Ask question

Ask question

All categories

3 years ago

8

When a baseball player swings a bat slowly as part of their warm up, that is an example of ____________________________ stretchi

ng.

1 answer:

Artist 52 [7]3 years ago

3 0

Static

<u>Explanation:</u>

The term warm-up in sport is defined as a period of preparatory exercise to enhance subsequent competition or training performance . A pre-game warm-up for team sports typically includes a period of sub maximal running, static stretching of the major muscle groups and sport specific movements incorporating various range of motion (ROM) exercises with skill-based drills executed at, or just below game intensity

You might be interested in

Comparativo entre a máquina a vapor de Neu carmen e James Watt!

Bas_tet [7]

Answer:

jame watt

Explanation:

James Watt tiña máis importancia que outro home

3 0

3 years ago

A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part

Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

<u>Polar moment of Inertia</u>

$(I_p)s = \frac{\pi(0.55)4}2$

= 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

$T_{max} = T_{allow} \frac{I_p}{r}$

= $(14 \times 10^3) \times (\frac{0.14374}{0.55})$

= 3658.836 lb.in

= $\frac{3658.836}{12}$ lb.ft

= 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

$T_{max} = T_{allow}( \frac{Ip}{r})$

= $(14 \times10^3) \times ( \frac{0.13101}{0.55})$

= 3334.8 lb.in

= $(\frac{3334.8}{12} )$ lb.ft

= 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

$P_{max} = 2 \pi f_T$

= $2\pi(2.1) \times 304.9$

= 4023.061 lb. ft/s

= $(\frac{4023.061}{550})$ hp

= 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

$P _{max,t} = 2\pi f_T$

= $2\pi(2.1) \times 277.9$

= 3666.804 lb.ft /s

= $(\frac{3666.804}{550})$ hp

= 6.667 hp

7 0

3 years ago

The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor

uysha [10]

Answer:

c. $a_m$

Explanation:

We are given that

Acceleration due to gravity on the moon= $a_m$

Acceleration due to gravity on the earth= $a_e$

$g_m=\frac{1}{6}g_e$

Net force due to am on an object on moon= $F_{net}=ma_m$

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth= $F=ma_e$

$F=F_{net}$ (given)

$ma_m=ma_e$

$a_e=a_m$

Hence, option c is true.

3 0

2 years ago

Gravitational forces is ------ force

mel-nik [20]

Answer:

Gravitational force is <u>noncontact</u> force

Explanation:

Contact force occurs due to the contact between two different objects. Non-contact force occurs due to either attraction or repulsion between two objects such that there is no contact between these objects. There is no field linked with the contact force. ... Gravitational force is an example of a non-contact force.

7 0

2 years ago

What is the separation distance, in meters, between masses m1 = 15 x 107 kg and m2 = 62 x 107 kg when the gravitational force be

nalin [4]

Answer:

94.1 m

Explanation:

From Coulombs law,

F = Gm1m2/r²................... Equation 1

where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.

Make r the subject of the equation,

r = √(Gm1m2/F)................. Equation 2

Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 2

r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)

r √(886.16×10)

r √(88.616×10²)

r = 9.41×10

r = 94.1 m.

Hence the distance of separation = 94.1 m

3 0

3 years ago