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2 years ago

To improve your cardiovascular fitness:

Physics

2 answers:

RoseWind [281]2 years ago

4 0

C.Perform push ups and curl-ups daily

matrenka [14]2 years ago

4 0

Answer:

A. Include a minimum of 20-30 minutes of aerobic activity 3-4 times per week

Explanation:

Your cardiovascular fitness, also called cardiovascular fitness, says a lot about your health and how much other conditions can affect it. Basically, cardiovascular fitness measures how well your body absorbs oxygen and distributes it to your muscles and organs during prolonged periods of exercise. The best way to improve your cardiovascular fitness is through exercise that forces and stimulates the heart to work harder. Aerobic exercise is the best alternative for this.

For this reason, we can say that the best answer to your question is the letter A.

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A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of

enot [183]

Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}$

In the old transformer the ratio of voltages was:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1$

In the new transformer the ratio of voltages is:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1$

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1$

If the old current output was 425 kV, the ratio of current was:

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{425}= 0.028\\\\I_2=0.028I_1$

Then, the ratio of the new output over the old output is:

$\frac{I_{2new}}{I_{2old}} =\frac{0.024\cdot I_1}{0.028\cdot I_1}= 0.86$

The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).

c) The power loss is expressed as:

$P_L=I^2\cdot R$

Then, the ratio of losses is (R is constant for both power losses):

$\frac{P_n}{P_o} =\frac{I_n^2R}{I_o^2R} =(\frac{I_n}{I_o} )^2=0.86^2=0.74$

The losses in the new line are 74% the losses in the old line.

7 0

2 years ago

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)

You calculate the total work:

$W_T=Fd-F_fd=(F-F_f)d$ (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

$F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N$

Next, you replace the values of all parameters in the equation (4):

$W_T=(130N-105.84N)(5.00m)=120.80J$

$\Delta K=120.80J$

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

$W_T=\frac{1}{2}M(v_2^2-v_1^2)$ (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

$v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}$

The final speed of the box is 2.58m/s

5 0

3 years ago

A boat heads directly across a river. Its speed relative to the water is 3 m/s. It takes it 539 seconds to cross, but it ends up

nikklg [1K]

Answer:

3.24 m/s

Explanation:

Suppose that the boat sails with velocity (relative to water) direction being perpendicular to water stream. Had there been no water flow, it would have ended up 0m downstream

Therefore, the river speed is the one that push the boat 662 m downstream within 539 seconds. We can use this to calculate its magnitude

$v_r = 662 / 539 = 1.23 m/s$

So the boat velocity vector relative to the bank is the sum of of the boat velocity vector relative to the water and the water velocity vector relative to the bank. Since these 2 component vectors are perpendicular to each other, the magnitude of the total vector can be calculated using Pythagorean formula:

$v = \sqrt{v_b^2 + v_r^2} = \sqrt{3^2 + 1.23^2} = \sqrt{9 + 1.5129} = \sqrt{10.5129} = 3.24$ m/s