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3 years ago

Provide 2 important pitching tips you can tell someone else. softball

1 answer:

5 0

I know you only need two, but here are some options to choose from.

Tip 1 - throwing a fastball. Use two points to keep your arm properly aligned your biceps brush your ear at the top of the backswing, and your pitching hand brushes your hip at release.

Tip 2 - throwing a curveball. Pictures should know that in order to maximize your curves, you should visualize a series of dots from the mound to the outside corner of the plate. Pitch along those dots.

Tip 3 - throwing a fast pitch rise. It’s possible that your wrist snap may be sideways. Play with different grips or finger pressures and try to relax them.

Tip 4 - throwing a change-up. Don’t always throw the change-up in a given situation. Make sure to change your pitch selection.

Tip 5 - throwing a drop-ball. Keep your pitching arm close to your body to avoid injury.

You might be interested in

A light wave travels through air in equals 1.00 at an angle of 35 degrees what angle does it have when it passes from the air in

lyudmila [28]

Answer: Angle 59 degree

Explanation: Given that the

n1 = 1.0

n2 = 1.5

Øi = 35 degree

From Snell law, which says that

n1/n2 = sinØ1/ sinØ2

Substitute all the parameters into the formula

1/1.5 = sin 35/sinØ2

Cross multiply

Sin Ø2 = 1.5 sin35

SinØ2 = 1.5 × 0.573 = 0.860

Ø2 = sin^-1(0.860)

Ø2 = 59.36 degree

Ø2 = 59 degree ( approximately)

It has angle 59 degree when passing from air to glass

5 0

3 years ago

A block of a plastic material floats in water with 42.9% of its volume under water. What is the density of the block in kg/m3?

adell [148]

To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.

By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore

$\sum F = 0$

$F_b -W = 0$

$F_b = W$

$F_b = mg$

Here,

m = mass

g =Gravitational energy

The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

$\rho_w V_{displaced} g = mg$

Remember the expression for which you can determine the relationship between mass, volume and density, in which

$\rho = \frac{m}{V} \rightarrow m = V\rho$

In this case the density would be that of the object, replacing

$\rho_w V_{displaced} g = V\rho g$

Since the displaced volume of water is 0.429 we will have to

$\rho_w (0.429V) = V \rho$

$0.429\rho_w= \rho$

The density of water under normal conditions is $1000kg / m ^ 3$ , so

$0.429(1000) = \rho$

$\rho = 429kg/m^3$

The density of the object is $429kg / m ^ 3$

7 0

3 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago

PLEASE HELP AND HURRY!!!!!!!!!

never [62]

Answer:

I believe the answer would be A. point x

4 0

4 years ago

You kick a soccer ball across a field the ball travels across the field and slowly comes to a stop. Since the ball slowly comes

marysya [2.9K]

Answer:

It slowly decreases and the friction acting on it slowing it down becomes the bigger net force, if that makes sense :)

Explanation:

6 0

3 years ago