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gtnhenbr [62]
3 years ago
5

If a car is moving at 90 km/h and it rounds a corner, also at 90 km/h, does it maintain a constant speed? A constant velocity? I

f
Physics
1 answer:
valentina_108 [34]3 years ago
6 0

Answer:

Constant speed but different velocity

Explanation:

Since the car speed at the corner is 90km/h which is the same as before. The speed, or magnitude of the velocity vector, remains unchanged. The direction of the velocity vector, on the other hand, is changing as the car turns around the corner. Therefore, the car maintains a constant speed, but not a constant velocity.

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A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in
Gekata [30.6K]

A.

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

Substituting the values of the variables into the  equation, we have

Q = mcΔT

Q = ρVcΔT

Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C

Q = 482630400 J

So, the energy of the hot water is 482630400 J

B.

The elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

Substituting the values of the variables into the equation, we have

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Learn more about heat energy here:

brainly.com/question/11961649

5 0
2 years ago
When a plant is entering the calvin cycle of photosythesis______.
Korolek [52]
The Calvin cycle<span> refers to the light-independent reactions in photosynthesis that take place in three key steps. Although the </span>Calvin Cycle<span> is not directly dependent on light, it is indirectly dependent on light since the necessary energy carriers (ATP and NADPH) are products of light-dependent reactions.

So basically it indirectly needs the light, even it's called light-independant reaction.

So the answer is the last one.</span>
5 0
3 years ago
Read 2 more answers
A uniform disk with a 25 cm radius swings without friction about a nail through the rim. If it is released from rest from a posi
ValentinkaMS [17]

Answer:

Explanation:

During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy

mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .

mgr  = 1/2 x ( 1/2 m r²+ mr²) x ω²

gr  = 1/2 x 1/2  r² x ω² + 1/2r² x ω²

g = 1 / 4 x ω² r + 1 / 2 x ω² r

g = 3  x ω² r/ 4

ω² = 4g /3 r

= 4 x 9.8 /  3 x  .25

= 52.26

ω = 7.23  rad / s .

6 0
2 years ago
Which of the following tend to react by gaining 1 electron
Andrei [34K]
The correct answer should be C
8 0
2 years ago
Read 2 more answers
The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the
natulia [17]

Answer:

The value is   A =  2.80 *10^{-4} \  m^2

Explanation:

From the question we are told that

The  operating temperature is  T  =  2450 \  K

The emissivity is  e =  0.350

 The  power rating is  P  =  200 \  W

Generally the area is mathematically represented as

      A = \frac{P}{ e *  \sigma  *  T^2}

Where  \sigma is the Stefan Boltzmann constant  with value  

      \sigma  =  5.67 *10^{-8} \  W/m^2\cdot K^4

So

     A =  \frac{200}{0.350 *  5.67*10^{-8} *  2450^{4}}

     A =  2.80 *10^{-4} \  m^2

8 0
3 years ago
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