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4 days ago

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A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th

at willbreak under a tension of 54N or more. He is eager to the fish to the dock in the shortest time possible. If the fish is at rest at the water's surface, what is the least amount of time in which the fisherman can raise the fish to the dock without losing it?

1 answer:

Ostrovityanka [2.2K]4 days ago

7 0

The time taken, t = 2 seconds, can be deduced from the scenario involving the fish being lifted.

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Assume that charge −q is placed on the top plate, and +q is placed on the bottom plate. What is the magnitude of the electric fi

Ostrovityanka [2204]

Let A represent the area of each plate. According to Gauss's Law, the electric field present between the plates can be derived.

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3 days ago

If a woman weighs 125 lb, her mass expressed in kilograms is x kg, where x is

Sav [2226]

To tackle this issue, it's essential to understand the conversion of pounds to kilograms:
1 lb = 0.45 Kg
By applying a straightforward rule of three
1 lb ---> 0.45 Kg
125 lb ---> x
Solving for x yields:
x = ((125) / (1)) * (0.45) = 56.25 Kg.
Response
her mass in kilograms is 56.25 Kg.

3 0

10 days ago

Read 2 more answers

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Sav [2226]

Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

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Additional explanation

The Ideal Gas Law that should be remembered is:

$\large {\boxed {PV = nRT} }$

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

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Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

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Next, we will determine the work performed on the gas:

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

Step B:

By utilizing the methodology mentioned earlier:

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

Next, we will ascertain the work completed on the gas:

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

$\texttt{ }$

Ultimately, we can calculate the total work done and heat supplied as follows:

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

$\texttt{ }$

Learn more

Minimum Coefficient of Static Friction:
The Pressure In A Sealed Plastic Container:
Effect of Earth’s Gravity on Objects:

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

2 days ago

Read each scenario below. Then select the answer that best completes each sentence.

Ostrovityanka [2204]

Answer:

The power used by raul's microwave must match the power consumed by katrina's because both microwaves took different durations to accomplish the same heating task.

Explanation:

The power output from a car engine is equivalent to that of a bicycle since both perform the same amount of work over time. Both raul and katrina shared a frozen meal, heating each portion in different microwaves. Katrina's portion was warm in one minute, whereas raul's portion required two minutes. Therefore, the power utilized by raul's microwave aligns with that of katrina's, given that it took longer to achieve the same result.

7 0

26 days ago

Read 2 more answers

A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag

Softa [2029]

Answer:

57.94°

Explanation:

We understand that the formula for flux is

$\Phi =E\times S\times COS\Theta$

where Ф represents flux

E indicates electric field

S denotes surface area

θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78 $\frac{Nm^{2}}{sec}$

E= $1.44\times 10^{4}\frac{Nm}{C}$

S= $\pi \times 0.057^{2}$

$COS\Theta =\frac{\Phi }{S\times E}$

= $\frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}$

=0.5306

θ=57.94°

4 0

19 days ago