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DaniilM [7]
2 years ago
13

Heyy! i’ll give brainliest please help

Physics
1 answer:
taurus [48]2 years ago
7 0
The answer is south
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Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến
Mila [183]

Answer:

P = 2439.5 W = 2.439 KW

Explanation:

First, we will find the mass of the water:

Mass = (Density)(Volume)

Mass = m = (1 kg/L)(10 L)

m = 10 kg

Now, we will find the energy required to heat the water between given temperature limits:

E = mCΔT

where,

E = energy = ?

C = specific heat capacity of water = 4182 J/kg.°C

ΔT = change in temperature = 95°C - 25°C = 70°C

Therefore,

E = (10 kg)(4182 J/kg.°C)(70°C)

E = 2.927 x 10⁶ J

Now, the power required will be:

Power = P = \frac{E}{t}

where,

t = time = (20 min)(60 s/1 min) = 1200 s

Therefore,

P = \frac{2.927\ x\ 10^6\ J}{1200\ s}

<u>P = 2439.5 W = 2.439 KW</u>

7 0
2 years ago
What do these letters stand for<br> P=mv
victus00 [196]

Answer:

The equation for momentum of a piece of matter.

In either case, the momentum would be less than a linebacker hitting you at full speed. The equation for momentum is written: p = mv where p stands for momentum. That is, mass times velocity equals momentum.

Explanation:

Hope This Helps

Have A Great Day

4 0
2 years ago
Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The
Galina-37 [17]

Answer:

220 A

Explanation:

The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.

So, F = BI₁L

F = (μ₀I₂/2πd)I₁L

F = μ₀I₁I₂L/2πd

Given that the current in the rods are the same, I₁ = I₂ = I

So,

F = μ₀I²L/2πd

Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²

So, F = W

μ₀I²L/2πd = mg

making I subject of the formula, we have

I² = 2πdmg/μ₀L

I = √(2πdmg/μ₀L)

substituting the values of the variables into the equation, we have

I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])

I = √(0.0049 × 10⁷kgm²/s²H)

I = √(0.049 × 10⁶kgm²/s²H)

I = 0.22 × 10³ A

I = 220 A

7 0
3 years ago
A 200g block on a 50cm long string swings in a circle, it's frictionless and 75rpm. What is its speed and tension on string
krek1111 [17]
Angular velocity = (75x2pie)/60
                          =2.5pie ras^-1 
linear velocity(or speed) at end of string, v = radius x angular velocity
                                                           v= 0.5 x 2.5pie
                                                           v=3.93 ms^-1

tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
                                                                                      F= (0.2 x 3.93^2)/0.5
                                                                                      F=6.18 N
(sorry if wrong)
3 0
3 years ago
Read 2 more answers
Figure 24-40 shows a thin rod with a uniform charge density of 2.40 μC/m. Evaluate the electric potential (in V) at point P if d
LuckyWell [14K]
Can you send a picture???
7 0
3 years ago
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