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Maslowich
3 years ago
7

With a 900 V source, voltage is divided across 3 series resistors of 300 V, 280 V, and:

Physics
1 answer:
telo118 [61]3 years ago
6 0

Answer:

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Referring to the map of the temperature field in the room, where was the warmest field found?
klasskru [66]
Is there supposed to be an image if so there is none
8 0
3 years ago
Read 2 more answers
A net force of 200 newtons is applied to a wagon for 3 seconds. This cause the wagons to undergo a change in momentum of ?
suter [353]
Physics- damon, Monday, December 1, 2014 at 3:27 pm force =change in momentum\ change in time or m a if m is constant 

change in momentum/3=200

change in momentum =3*200 kg m/s





8 0
3 years ago
a 3000 kg and a 7000 kg Mass attract each other with a force of 0.0015 N. What distance separates the two objects (Radius) (Plea
frutty [35]

Answer:

<h3> 3.057m</h3>

Explanation:

According to law of gravitation;

F = GMm/d²

G is the universal gravitation

M and m are the masses

d is the distance between the masses

d² = GMm/F

d² = 6.67408 × 10-11 *3000*7000/0.0015

d² = 140.15568*10^-5/0.0015

d² = 1.4016*10^-3/0.0015

d² = 1.4016*10^-3/1.5*10^-3

d²  = 0.9344*10

d² = 9.344

d = √9.344

d = 3.057m

Hence the distance between the two objects is  3.057m

3 0
3 years ago
If the car goes exits a freeway and goes from 65<br> mph to 35 mph is it accelerating?
Zolol [24]

Answer:

No, the car is decelerating  

Explanation:

No the car is decelerating if it exits a freeway and goes from 65

mph to 35 mph since the change in velocity is negative.

change in velocity = final - initial

change in velocity  = 35 - 65

change in velocity = -30mph

Since the change in velocity is negative, hence the car is decelerating. Deceleration is a negative acceleration

8 0
3 years ago
A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,
gavmur [86]

Answer:

a. ρ_\beta=1.996J/m^3

b. U_E=9.445x10^{-15} J/m^3

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

i=14A, d=2.5mm, R=3.3Ω, l=1 km, E_o=8.85x10^{-12}F/m, u_o=4*x10^{-7}H/m

ρ_\beta=\frac{\beta^2}{2*u_o}

ρ_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2

ρ_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}

ρ_\beta=1.996J/m^3

b. The electric field can be find using the equation:

U_E=\frac{1}{2}*E_o*E^2

E=(\frac{i*R}{l})^2

U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2

U_E=9.445x10^{-15} J/m^3

4 0
3 years ago
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