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geniusboy [140]
2 years ago
10

A book is moved once around the edge of a tabletop with dimensions 1.75 m à 2.25 m. If the book ends up at its initial position,

what is its displacement? If it completes its motion in 23 s, what is its average velocity? What is its average speed?
Physics
1 answer:
uysha [10]2 years ago
6 0

Explanation:

The dimension of the book is 1.75 m × 2.25 m. If the book ends up at its initial position. The displacement of the book is equal to zero as the object reaches to its initial position.

If it completes its motion in 23 s, t = 23 s

Total displacement of the book is equal to its perimeter. It is given by :

d=2(1.75+2.25)=8\ m

The net displacement divided by total time taken is called the average velocity of an object. Here, the displacement is 0. So, average velocity is 0.

The average speed of an object is given by :

v=\dfrac{d}{t}

v=\dfrac{8\ m}{23\ s}

v = 0.347 m/s

So, the average speed of the book is 0.347 m/s. Hence, this is the required solution.

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The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

7 0
1 year ago
A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitu
Mumz [18]

Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

T=\dfrac{2\pi}{\omega}

0.74=\dfrac{2\pi}{\omega}

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude  = 0.13 m

8 0
2 years ago
"if a stream flow measures 12 meters in 60 seconds, what is the stream's average rate of flow?"
Vlad [161]
<span>Discharge is the volume of water moving down a stream or river per unit of time, commonly expressed in cubic feet per second or gallons per day. In general, river discharge is computed by multiplying the area of water in a channel cross section by the average velocity of the water in that cross section: discharge = area * velocity. In this case, the answer is 0.2 m/s.</span>
7 0
2 years ago
To compare the hearing capacities of 5 of his friends, Ravi designs a simple experiment. He places a CD player at the end of a l
aliya0001 [1]

Answer:

A) having each person listen to just 1 song instead of 5.

Explanation:

Ravi could accurately use the outcome to compare the hearing capacities of his friends by this experiment if appropriate precautions are observed.

Playing 5 well known songs simultaneously could result to the interference of sounds when each participant moves away from the player. Which could naturally cause the variation in volume of the songs played with respect to the increasing distance.

Therefore, the reliability of the result from this experiment can be increased if each person listen to just 1 song instead of 5 at a predetermined volume. Each participant would be able to focus on hearing a song during the experiment.

8 0
2 years ago
A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to t
arlik [135]

Answer:

The magnitude of the torque on the loop due to the magnetic field is 4.7\times10^{-4}\ N-m.

Explanation:

Given that,

Diameter = 10 cm

Current = 0.20 A

Magnetic field = 0.30 T

Unit vectorn=-0.60\hat{i}-0.080\hat{j}

We need to calculate the torque on the loop

Using formula of torque

\tau=NIAB\sin\theta

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}

\tau=4.7\times10^{-4}\ N-m

Hence, The magnitude of the torque on the loop due to the magnetic field is 4.7\times10^{-4}\ N-m.

5 0
2 years ago
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