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tankabanditka [31]
3 years ago
4

Three different experiments are conducted that pertain to the oscillatory motion of a pendulum. For each experiment, the length

of the pendulum and the mass of the pendulum are indicated. In all experiments, the pendulum is released from the same angle with respect to the vertical. If the students collect data about the kinetic energy of the pendulum as a function of time for each experiment, which of the following claims is true?
a. The data collected from Experiment 1 will be the same as the data collected from Experiment 2.
b. The data collected from Experiment 1 will be the same as the data collected from Experiment 3.
c. The data collected from Experiment 2 will be the same as the data collected from Experiment 3.
d. The data collected from each experiment will be different.
Physics
1 answer:
ryzh [129]3 years ago
0 0

Answer:

d. The data collected from each experiment will be different.

Explanation:

The time period of simple pendulum depends on length of pendulum and independent of mass of pendulum.

For three experiments data is collected. Kinetic energy depends on mass and square of velocity.

SO FOR THREE EXPERIMENTS THE DATA COLLECTED WILL BE DIFFERENT.

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How can we reduce the ripple factor of a bridge recitifier from 0.48 to 0.1 approximately ?
padilas [110]
U can try to use capacitor , the value of capacitor depends on circuit

ripple factors signifies the ac components, by def its ratio of rms value of ac component to value of dc component
so in order to reduce use of a capacitor which denies the sudden changes in voltage, which is the charracteristic of Ac signals

hope this helps
4 0
4 years ago
Which is colder, 0°C or 20°F?
Mashutka [201]

In order to compare the two temperatures, we need to convert 20°F into Celsius. The formula we need to use is:

T(^{\circ}C)=\frac{5}{9}(T(^{\circ}F)-32)

Substituting 20°F, we find

T(^{\circ}C)=\frac{5}{9}(20^{\circ}F-32) =-6.7^{\circ}C


And this is less than 0°C, so the answer is

20°F is colder than 0°C.


7 0
4 years ago
Read 2 more answers
A resin tube and a PVC rod are both rubbed against styrofoam. Do
Ivan

The concepts of electrostatics allow us to find the result for the question about the electric forces in the two tubes is:

  • The tubes are attracted by having charges of different signs.

<h3>Electrostatics</h3>

Electrostatics studies the transfer of charge between bodies when they rub together, in generating the charges of the insulating bodies they are not mobile, so to transfer them from one body to another they must be in contact.

The force electric between charges is of two types:

  • Attractive. If the charges are of different signs.
  • Repulsive.  If  the charges are of the same sign.

The transferred charge depends on the bodies involved in the tables showns that the resin is positively charged when rubbed, the PVC acquires very little charge and polystyrene acquires a negative charge due to rubbing.

They indicate that the resin is rubbed with polystyrene, therefore the resin must acquire a positive charge and the polystyrene a negative charge.

Then the PVC is rubbed with the polystyrene, which initially we will assume neutral, the PVC does not acquire a charge and the polystyrenes acquires a negative charge, if the polystyrene is isolated from the ground, it shares this negative charge with the PVC, therefore the two materials remain with half of the negative charge.

Finally, we bring the resin, which is positively charged, closer to the PVC, which has a slight negative charge, therefore the two bodies must attract each other.

In conclusion using the concepts of electrostatics we can find the result for the question about the electric forces in the two tubes is:

  • The tubes are attracted by having charges of different signs.

Learn more about electrostatics here: brainly.com/question/17692887

6 0
2 years ago
In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and part
lesya [120]

Answer:

Charge of particle 2, q_2=-7.13\ \mu C

Explanation:

Given that,

Charge 1, q_1=3.11\ \mu C=3.11\times 10^{-6}\ C

The distance between charges, r = 0.241 m

Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

F=k\dfrac{q_1q_2}{r^2}

q_2=\dfrac{Fr^2}{kq_1}

q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}

q_2=7.13\times 10^{-6}\ C

or

q_2=7.13\ \mu C

So, the magnitude of electric charge 2 is q_2=7.13\ \mu C. Since, the force is attractive then the magnitude of charge 2 must be negative.

5 0
3 years ago
If the discovery of oxygen wasn’t made in 1774, do you think it would have been discovered later on?
Scilla [17]
I don’t know sorry I’m new
8 0
2 years ago
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