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Aleonysh [2.5K]
3 years ago
5

Which of these is NOT a possible type of energy transformation?

Physics
1 answer:
Orlov [11]3 years ago
6 0

to be franc i really think the answer is B

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Define pressure and enlist its various units
const2013 [10]

Answer:

pressure is an affect which occurs when a force is applied on a surface area

Explanation:

its unit are barr,millbarr,torr,mmhg

5 0
2 years ago
If a machine has an efficiency of 94% and you apply 574J of work, how much work do you get out of the machine
Mariana [72]
<h3>Answer:</h3>

539.56 Joules

<h3>Explanation:</h3>
  • Efficiency of a machine is the ratio of work output to work input expressed as a percentage.
  • Efficiency = (work output/work input) × 100%
  • Efficiency of a machine is not 100% because so energy is lost due to friction of the moving parts and also as heat.

In this case;

Efficiency = 94%

Work input = 574 Joules

Therefore, Assuming work output is x

94% = (x/574 J) × 100%

0.94 = (x/574 J)

<h3>x = 539.56 J</h3>

Thus, you get work of 539.56 J from the machine

7 0
3 years ago
The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c
maria [59]

To solve this problem we will start using the concepts related to the electric field, from there we will find the load exerted on the body. Through this load it will be possible to make a sum of forces in balance to find the load that a human supports. Finally with these values it will be possible to find the repulsive force. We will proceed as follows,

The electric field is

E= \frac{kQ}{R^2}

Here,

k = Coulomb's Constant

Q = Charge

R = Distance (At this case from the center of mass of the earth to the surface)

Rearranging to find the charge,

Q = \frac{ER^2}{k}

Replacing,

Q = frac{(150)(6.38*10^6)}{8.99*10^9}

Q = 6.79*10^5 C

Since the electric field is directed towards the center of earth, the charge is negative.

PART A) Once the load is found we can proceed to apply the balance of Forces, for which the electrostatic force must be equivalent to the weight, this in order to satisfy the balance, therefore

F_w = F_e

mg = \frac{kQq}{R^2}

Replacing,

(62)(9.8) = \frac{(8.99*10^9)(q)(-6.79*10^5)}{(6.38*10^6)^2}

Solving for q,

q = -4.056C

PART B) Finally using the given distance and the values of the found load we can find the repulsive Force, which is

F =\frac{kq^2}{d^2}

F = \frac{(8.99*10^9)(-4.056)^2}{110^2}

F = 1.22*10^7N

PART C) The answer is no. According to the information found, we can conclude that traveling through an electric field is not viable because there is a repulsive force of great magnitude acting on the body.

3 0
3 years ago
How much water
nevsk [136]

Answer:

The mass is  m  =  3.75 \  kg

Explanation:

From the question we are told that

     The initial temperature is  T_1 =  15^oC

      The final  temperature is  T_2 =  100 ^o C  \ (boiling point )

Generally the maximum heat produced by  1  Liter   of  natural gas is  9000 \ cal

 So the amount of heat produced by 100 L is  

              E = 9000 * 100

=>           E = 9000 00 \  cal

Generally given that the efficiency is  \eta =  0.35

Then actual heat received by the water is

         H  =  0.35 *  E

=> H  =  0.35 *   9000 00

=> H  =  315000 \  cal

Converting to kcal  

=> H  =  315000 \  cal = \frac{315000}{1000} =  315 \ kcal

Generally the specific heat of water is  

       c_w  =  1 kcal/ kg \cdot ^oC

Generally the heat received by the water is mathematically represented as

       H  =  m *  c_w  *  (T_2 - T_1)

=>     315  =  m *  1   *  ( 100  - 15 )

=>     m  =  3.75 \  kg

     

6 0
3 years ago
During take-off a 5kg model rocket, initially at rest, burns fuel for 4.6s causing its speed to increase from rest to 39m/s duri
MrRa [10]

Answer:

The height is "89.61 m". A further explanation is given below.

Explanation:

According to the question,

Mass of rocket,

m = 5 kg

Time,

t = 4.6 s

Initial speed of rod

u = o m/s

Final speed,

v = 39 m/s

drag,

= 60 N

So,

⇒ acceleration, a=\frac{v-u}{t}

                              =\frac{39-0}{4.6}

                              = 8.478 \ m/s^2

⇒ F_{net}= Net \ force

           =ma

           =5\times 8.478

           =42.89 \ N

Now,

The thrust will be:

⇒ Thrust-weight-drag=Net \ force

⇒ Thrust-(5\times 9.8)-60=42.89

⇒           Thrust-49-60=42.89

⇒                 Thrust-109=42.89

⇒                           Thrust=42.89+109

⇒                           Thrust=151.89 \ N

The height will be:

⇒ h=\frac{1}{2} at^2

      =\frac{1}{2}\times 8.47\times (4.6)^2

      =\frac{1}{2}\times 8.47\times 21.16

      =89.61 \ m

5 0
3 years ago
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