 0. Then using the equation of velocity:

v = v0 + g · t

0 = 26 m/s - 9.8 m/s² · t

-26 m/s / -9.8 m/s² = t

t = 2.65 s

The height at that time will be the maximum height:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

y = 26 m/s · 2.65 s - 1/2 · 9.8 m/s² · (2.65 s)²

y = 34.5 m

The maximum height of the ball is 34.5 m

If it takes the ball 2.65 s to reach the maximum height it will take another 2.65 s to return to the initial position. Then, the time it will be in the air is (2.65 s + 2.65 s) 5.30 s. However, let´s calculate the time it takes the ball to reach the initial position using the equation for height.

At the initial position y = 0. Then:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

0 = 26 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (26 m/s - 1/2 · 9.8 m/s² · t) (t = 0 when the ball is hit)

0 = 26 m/s - 1/2 · 9.8 m/s² · t

-26 / -4.9 m/s² = t

t = 5.31 s ( the difference with the 5.30 s obtained above is due to rounding the time to 2.65 s).

The ball is 5.31 s in the air.

Have a nice day!

5 0

3 years ago

If the magnitude of the electric field in air exceeds roughly 3 ✕ 106 N/C, the air breaks down and a spark forms. For a two-disk

Westkost [7]

Answer:

1.843 x 10^-5 C

Explanation:

Givens:

It is given that the air starts ionizing when the electric field in the air exceeds a magnitude of 3 x 10^6 N/C, which means that the max electric field can stand without forming a spark is 3 x 10^6 N/C.

Also it is given that the radius of the disk is 50 cm, it is required to find out the max amount of charge that the disk can hold without forming spark, which means the charge that would produce the max magnitude of the electric field that air can stand without forming spark, and since we know that the electric field in between 2 disk "Capacitor" is given by the following equation

E = (Q/A)/∈o (1)

Where,

Q: total charge on the disk.

A: the area of the disk.

Calculations:

We want to find the quantity of charge on the disk that would produce an electric field of 3 x 10^6 N/C, knowing the radius of the disk we can find the cross-section of the disk, thus substituting in equation (1) we find the maximum quantity of charge the disk can hold

Q = EA∈o

= (3 x 10^6) x (π*0.50) x (8.85 x 10^-12)

= 1.843 x 10^-5 C

note:

calculations maybe wrong but method is correct

8 0

3 years ago

A cheetah can accelerate from rest at the rate of 4m /s

lesya [120]

Acceleration of cheetah (a) = 4m/s²
time = 10s
initial velocity(u) = 0
final velocity = v
distance travelled = s

v = u +at = 0 + 10×4 = 40m/s
s = (v²-u²)/2a = 40²/(2×4) = 1600/8 = 200m

6 0

3 years ago