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3 years ago

All of the following affect the climate of an area except

Physics

2 answers:

Ksivusya [100]3 years ago

7 0

C. A cold front that came through does not affect the climate of an area.

stich3 [128]3 years ago

6 0

C Is Gone Be The Answer

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Refrigerant-134a enters the expansion valve of a refrigeration system at 160 psia as a saturated liquid and leaves at 30 psia. D

KatRina [158]

Answer:

Temperature : 92.9 F

Internal Energy change: -2.53 Btu/lbm

Explanation:

mh1=mh2

h1=h2

In table A-11 through 13E

p2=120Psi, h1= 41.79 Btu/lbm,

u1=41.49

So T1=90.49 F

P2=20Psi

h2=h1= 41.79 Btu/lbm

T2= -2.43F

u2= 38.96 Btu/lbm

T2-T1 = 92.9 F

u2-u1 = -2.53 Btu/lbm

3 0

3 years ago

A box weighs 25N. How much mass does it have?

Rudiy27

Explanation:

If box weight 25N on ground

MA=F

M(10)=25

M=2.5Kg

3 0

3 years ago

Two wires of the same metal having the same cross-

USPshnik [31]

The answer is true

Step by step explanation:

3 0

3 years ago

Read 2 more answers

In a certain city, electricity costs $0.20 per kw·h. what is the annual cost for electricity to power a lamp-post for 8.00 hours

UNO [17]

Part a)

per day electricity power consumed when 100 W bulb is used for 8 hours

$P = 8 * 100 = 800 Wh$

for one year consumption

$E = 365 * 800 = 292 kWH$

now the cost will be given

$cost = 0.20 * 292 = $ 58.4$

now when other energy efficient light is used

$P = 8 * 25 = 200 Wh$

for one year consumption

$E = 365 * 200 = 73 kWH$

now the cost will be given

$cost = 0.20 * 73 = $ 14.6$

8 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

4 years ago