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Novosadov [1.4K]
3 years ago
15

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

Physics
1 answer:
sveticcg [70]3 years ago
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
As a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change
Umnica [9.8K]
Refer to the diagram shown below.

m =  the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A =  the amplitude ( the maximum distance) of the mass from the equilibrium
        position

The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω =  the circular frequency of the motion
T =  the period of the motion so that ω = (2π)/T

The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)

In the equilibrium position,
x is zero;
v is maximum;
a is zero.

At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.

In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.

6 0
3 years ago
A box, compressing a spring by 4 m, is loaded and ready. Moments later the 8.0 kg box has been launched with
Mrac [35]

Answer:

50 N/m

Explanation:

Elastic energy = kinetic energy

EE = KE

½ kx² = ½ mv²

½ k (4 m)² = ½ (8.0 kg) (10.0 m/s)²

k = 50 N/m

3 0
3 years ago
When and where did the first nuclear reactor generate electricity
masha68 [24]
<h2>Answer: On December 20th, 1951 in Idaho, United States. </h2>

The world's first experimental nuclear power plant was the Experimental Breeder Reactor Number One (EBR-I), which was built in a desert in Idaho, United States.

This reactor made history when, on December 20th, 1951, four 200-watt light bulbs were illuminated by means of atomic energy, specifically by nuclear fission reaction.

5 0
3 years ago
Yung's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a sc
Lady bird [3.3K]

Answer:

1.082 mm

Explanation:

From the question, we can see that we were given The following

Wavelength of the atoms, λ = 502 nm = 502*10^-9 m

Radius of the screen away from the double slit, r = 1.1 m

We know that Y(20) = 10.2 mm = 10.2*10^-3 m

d = (20 * R * λ) / Y(20)

d = (20 * 1.1 * 502*10^-9)/10.2*10^-3

d = 1.1*10^-5 / 10.2*10^-3

d = 1.082 mm

Therefore, we can say that the distance of separation between the two slits is 1.082 mm

4 0
4 years ago
A residential subdivision encompasses 1100 acres with a housing density of four houses per acre. Assume that a high-value reside
aleksandrvk [35]

Answer:

(a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

Explanation:

Given that,

Area = 1100 acres

Number of house in 1 acres = 4

\text{Number of house in 1100 acres} = 4\times1100

\text{Number of house in 1100 acres} = 4400

Per house water demand = 800 g/day/house

(a). We need to calculate the average daily demand of this subdivision

Using formula for average daily demand

\text{average daily demand}=house\times\text{Per house water demand}

\text{average daily demand}=4400\times800\ gallon/day

\text{average daily demand}=3520000\ gallon/day

\text{average daily demand}=\dfrac{3520000}{24\times60}\ gallon/min

\text{average daily demand}=2444.44\ gallon/min

The average daily demand of this subdivision is 2444.44 gallon/min.

(b). We need to calculate the design-demand used to design the distribution system

Using formula for the design-demand

\text{design demand}=(Q_{max})daily\times\text{fire flow}

\text{design demand}=1.64\times2444.4\times1000

\text{design demand}=4008816\ gallon/m

Hence, (a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

8 0
3 years ago
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