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Marysya12 [62]
3 years ago
9

Soil conservation involves protecting _______ and preventing _______.

Physics
2 answers:
Naddika [18.5K]3 years ago
7 0
<span>Soil conservation involves protecting soil quality and preventing erosion </span>
navik [9.2K]3 years ago
6 0
Soil conservation involves protecting soil quality and preventing erosion.
Answer: A
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A 912-kg car is being driven down a straight, level road at a constant speed of 31.5 m/s. When the driver sees a police cruiser
sergejj [24]

Answer:

786.6 N

Explanation:

mass of car, m = 912 kg

initial velocity of car, u = 31.5 m/s

final velocity of car, v = 24.6 m/ s

time, t = 8 s

Let a be the acceleration of the car

Use first equation of motion

v = u + a t

24.6 = 31.5 + a x 8

a = - 0.8625 m/s^2

Force, F = mass x acceleration

F = 912 x 0.8625

F = 786.6 N

Thus, the force on the car is 786.6 N.

8 0
2 years ago
A cassette recorder uses a plug-in transformer to convert 120 V to 12.0 V, with a maximum current output of 200 mA. What is the
goldfiish [28.3K]

Answer:

20 mA

Explanation:

We have given that the transformer convert 120 V to 12 V means it is a step down transformer

The turn ratio of the transformer is given as a=\frac{N_1}{N_2}=\frac{V_1}{V_2}=\frac{120}{12}=10 here V_! is primary voltage and V_2 is secondary voltage

As the voltage decreases from primary to secondary so current will increase from primary to secondary

We have given output current that is secondary current so primary that is input current will be less

So input current =\frac{200}{10}=20mA

4 0
3 years ago
What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0
2 years ago
Read 2 more answers
What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a
viktelen [127]

Complete Question

A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.

What is the refrigerator's coefficient of performance? COP

(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP

Answer:

a

 COP = 8.49

b

  COP_1 = 9.49  

Explanation:

From the question we are told that

     The lower operation temperature of refrigerator is  T_1 =  -8.00^oC =  265 \  K

     The upper operation temperature of the refrigerator is   T_2 =  23.2 ^oC =  296.2 \  K

Generally the refrigerators coefficient of performance is mathematically represented as

        COP =  \frac{T_1}{T_2 - T_1  }

=>     COP =  \frac{265}{296.2 - 265  }

=>     COP = 8.49

Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as

            COP_1 =  \frac{T_2}{ T_2 - T_1}  

=>         COP_1 =  \frac{296.2}{ 296.2 - 265 }  

=>         COP_1 = 9.49  

8 0
3 years ago
2. An athlete of average size is hanging from the end of a 20 m long rope, which has a mass of 4 kg and is attached to a hook in
a_sh-v [17]

Answer:

  t = 0.319 s

Explanation:

With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by

             v = √T/λ

Linear density is

           λ = m / L

           λ = 4/20

           λ = 0.2 kg / m

The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg

           T = W = mg

           T = 80 9.8

           T = 784 N

The pulse rate is

          v = √(784 / 0.2)

          v = 62.6 m / s

The time it takes to reach the hook can be searched with kinematics

          v = x / t

          t = x / v

          t = 20 / 62.6

          t = 0.319 s

7 0
3 years ago
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