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mash [69]
3 years ago
10

Calculate potential energy

Physics
1 answer:
IrinaVladis [17]3 years ago
4 0
The potential energy of an object is defined by the equation: PE = mgh, where m = the mass of the object, g = the gravitational acceleration and h = the object's height above the ground.
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a wire is carrying a 2.45 A current. at what distance from the wire is the magnetic field 1.00x10^-6t
balu736 [363]

Answer:

0.49m

Explanation:

So you need to change the original equation for finding fields to find distance, and then just plug in the numbers

Which equals 0.49meters

Also it was right on Acellus :)

Hope this helps

5 0
3 years ago
Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

8 0
3 years ago
A block is sent up a frictionless ramp along which an x axis extends upward. The figure below gives the kinetic energy of the bl
ss7ja [257]
Kinetic energy =1/2 mv^2 

<span>m=2ke/v^2 </span>

<span>m=2(34)/3.6^2 </span>

<span>m=5.24 </span>

<span>force normal = mg </span>
<span>=5.24 x 9.8 </span>
<span>force normal = 51.4N

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.


</span>
5 0
3 years ago
Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?
Debora [2.8K]

493 \; \text{W}\cdot \text{m}^{-2}.

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

\dfrac{P}{A} = \sigma \cdot T^{4}

where,

  • P the total power emitted,
  • A the surface area of the body,
  • \sigma the Stefan-Boltzmann Constant, and
  • T the temperature of the body in degrees Kelvins.

\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}.

T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}.

\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}.

Keep as many significant figures in T as possible. The error will be large when T is raised to the power of four. Also, the real value will be much smaller than 493\; \text{W}\cdot \text{m}^{-2} since the emittance of a human body is much smaller than assumed.

5 0
3 years ago
A student initially 10.0 m East of his school walks 17.5 m West. The magnitude of the student's displacement, relative to the sc
Fantom [35]

Answer:

1. 7.5 m

2. towards west side

explanation:

I hope it will help you

3 0
2 years ago
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