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iVinArrow [24]
3 years ago
7

A.

Physics
1 answer:
CaHeK987 [17]3 years ago
6 0
C the way that the data is organized
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A hot-water radiator has a surface temperatue of 80 o C and a surface area of 2 m2 . Treating it as a blackbody, find the net ra
faust18 [17]

Answer:

925.04 J/s

Explanation:

T = 80 C = 80 + 273 = 353 K

To = 20 c = 20 + 273 = 293 K

A = 2 m^2

Use the formula for Stefan's law

Energy radiated per second

E = \sigma  A \left ( T^{4}-T_{0}^{4} \right )

E = 5.67 \times 10^{-8}\times 2\left ( 353^{4}-293^{4} \right )

E = 925.04 J/s

3 0
3 years ago
Two charges are located in the xx – yy plane. If ????1=−4.25 nCq1=−4.25 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.0
Sati [7]

Answer:

Ex=  -17.1 N/C

Ey =  +26.9 N/C

Explanation:

We apply formula of electric field:

Ep=k*q/d²

Ep:  Electric field at point ( N/C)

q: Electric charge (C)

k: coulomb constant (N.m²/C²)

d: distance from charge q to point P (m)

In the attached graph we observe the directions of the electric field at P(0,0) due to q1 and q2

Calculation of the field at point P due to the load q₁

E₁=k*q₁/d₁² = 9*10⁹*4.25*10⁻⁹/1.080²= 32.8 N/C : Magnitude of E1

Direction of E₁ :Because the charge q₁ is negative the field enters the charge (+ y)

Calculation of the field at point P due to the load q₂

d_{2} = \sqrt{1.30^{2}+0.450^{2}  }

d₂=1.375 m

E₂=k*q₂/d₂² = 9*10⁹*3.80*10⁻⁹/ 1.375² = 18.09 N/C Magnitude of E₂

Direction of E₂ :Because the charge q₂ is positive the field leaves the charge in direction of angle β

, then,E₂ tiene componentes x-y  en P.

E₂x=-E₂cos β= -18.09*(1.3/1.375)= -17.1 N/C

E₂y=-E₂sin β= -18.09*(0.45/1.375)= -5.9 N/C

Calculation of the electric field at point P located at the origin(0,0)

Ex=E₂x= -17.1 N/C

Ey=E₁y+E₂y =32.8 N/C -5.9 N/C = 26.9 N/C

4 0
3 years ago
Most of the mass of the milky way exists in the form of.
Andrews [41]

Answer: Dark matter.

Explanation: Hope it helps :)

7 0
3 years ago
Read 2 more answers
John(body mass=160pounds) is taking off for a long jump. The average ground reaction force Fg at takeoff is 1400 N pointing forw
slega [8]

Answer:

The free body diagram of John is shown in the attached figure (in the FBD john's mass is supposed to be concentrated at his center of mass and FBD is made of center of mass)

b) As shown in the FBD the ground reaction forces are:

i) In X direction F_{x}=1400cos(35^{o})=1146.81N

ii) In Y direction F_{y}=1400sin(35^{o})=803.0N

c) The respective accelerations in x and y direction's is calculated by newton's second law as indicated under

\sum F_{x}=ma_{x}\\\\\therefore a_{x}=\frac{\sum F_{x}}{m}=\frac{1146.8N}{72.57kg}=15.80m/s^{2}\\\\\sum F_{y}=ma_{y}\\\\\therefore a_{y}=\frac{\sum F_{y}}{m}=\frac{803.00-72.57\times 9.81}{72.57}=1.255m/s^{2}

4 0
3 years ago
Explain the following (using the concepts of moment, center of gravity and stability.) a) Racing cars are low, with wide wheels.
LUCKY_DIMON [66]

Answer:

b

Explanation:

3 0
3 years ago
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