The product nuclei may or may not be ______

You’ve been in a car traveling at 80 km/hr. How many hours will it take to travel 640 km?

iren2701 [21]

I’m pretty sure it’s 8 hours

5 0

3 years ago

the natural process where the sun's energy is trapped by the atmosphere to make the earth warm enough for life is known as globa

alexira [117]

The natural process is called greenhouse effect. This is when the Sun's light enters the Earth, and converts to heat. Light passes through the Earth's atmosphere much easier compared to heat. Because of this, heat is trapped on Earth because it cannot exit the atmosphere as easily as light does.

6 0

4 years ago

Is work done if you carry a book across the room at a constant velocity?

Stella [2.4K]

<span>No. Work is not done if you carry a book across the room
at a constant velocity?

The force applied is perpendicular to the direction of motion. (C)</span>

4 0

4 years ago

Read 2 more answers

A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is used to heat a house. The mass flow rate

Savatey [412]

(a)

Use â€œsaturated refrigerant-134a pressure tableâ€, to find $ , h , and of refrigerant-134a, at 320kPa (P) .

$h_{1} h_{g}$ = 251.93kJ/kg

$s_{1} (s_g}$ ) =0.93026 kJ/kg-K

$v_{1}(v_{g})$ =0.063681 m/kg

we solve $h_{6}$ ,

$h_{6}$ = 282.62 kj/kg

$h_{3} (h_{f)}$ = 127.25 kJ/kg

We use this formula for finding $Q_{n}$

$Q_n} =( h_{2} - h_{3} )$

= 38.8 kw

(b)

Find the COP of the heat pump ( $COP_{R}$ ) .

( $COP_{R}$ ) = $q_{L} /w_{in}$

$COP_{R}$ = (282.62 -127.25) / (282.62-251.93)

= 5.06

What is evaporator pressures ?

Valves for regulating evaporator pressure

Although the compressor suction pressure may be lower, evaporator pressure regulation (EPR) valves can be employed in the suction line to prevent the evaporator pressure from dropping below a set or controlled value.

An EPR valve is used for the following tasks:

1. Prevent potential damage to a liquid chilling evaporator from the liquid freezing.

2. Prevent frost from accumulating on an air-cooling evaporator when it is near freezing point or when operation cannot be interrupted by a brief failure.

3. Permit the operation of two or more evaporators with the same compressor at various load temperatures.

4. Vary the evaporator pressure in accordance with a fluctuating load that is managed by the load temperature.

Learn more about evaporator pressures visit this link:

brainly.com/app/ask?q=What+is+evaporator+pressures+

#SPJ4

7 0

1 year ago

A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back

Kobotan [32]

Answer:

a) 5 m/s

b) 17.8542 m/s

c) 24.7212 m/s

0.229

Explanation:

t = Time taken

u = Initial velocity = 5 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=5-9.81\times t\\\Rightarrow \frac{-5}{-9.81}=t\\\Rightarrow t=0.51 \s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=5\times 0.51+\frac{1}{2}\times -9.81\times 0.51^2\\\Rightarrow s=1.27\ m$

So, the stone would travel 1.27 m up

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.27+0^2}\\\Rightarrow v=5\ m/s$

Velocity as the rock passes through the original point is 5 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.27\times 2}{9.81}}\\\Rightarrow t=0.51\ s$

Time taken to reach the original point is 0.51+0.51 = 1.02 seconds

So, total height of the rock would fall is 30+1.27 = 31.27 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 16.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{16.27\times 2}{9.81}}\\\Rightarrow t=1.82\ s$

Time taken by the stone to reach 15 m above the ground is 1.82+0.51 = 2.33 seconds

$v=u+at\\\Rightarrow v=0+9.81\times 1.82\\\Rightarrow v=17.8542\ m/s$

Speed of the ball at 15 m above the ground is 17.8542 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 31.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{31.27\times 2}{9.81}}\\\Rightarrow t=2.52\ s$

$v=u+at\\\Rightarrow v=0+9.81\times 2.52\\\Rightarrow v=24.7212\ m/s$

Speed of the stone just before it hits the street is 24.7212 m/s

F = Force

m = Mass = 100 kg

g = Acceleration due to gravity = 9.81 m/s²

s = Displacement = 4 km = 4000 m

P = Power = 1 hp = 745.7 Watt

t = Time taken = 20 minutes = 1200 seconds

μ = Coefficient of sliding friction

F = μ×m×g

⇒F = μ×100×9.81

W = Work done = F×s

P = Work done / Time

⇒P = F×s / t

$745.7=\frac{\mu \times 981\times 4000}{1200}\\\Rightarrow \mu=\frac{747.5\times 1200}{981\times 4000}\\\Rightarrow \mu=0.229$

Coefficient of sliding friction is 0.229

8 0

3 years ago

The product nuclei may or may not be _______