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rosijanka [135]
2 years ago
7

A physics student hurries through their lab, releasing the bob of a simple pendulum from a height, and allowing it to swing. He

measures the period of the pendulum and calculates that g = 8 m/s2. What might the student have done wrong? Select all that apply. The length of the pendulum string was too long, so the equation for the period of a pendulum was no longer valid in this case. The string wasn't taut when he released the bob, causing the bob to move erratically. The student only timed one cycle, introducing a significant timing error. The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. The mass of the bob was too large, so the equation for the period of a pendulum was no longer valid in this case. Instead of releasing the bob from rest, the student threw the bob downward.
Physics
1 answer:
MakcuM [25]2 years ago
3 0

Answer:

The string wasn't taut when he released the bob, causing the bob to move erratically.

The student only timed one cycle, introducing a significant timing error.

The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case.

Explanation:

The period of a simple pendulum is given by the formula

T=2\pi \sqrt{\frac{L}{g}}

where

T is the period of oscillation

L is the length of the pendulum

g is the acceleration due to gravity

Therefore, it is possible to measure the value of g in an experiment, by taking a pendulum, measuring its length L, and measuring its period of oscillation T. Re-arranging the equation above, we get the value of g as:

g=(\frac{2\pi}{T})^2 L

Here the value of g measured in the experiment is 8 m/s^2 instead of 9.8 m/s^2. Let's now analyze the different options:

The length of the pendulum string was too long, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. There is no constraint on the length of the pendulum.

The string wasn't taut when he released the bob, causing the bob to move erratically. --> TRUE. This is possible, as if the string is not taut, the pendulum would not start immediately its oscillation, so the period would be larger causing a smaller value measured for g.

The student only timed one cycle, introducing a significant timing error. --> TRUE. This is also impossible: in fact, we can get a more accurate measurement of the period if we measure several oscillations (let's say 10), and then we divide the total time by 10.

The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. --> TRUE. The formula written above for the period of the pendulum is valid only for small angles.

The mass of the bob was too large, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. The equation that gives the period of the pendulum does not depend on the mass.

Instead of releasing the bob from rest, the student threw the bob downward. --> FALSE. In fact, this force would have been applied only at the very first moment, but then later the only force acting on the pendulum is the force of gravity, so the formula of the period would still be valid.

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23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n
PSYCHO15rus [73]

Answer:

a)  (95.4 i^ + 282.6 j^) N , b) 298.27 N  71.3º and c)   F' = 298.27 N   θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

      sin θ = Fjy / Fj

      cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

          T = 90 -32 = 58º

         Fjy = Fj sin 58

         Fjx = FJ cos 58

         Fjx = 180 cos 58 = 95.4 N

         Fjy = 180 sin 58 = 152.6 N

Andrea's force is

         Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

       Fx = Fjx

       Fx = 95.4 N

Axis y

       Fy = Fjy + Fa

       Fy = 152.6 + 130

       Fy = 282.6 N

       F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

       F² = Fx² + Fy²

       F = √ (95.4² + 282.6²)

       F = √ (88963)

       F = 298.27 N

       tan θ = Fy / Fx

       θ = tan-1 (282.6 / 95.4)

       θ = tan-1 (2,962)

       θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

       F' = 298.27 N

       θ' = 180 + 71.3

       θ = 251.4º

4 0
2 years ago
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
2 years ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
2 years ago
Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of
schepotkina [342]
If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!
5 0
2 years ago
Gbenga needs to get glasses to correct his farsightedness. His eyes currently cannot focus on objects that are within 2 ft (or 6
Scilla [17]

Answer:

Focal Length = 38.61cm, Power = 2.59 Diopter, Converging lens.

Explanation:

When an object is placed 25cm from Gbenga's eye, the glasses lens must produce an image 61cm away (Gbenga's eye near point).

An image 61cm from the eye will be (61cm - 1.6cm) from the glasses.

i.e. d_{i}=61cm-1.6cm=59.4cm

and d_{o} = 25cm - 1.6cm = 23.4cm

note d_{i} will be negative because the image is formed on the same side as the object.

finally, d_{i}=-59.4cm\\d_{o}=23.4cm

the formula for finding the focal length f is given as

f=\frac{d_{i}d_{o}  }{d_{o}+d_{i}  }

f=\frac{-59.4*23.4}{23.4-59.4} \\

f=\frac{-1389.96}{-36}

f=38.61cm

The focal length is positive which indicates converging lens

power p=\frac{1}{f}

but f\\ must be in metres

Therefore, f=38.61cm=0.3861m

p=\frac{1}{0.3861}

p=2.59 Diopter

3 0
2 years ago
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