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3 years ago

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For a ship moving against the current, it takes 9 hours to cover a distance of 113.4 miles. how much does it take this ship to r

eturn if the rate of the current is 1.9 mph?

2 answers:

Digiron [165]3 years ago

3 0

Let x mph be the speed of the ship in still water.
Against the current, the net speed is, (x-1.9) mph.
Time, t= distance, d/ Speed, v => 9= 113.4/(x-1.9) => 9(x-1.9) = 113.4 => x-1.9 = 113.4/9 => x = 12.6+1.9 = 14.5 mph

Moving with the current, the net speed of the ship = 14.5+1.9 =16.4 mph

Time take, t = d/v = 113.4/16.4 = 6.915 hours

Vanyuwa [196]3 years ago

3 0

Answer:

The answer is it takes 6.91 hours for the ship to return.

Hope this helped!

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Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on

Romashka [77]

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$\frac{1}{4\pi\spsilon_0}=9\times10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35\times10^{-4}$ N

So, from equation (i)

$1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}$

$\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$\frac{q_1+q_2}{2}$

d=20cm=0.2m, and $F_c=1.406\times10^{-4}$ N

So, from equation (i)

$1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}$

$\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}$

$\Rightarrow q_1+q_2=\pm5\times 10^{-8}$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)$

$\Rightarrow q_1=5\times 10^{-8}-q_2$

The equation (ii) become:

$(5\times 10^{-8}-q_2)q_2=6\times10^{-16}$

$\Rightarrow -(q_2)^2+5\times 10^{-8}q_2-6\times10^{-16}=0$

$\Rightarrow q_2=3\times10^{-8}, 2\times10^{-8}$

From equation (iii)

$q_1=2\times10^{-8}, 3\times10^{-8}$

So, the magnitude of initial charges on both the sphere are $3\times10^{-8}$ Coulombs $=0.03 \mu C$ and $2\times10^{-8}$ Colombs or $0.02 \mu C$ .

Considerion the nature of charges too,

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

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3 years ago

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Answer:

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n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend

Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

$p_f=p_i\\mv_1+Mv_2=(m+M)v$

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

$(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}$

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

$E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2$

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

$(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m$

hence, the heigth is 68cm

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Calculate the amount of heat needed to increase the temperature of 200 grams of water at 10°C to 95°C.

Alexus [3.1K]

the amount of heat need is 71060 Joules

Explanation

The equation for the amount of heat needed to increase the temperature of water is as follows:

$\begin{gathered} E=mC(T_2-T_1) \\ where \\ \\ \end{gathered}$

where E is the amount of heat

C is the specific heat capacity of water having a standard value of 4180 J/kg*K

T is the final ans initial temperature in kelvin

m is the mass

so

Step 1

given

$\begin{gathered} mass=m=200\text{ gr=0.2 kg} \\ T_1=10\text{ \degree C} \\ T_2=95\text{ \degree C} \\ C=4180\text{ J/kh*K} \end{gathered}$

a) convert the temperature into kelvin

$\begin{gathered} 10\text{ \degree C=\lparen10+273\rparen K=283 K} \\ 95\text{ \degree C=\lparen95+273 \rparen K=368 K} \end{gathered}$

now, replace in the formula

$\begin{gathered} E=mC(T_{2}-T_{1}) \\ E=0.2\text{ kg*4180L/kg*K*\lparen368K-283 k\rparen} \\ E=71060\text{ J} \end{gathered}$

therefore, the amount of heat need is 71060 Joules

I hope this helps you

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Answer:

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