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Pavel [41]
3 years ago
10

A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3000 m in 210 seconds. How fast is the car moving

after this time?
Physics
1 answer:
dedylja [7]3 years ago
6 0

Answer:

I dont know i just need the points

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Compare the time period of two simple pendulums of length 4m and 16m at a place.
Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L)  is defined as:

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

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0
2 years ago
An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 m/s, what is the frequency of the
bazaltina [42]

Answer:

the frequency of the second harmonic of the pipe is 425 Hz

Explanation:

Given;

length of the open pipe, L = 0.8 m

velocity of sound, v = 340 m/s

The wavelength of the second harmonic is calculated as follows;

L = A ---> N   +  N--->N   +   N--->A

where;

L is the length of the pipe in the second harmonic

A represents antinode of the wave

N represents the node of the wave

L = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{4} \\\\L = \lambda

The frequency is calculated as follows;

F_1 = \frac{V}{\lambda} = \frac{340}{0.8} = 425 \ Hz

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.

5 0
3 years ago
Peristalsis is the action caused by contractions of muscles in this structure in the digestive system...
ollegr [7]

Answer:

I believe the answer is esophagus

8 0
2 years ago
The magnitude of electrical force between a pair of charged particles is ____ as much when the particles are moved half as far a
Gnesinka [82]

The magnitude of the electrical force between a pair of charged particles is 4 Times as much when the particles are moved half as far apart.

This can be easily understood by Columb's law,

F_{new} = \frac{kQ_{1}Q_{2}}{r^{2}}

which state's that the amount of electrical force experienced by two charged particles is inversely proportional to the square of the distance between them.

∴ \frac{F_{new} }{F_{old} } = \frac{Distance_{new}^{2}  }{Distance_{old}^{2}  }

Now, we know the new distance is half the original distance,

F_{new} = \frac{kQ_{1}Q_{2}}{\frac{r}{2}^{2} } \\F_{new} = 4\frac{kQ_{1}Q_{2}}{r^{2}}

F_{new} = 4F_{old}

The electrical force of attraction or electrostatic force of attraction between two charged particles refers to the amount of attractive or repulsive force that exists between the two charges. This can be calculated by Columb's Law.

A charged particle in physics is a particle that has an electric charge. It might be an ion, such as a molecule or atom having an excess or shortage of electrons in comparison to protons. The same charge is thought to be shared by an electron, a proton, or another primary particle.

Learn more about electrical force here

brainly.com/question/2526815

#SPJ4

8 0
1 year ago
The gamma ray released by each decay carries 140kev of energy. find the total energy e released by decays in the 2 hours.
olya-2409 [2.1K]
First, we would need to know the decaying isotope.
Next, we use the decay formula
A = Ao e^(-kt)
After determining the remaining amount after two hours, the decay reaction can be used to determine the number of gamma rays released. If the given is in terms of mole, then the total energy is
E = 140n KeV where n is the number of moles of gamma rays released
8 0
3 years ago
Read 2 more answers
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