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stepladder
12 hours ago
9

A 0.675 kg mass is attached to a

Physics
1 answer:
Sav [2.2K]12 hours ago
5 0

Answer:

frequency = 6.28Hz

Explanation:

all necessary information is attached to the sheet

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A cliff diver on an alien planet dives off of a 32 meter tall cliff and lands in a sea of hydrochloric acid 1.20 seconds later.
inna [2218]

Answer:

44.4m/s^2

Explanation:

Utilize the equation...S = ut + 1/2at^2

where...S = 32m...u = 0m/s....t = 1.20s

32 = (0)(1.20) + 0.5(1.20^2)a

; The acceleration due to gravity is 44.4m/s^2

3 0
28 days ago
Mrs. Gonzalez is about to give birth and Mr. Gonzalez is rushing her to the hospital at a speed of 30.0 m/s. Witnessing the spee
Ostrovityanka [2214]

Answer: The frequency is 1714.3 Hz

Explanation: The calculation is derived from the Doppler effect formula.

Since the source is approaching the observer, the observer's velocity is considered positive.

Refer to the attached document for the detailed derivation

3 0
1 month ago
Which, if any, of the following statements concerning the work done by a conservative force is NOT true? All of these statements
Yuliya22 [2446]

Answer:

If the starting and ending points are identical, the overall work equals zero.

Explanation:

Option (D) is correct.

A force is considered conservative when the work performed by it while moving an object from point A to point B does not rely on the path taken and remains consistent across all paths. The work done is determined solely by the initial and final locations of the particle. Thus, when the initial and final positions in a conservative field coincide, the work is said to be zero.

8 0
1 month ago
Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+βF(x)=αx3+β.
Sav [2230]

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To calculate the kinetic energy variation, we can utilize the work-energy theorem.

W = ΔK

∫ F .dx = K - K₀

If the object starts from rest, then K₀ = 0.

So, ∫ F dx cos θ = K.

As the force and displacement directions align, the angle is zero, and hence the cosine is 1.

Now we can substitute and perform integration:

α ∫ x³ dx + β ∫ dx = K.

Thus, α x⁴ / 4 + β x = K.

Next, we evaluate from the limits F = 0 to F:

α (x⁴ / 4 - 0) + β (x - 0) = K.

Consequently, K = αX⁴ / 4 + β x.

This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.

To finalize the computation, we need to ascertain the displacement.

8 0
1 month ago
An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:
Sav [2230]

Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

\texttt{ }

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!

\texttt{ }

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}

\texttt{ }

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:

\texttt{ }

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

5 0
2 days ago
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