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posledela
3 years ago
9

Your performance on the pacer is used to measure which components of health- related fitness?

Physics
1 answer:
77julia77 [94]3 years ago
8 0
Your performance on the pacer test is used to measure cardiovascular components of health-related fitness
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Help pls!
REY [17]

Answer:

Geothermal power can provide consistent electricity throughout the day and year - continuous baseload power and flexible power to support the needs of variable renewable energy resources, such as wind and solar. Sustainable Investment.

Explanation:

THIS IS WHY WE SHOULD USE GEOTHERMAL ENERGY IN FUTURE

YOU CAN MARK ME AS BRAINIEST IF YOU WANT

3 0
3 years ago
Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electri
Vilka [71]

Answer:

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

Explanation:

For this exercise let's use the electric field expression

         E = k q / r²

where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee

let's calculate the field for each charge

 

Q = 24 pC = 24 10⁻¹² C

         E₁ = 9 10⁹ 24 10⁻¹² / 0.20²

         E₁ = 5.4 N / C

Q = 32 pC = 32 10⁻¹² C

         E₂ = 9 10⁹ 32 10⁻¹² / 0.2²

         E₂ = 7.2 N / C

let's find the difference between these two fields

         ΔE = E₂ -E₁

         ΔE = 7.2 - 5.4

         ΔE = 1.8 N / C

the minimum detection field is

         E_minimum = 0.77 N / C

        ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

8 0
3 years ago
What is the motion of the object?
aleksley [76]

Answer:

<em>Thus, the object is accelerating to the left</em>

Explanation:

<u>The Net Force</u>

The net force is the result of adding all the forces as vectors acting on a body.

\vec F=\vec F_1+\vec F_2+...+\vec F_n

Each vector can be expressed in its rectangular components Fx and Fy, and the sum is the sum of the rectangular components separately.

Second Newton's law gives the relation between the net force and the acceleration of the body:

\vec F = m.\vec a

We can see the acceleration is a vector with the same direction as the net force.

The diagram shows two vertical forces and two horizontal forces.

The vertical forces are acting in opposite directions and with the same magnitude, thus they cancel out, leaving zero net force in the y-axis.

The horizontal forces are opposite and with different magnitudes. Since the force acting to the left (F3) has a greater magnitude than the force acting to the right (F4), there is a net force directed to the left with a magnitude of 60 N - 20 N = 40 N

Thus, the object is accelerating to the left

4 0
3 years ago
What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i
babymother [125]

Answer:

\Delta S=1.69J/K

Explanation:

We know,

\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}      ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

0.28=1-\frac{Q_2}{3.78\times 10^{3}}

or

\frac{Q_2}{3.78\times 10^3}=0.72

or

Q_2=3.78\times 10^3\times0.72

⇒ Q_2 =2.721\times 10^3 J

Now,

The entropy change (\Delta S) is given as:

\Delta S=\frac{\Delta Q}{T_1}

or

\Delta S=\frac{Q_1-Q_2}{T_1}

substituting the values in the above equation we get

\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}

\Delta S=1.69J/K

7 0
3 years ago
Technician A says that the cooling system should be tested for leaks using a pressure-operated pressure pump. Technician B says
mart [117]

Answer:

The correct option is;

Technician A only

Explanation:

Leaks in the cooling system parts such as the radiator cap can be tested with a pressure operated hand pump by attaching the pump cap to the pump with an adaptor and apply pressure enough for there to be a release of pressure at the cap.

The pressure added to the system should be maintained for up to two minutes. If the pressure drops quicker, then there is likely to be leaks in the system

The freezing and boiling point of the coolant is measured with an antifreeze tester which carries the appropriate apparatus to measure the coolants freezing and boiling points.

3 0
2 years ago
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