Which of the following are binary ionic compounds?

Determine the kinetic energy of the ball immediately after it is hit. (You must provide an answer before moving to the next part

earnstyle [38]

The question is incomplete. Here is the complete question.

A baseball palyer hits a 5.1 oz baseball with an initial velocity of 140ft/sat an angle of 40° with the horizontal as shown. Determine

a) The kinetic energy of the ball immediately after it is hit

b) The kinetic energy of the ball when it reaches its maximum height

c) The maximum height above the ground reached by the ball.

Answer: a) KE = 131.64 J

b) KE = 0

c) h = 126 ft

Explanation: Kinetic energy is the energy an object posses due to its motion. It can be calculated as $KE=\frac{1}{2}mv^{2}$

a) Kinetic energy's unit is Joule. So, we have to transform ounce in kg and ft/s in m/s for the units to correspond:

m = 5.1(0.02835)

m = 0.1445 kg

v = 140 ft = 42.67 m/s

Then, kinetic energy is

$KE=\frac{1}{2}(0.1445)(42.67)^{2}$

KE = 131.64 J

Kinetic energy immediately after the ball is hit is 131.64 J.

b) At its maximum height, the ball has its highest potential energy. Because of the law of conservation of energy, when potential energy is maximum, kinetic energy is minimum and vice-versa. So, at the maximum height, kinetic energy is 0.

c) This type of motion is projectile motion. The maximum height on a projectile motion can be determined by

$v_{y}^{2}=v_{0y}^{2}-2g\Delta y$

When h is maximum, $v_{y}=0$

Velocity of the ball has an angle with the horizontal, so initial velocity at the y-axis is

$v_{0y}=v_{0}sin(\theta)$

Substituting and solving

$v_{y}^{2}=v_{0}^{2}sin^{2}(\theta)-2gh$

$0=(42.67)^{2}sin^{2}(40)-2(9.8)h$

$19.6h=(42.67)^{2}(0.643)^{2}$

$h=\frac{(1820.73)(0.4132)}{19.6}$

h = 38.4 m

Transforming into ft: h = 126 ft

The maximum height above the ground reached by hte ball is 126 feet.

6 0

3 years ago

A shopping cart with two nice kids, rolls off a horizontal roof ledge that is 50

Alisiya [41]

Answer:

9.39 m/s

Explanation:

Using the y-direction, we can solve for the time t it takes for the cart to reach the ground.

Assume the up direction is positive and the down direction is negative.

v₀ = 0 m/s
a = -9.8 m/s²
Δy = -50 m
t = ?

Find the constant acceleration equation that contains these four variables.

Δy = v₀t + 1/2at²

Substitute known values into this equation.

-50 = (0)t + 1/2(-9.8)t²

Multiply and simplify.

-50 = -4.9t²

Divide both sides of the equation by -4.9.

10.20408163 = t²

Square root both sides of the equation.

t = 3.194382825

Now we can use this time t and solve for v₀ in the x-direction. Time is most often our link between vertical and horizontal components of projectile motion.

List out known variables in the x-direction.

v₀ = ?
t = 3.194382825 s
a = 0 m/s²
Δx = 30 m

Find the constant acceleration equation that contains these four variables.

Δx = v₀t + 1/2at²

Substitute known values into the equation.

30 = (v₀ · 3.194382825) + 1/2(0)(3.194382825)²

Multiply and simplify.

30 = v₀ · 3.194382825

Divide both sides of the equation by 3.194382825.

v₀ = 9.391485505

The cart was rolling at a velocity of 9.39 m/s (initial velocity) when it left the ledge.

8 0

3 years ago

Read 2 more answers

7. At room temperature, a substance with a melting

Lorico [155]

Answer:

The answer is Solid

Explanation:

Hope this helps!

5 0

3 years ago

A dog can hear sounds in the range from 15 to 50,000 hz. what wavelength corresponds to the lower cut-off point of the sounds at

Ira Lisetskai [31]

The frequency corresponding to the lower cut-off point is
$f=15 Hz$
The speed of sound at a temperature of $20^{\circ}C$ is approximately
$v=343 m/s$

Therefore we can use the relationship between speed of a wave, frequency and wavelength to calculate the corresponding wavelength:
$\lambda= \frac{v}{f}= \frac{343 m/s}{15 Hz}=22.87 m$
So, the corresponding wavelength is 22.87 m.

3 0

3 years ago

A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0

viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
$KE = \frac{1}{2}mv^2 \\\\PE = mgh$

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

$E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)$

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

$E_f = \frac{1}{2}mv_f^2$

And:
$W_A = E_i - E_f$

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
$W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) - \frac{1}{2}mv_f^2$

Solving for the work done by air resistance:
$W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) - \frac{1}{2}(.450)(19.89^2)$

$W_A = \boxed{42.552 J}$

8 0

2 years ago