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marusya05 [52]
2 years ago
13

Rubber rods charged by rubbing with cat fur repel each other. Glass rods charged by rubbing with silk repel each other. A rubber

rod and a glass rod charged respectively as above attract each other. A possible explanation is that a. Any two rubber rods charged this way have opposite charges on them.b. Any two glass rods charged this way have opposite charges on them.c. A rubber rod and a glass rod charged this way have opposite charges on them.d. All rubber rods always have an excess of positive charge on them.e. All glass rods always have an excess of negative charge on them.
Physics
1 answer:
puteri [66]2 years ago
5 0

Answer:

C. A rubber rod and a glass rod charged this way have opposite charges on them.

Explanation:

When a rubber rod is rubbed against cat fur, it acquires a negative charge, it becomes negatively charged.

When you then try to bring two rubber rod's together, they repel because like charges repel.

Meanwhile, when you rub a glass rod against silk, it loses electrons to the silk material and becomes positively charged.

When you bring two positively charged glass rod's together, they repel, because like charges repel.

However, when you bring the rubber rod and a glass rod together, the attract each other because unlike/opposite charges attract.

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A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy
GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

v=\sqrt{v_1^2+v_2^2}

v=\sqrt{5^2+8^2}=9.43\ m/s

Let E is the kinetic energy of the plane. It is given by :

E=\dfrac{1}{2}mv^2

E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0
1 year ago
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568 muons were counted by a detector on the top of Mount Washington in a one hour period of time. Assuming moving muons keep tim
AlladinOne [14]
The answer to this question is:

C-"That moving clocks run slower"

Your Welcome :)
6 0
2 years ago
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The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviatio
enot [183]

Answer:

2.25 %

Explanation:

65-95-99.7 is a rule to remember the precentages that lies around the mean.

at the range of mean (\mu) plus or minus one standard deviation (\sigma), P([\mu-\sigma \leq X \leq \mu+\sigma])\approx 68.3%

at the range of mean plus or minus two standard deviation, P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%

at the range of mean plus or minus three standard deviation, P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%

So, note that  they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) (P(X \leq \mu+2\sigma))  

So we know that the 95.5% is between \mu - 2\sigma = 0.3 -2*0.1 = 0.28 and \mu + 2\sigma = 0.3 +2*0.1 = 0.32, hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than  0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.

So P(X \leq \mu+2\sigma) \approx 2.25%

8 0
1 year ago
A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init
Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

h=\frac{gt^2}{2}

For second stone

h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}---2

Equating 1 &2 we get

\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}

\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

26.46t-35.721-52.92t+142.884=0

t=4.05 s

4 0
1 year ago
An experiment consists of determining the speed of automobiles on a highway by the use of radar equipment. The random variable i
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The random variable in this experiment is a Continuous random variable.

Option D

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The continuous random variable is random variable where the data can take infinite variables. For example random variable is taken for measuring "speed of automobiles" on the highways. The radar instrument depicts time taken by automobile in particular what speed. They are the generalization of discrete random variables not the real numbers as a random data is created. It gives infinite sets of all possible outcomes. It is obvious that outcomes of the instrument depend on some "physical variables" those are not predictable as depends on the situation.

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