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Basile [38]
3 years ago
7

A carnot engine has an efficiency of 50% when its sink temperature is 27°C. The temperature of source is: a) 300°C b) 327°C c) 2

73°C d) 373°C​
Physics
1 answer:
hram777 [196]3 years ago
5 0
<h2>Answer:</h2>

327 degrees

<em>Have a great day! </em>

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HELLO , PLZ HELP .
sergey [27]

Answer:

the pressure due to the water on the diver is 200,000 pascal

pressure = height × density × acceleration due to gravity

p = 20×1000×10

p=200,000 pascal

6 0
3 years ago
A 14.8-g piece of Styrofoam carries a net charge of -0.742 µC and is suspended in equilibrium above the center of a large, horiz
MAXImum [283]
Base on your question where a 14.8g of piece of Styrofoam carries a net charge of -0.742C and is suspended in equilibrium above the center of a large, horizontal sheet of plastic so the ask of the problem is to calculate  the charge per unit area on the plastic sheet. The answer would be 21.96
3 0
3 years ago
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0
3 years ago
If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
3 years ago
Dr. james van allen and his group of united states scientists discovered a set of ___________. magnetic equators planets magneti
Fofino [41]

RADIATION BELTS....... I think but it should be radiation belt

3 0
3 years ago
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