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Mazyrski
14 days ago
6

A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object o

f mass M and secures the other end of the string so that the object is at rest as it hangs from the string. When the student raises the object to a height above its lowest point and releases it from rest, the object undergoes simple harmonic motion. As the student collects data about the time it takes for the pendulum to undergo one oscillation, the student observes that the time for one swing significantly changes after each oscillation. The student wants to conduct the experiment a second time. Which two of the following procedures should the student consider when conducting the second experiment?
Physics
1 answer:
ValentinkaMS [1.1K]14 days ago
7 0

Answer:

-utilize precisely the same apparatus

-maintain identical measures (release height)

Explanation:

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A bowling pin is thrown vertically upward such that it rotates as it moves through the air, as shown in the figure. Initially, t
Ostrovityanka [942]

Answer:

Explanation:

The equation used to determine the maximum height of the bowling pin during its trajectory is given by;

H = u²/2g

where u, the initial speed/velocity, equals 10m/s

g stands for gravitational acceleration = 9.81m/s²

Substituting in the values gives us

H = 10²/2(9.81)

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3 0
14 days ago
the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point
serg [1198]

Answer:

The pressure measured at this moment is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

P=1.81\ kW

We are tasked with finding the pressure here

Applying the pressure formula

P=F\times v

P=A_{1}P_{1}\times v_{1}

Here, A_{1}v_{1}=Q_{1}

Where, v refers to velocity

Insert the values into the equation

1.81 =P_{1}\times0.124\times\dfrac{1}{60}

P_{1}=\dfrac{1.81\times60}{0.124}

P_{1}=875.80\ kPa

P_{1}=0.875\ mPa

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5 0
11 days ago
A uniform magnetic field makes an angle of 30o with the z axis. If the magnetic flux through a 1.0 m2 portion of the xy plane is
Yuliya22 [1161]

Response:

(b) 10 Wb

Clarification:

Given;

angle of the magnetic field, θ = 30°

initial area of the plane, A₁ = 1 m²

initial magnetic flux through the plane, Φ₁ = 5.0 Wb

The equation for magnetic flux is;

Φ = BACosθ

where;

B denotes the magnetic field strength

A represents the area of the plane

θ is the inclination angle

Φ₁ = BA₁Cosθ

5 = B(1 x cos30)

B = 5/(cos30)

B = 5.7735 T

Next, calculate the magnetic flux through a 2.0 m² section of the same plane:

Φ₂ = BA₂Cosθ

Φ₂ = 5.7735 x 2 x cos30

Φ₂ = 10 Wb

<pHence, the magnetic flux through a 2.0 m² area of the same plane is 10 Wb.

Option "b"

3 0
5 days ago
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13 days ago
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