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3 years ago

9

Two charged objects have a repulsive force of .080 N. if the charge of both of the objects is doubled, then what is the new forc

e?

1 answer:

Viefleur [7K]3 years ago

4 0

This was kinda hard thinking it might be that the new force is going to 4 times bigger than the original one.

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Part A : A cylindrical water tank 10cm high and 520cm in diameter is filled with solution with a density of 0.75 g/cm^3

lisov135 [29]

Answer: $0.1066\ psi, 15.611\ kN$

Explanation:

Given

the height of the tank is $h=10\ cm$

The diameter of the tank is $d=520\ cm$

Density of solution $\rho=0.75\ g/cm^3\ or\ 750\ kg/m^3$

$(a)$ Water pressure at the bottom of the tank is

$P=\rho gh\\P=750\times 9.8\times 0.1\\P=735\ Pa$

$1\ Pa=0.000145038\ psi$

$\Rightarrow P=735\ Pa\ or\ 0.1066\ psi$

$(b) \text{Average force on the bottom is the product of pressure and area of the base}$

$F_{avg}=735\times \pi \cdot (\frac{520}{200})^2\\\\F_{avg}=735\times 3.142\times 6.76\\F_{avg}=15,611.34\ N\ or\ 15.611\ kN$

6 0

3 years ago

m_Cu * sh_CuA system consists of a copper tank whose mass is 13 kilogram , 4 kilogram of liquid water, and an electrical resisto

notsponge [240]

Answer:

T₂ = 49.3°C

Explanation:

Applying law of conservation of energy to the system we get the following equation:

Energy Supplied by Resistor = Energy Absorbed by Tank + Energy Absorbed by Water

E = mC(T₂ - T₁) + m'C'(T'₂ - T'₁)

where,

E = Energy Supplied by Resistor = 100 KJ = 100000 J

m = mass of copper tank = 13 kg

C = Specific Heat of Copper = 385 J/kg.°C

T₂ = Final Temperature of Copper Tank

T₁ = Initial Temperature of Copper Tank = 27°C

T'₂ = Final Temperature of Water

T'₁ = Initial Temperature of Water = 50°C

m' = Mass of Water = 4 kg

C' = Specific Heat of Water = 4179.6 K/kg.°C

Since, the system will come to equilibrium finally. Therefor: T'₂ = T₂

Therefore,

(100000 J) = (13 kg)(385 J/kg.°C)(T₂ - 27°C) + (4 kg)(4179.6 J/kg.°C)(T₂ - 50°C)

100000 J = (5005 J/°C)T₂ - 135135 J + (16718.4 J/°C)T₂ - 835920 J

100000 J + 135135 J + 835920 J = (21723.4 J/°C)T₂

(1071055 J)/(21723.4 J/°C) = T₂

<u>T₂ = 49.3°C</u>

8 0

3 years ago

A student is creating an electromagnet for an investigation. Which feature of the electromagnet will least influence the magneti

garri49 [273]

C the number of wire coils

5 0

3 years ago

Can someone please here me I’m confused

Naddika [18.5K]

I would say z, x, y, w

5 0

4 years ago

During a race, a runner runs at a speed of 6 m/s. Four seconds later, she is running at a speed of 14 m/s. What is the runner's

DedPeter [7]

Answer:

D. 2 m/s²

Step-by-step explanation:

Initial speed of the runner (u) = 6 m/s

Final speed of the runner (v) = 14 m/s

Time taken (t) = 4 s

By using equation of motion, we get:

$\bf \longrightarrow v = u + at \\ \\ \rm \longrightarrow 14 = 6 + 4a \\ \\ \rm \longrightarrow 14 - 6 = 6 - 6 + 4a \\ \\ \rm \longrightarrow 8 = 4a \\ \\ \rm \longrightarrow 4a = 8 \\ \\ \rm \longrightarrow \dfrac{4a}{4} = \dfrac{8}{4} \\ \\ \rm \longrightarrow a = 2 \: m {s}^{ - 2}$

$\therefore$ Acceleration of the runner (a) = 2 m/s²

3 0

3 years ago