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Alexxx [7]
2 years ago
14

According to the exercise principle of balance, a workout should __________.

Physics
2 answers:
xxMikexx [17]2 years ago
6 0
<span>a. address all components of fitness 

</span>According to the exercise principle of balance, a workout should address all components of fitness.
<span>NOT:
</span>b. focus on balance and coordination 
<span>c. target specific areas for improvement </span>
<span>d. vary exercise activities over time</span>

8090 [49]2 years ago
6 0
<span>According to the exercise principle of balance, a workout should:


</span><span>a. address all components of fitness </span><span>

Explanation:


</span>The Balance Principle dictates that each one training should be properly proportioned so as to realize best results. This broad principle operates at several<span> levels of human performance.
All things </span>moderately apply to sports coaching moreover general health and well being
This principle suggests that the proper mixture of coaching activities, diet, and healthy way habits are needed for the best functioning
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Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n
Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

<u>FOR FRIEND ON MOON</u>:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

<u>FOR ME ON EARTH</u>:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

<u>Therefore, the friend on moon will be richer.</u>

7 0
2 years ago
A negative charge of -0.550 μC exerts an upward 0.900-N force on an unknown charge that is located 0.300 m directly below the fi
allochka39001 [22]

Answer:

a.   q2 = 16.4μC, positive charge

b.   F = 0.900N

c.   downward

Explanation:

a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:

F_e=k\frac{q_1q_2}{r^2}            (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between the charges = 0.300m

q1: charge 1 = -0.550 μC = 0.550*10^-6C

q2: charge 2 = ?

Fe: electric force = 0.900N

The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.

You solve the equation (1) for the second charge ans replace the values of the other parameters:

q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C

The values of the second charge is 1.64 μC

b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.

The force exerted on the first charge is 0.900N

c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.

3 0
2 years ago
A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to
daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s                                                                                

Explanation:

It is given that,

Mass of cart 1, m_1=2\ kg

Mass of cart 2, m_2=4\ kg  

Initial speed of cart 1, u_1=3\ m/s          

Initial speed of cart 2, u_2=0 (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}        

V=\dfrac{2\times 3}{(2+4)}

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

J=F\times t=m(V-u)

J=m(V-u)

J=2\times (1-3)    

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0
2 years ago
Hoosier Manufacturing operates a production shop that is designed to have the lowest unit production cost at an output rate of 1
abruzzese [7]

Answer:

90.77%

its capacity utilization rate for the month is 90.77%

Explanation:

The capacity utilisation rate can be expressed mathematically as;

Capacity utilisation rate = capacity used/Best operating level × 100%

Given;

Total Number of production time = 205hours

Production output/capacity used = 21400 units

Best operation rate = 115units/hour

Best operation output for the month of July( at best operation level )

=115units/hour × 205 hours = 23575 units

Capacity utilisation rate = 21400/23575 × 100%

= 90.77%

3 0
2 years ago
An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start
FrozenT [24]

Answer:

  • The total distance traveled is 28 inches.
  • The displacement is 2 inches to the east.

Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector \hat{i} pointing in the west direction, the ant start at position

\vec{r}_0 = 16 \ inch \ \hat{i}

Then, moves to

\vec{r}_1 = 29 \ inch \ \hat{i}

so, the distance traveled here is

d_1 = |\vec{r}_1 - \vec{r}_0  | = | 29 \ inch   \ \hat{i} - 16 \ inch   \ \hat{i}  |

d_1 =  | 13 \ inch   \ \hat{i}  |

d_1 =  13 \ inch

after this, the ant travels to

\vec{r}_2 = 14 \ inch \ \hat{i}

so, the distance traveled here is

d_2 = |\vec{r}_2 - \vec{r}_1  | = | 14 \ inch   \ \hat{i} - 29 \ inch   \ \hat{i}  |

d_2 =  | - 15 \ inch   \ \hat{i}  |

d_2 =  15 \ inch

The total distance traveled will be:

d_1 + d_2 = 13 \ inch + 15 \ inch = 28 \ inch

The displacement is the final position vector minus the initial position vector:

\vec{D}=\vec{r}_2 - \vec{r}_1

\vec{D}= 14 \ inch   \ \hat{i} - 16 \ inch \ \hat{i}

\vec{D}= - 2 \ inch \ \hat{i}

This is 2 inches to the east.

6 0
2 years ago
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