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2 years ago

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

2 answers:

6 0

a. address all components of fitness

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

8090 [49]2 years ago

6 0

According to the exercise principle of balance, a workout should:

a. address all components of fitness 

Explanation:

The Balance Principle dictates that each one training should be properly proportioned so as to realize best results. This broad principle operates at several levels of human performance.
All things 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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A charge of 5.67 x 10-18 C is placed 3.5 x 10 m away from another charge of - 3.79 x 10 "C

miskamm [114]

Answer:

1. 579 x 10 ^-22N

Explanation:

F = kq1q2/r^2

= 9.0 x 10^9 x 5.67 x 10^-18 x 3.79 x 10^-18/ (3.5 x 10^-2)^2

= 1. 579 x 10 ^-22N

6 0

2 years ago

A 4500-kg spaceship is in a circular orbit 190 km above the surface of Earth. It needs to be moved into a higher circular orbit

Maslowich

Answer:

Explanation:

Total energy of a satellite in an orbit , h height away

= - GMm /2 ( R + h )

When h = 380 km

Total energy of a satellite = $\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+380)\times10^3}$

= - 13.25 x 10¹⁰ J

When h = 190 km

Total energy of a satellite =

$\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+190)\times10^3}$

= - 13.63 x 10¹⁰ J

Diff

= 38 x 10⁸ J Energy will be required.

8 0

2 years ago

4. A trolley of mass 2kg rests next to a trolley of mass 3 kg on a flat

nydimaria [60]

Answer:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. The total momentum of the trolleys after separation is zero

c. The momentum of the 2 kg trolley after separation is 12 kg·m/s

d. The momentum of the 3 kg trolley is -12 kg·m/s

e. The velocity of the 3 kg trolley = -4 m/s

Explanation:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0

c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s

d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s

e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley

∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s

The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s

3 0

2 years ago

B. A hydraulic jack has a ram of 20 cm diameter and a plunger of 3 cm diameter. It is used for lifting a weight of 3 tons. Find

lozanna [386]

Answer:

option (b)

Explanation:

According to the Pascal's law

F / A = f / a

Where, F is the force on ram, A be the area of ram, f be the force on plunger and a be the area of plunger.

Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm

A = π R^2 = π x 100 cm^2

F = 3 tons = 3000 kgf

diameter of plunger, d = 3 cm, r = 1.5 cm

a = π x 2.25 cm^2

Use Pascal's law

3000 / π x 100 = f / π x 2.25

f = 67.5 Kgf

4 0

1 year ago

Official (Closed) - Non Sensitive

Pavlova-9 [17]

Answer:

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi)/a

t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m)/(29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

t = 319.47 s

5 0

2 years ago