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Lerok [7]
3 years ago
8

What is the acceleration experiance by a car that takes 10s to reach 27m/s from rest

Physics
1 answer:
lina2011 [118]3 years ago
4 0

Magnitude of acceleration = (change in speed) / (time for the change).

Change in speed  =  (27 - 0)  =  27 m/s
Time for the change = 10 s

Magnitude of acceleration =  (27 m/s) / (10 s)  =  2.7 m/s²  .
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A string that is under 54.0 N of tension has linear density 5.20 g/m . A sinusoidal wave with amplitude 2.50 cm and wavelength 1
kicyunya [14]

Answer:

8.89288275 m/s

Explanation:

F = Tension = 54 N

\mu = Linear density of string = 5.2 g/m

A = Amplitude = 2.5 cm

Wave velocity is given by

v=\sqrt{\frac{F}{\mu}}\\\Rightarrow v=\sqrt{\frac{54}{5.2\times 10^{-3}}}\\\Rightarrow v=101.90493\ m/s

Frequency is given by

f=\frac{v}{\lambda}\\\Rightarrow f=\frac{101.90493}{1.8}\\\Rightarrow f=56.61385\ Hz

Angular frequency is given by

\omega=2\pi f\\\Rightarrow \omega=2\pi 56.61385\\\Rightarrow \omega=355.71531\ rad/s

Maximum velocity of a particle is given by

v_m=A\omega\\\Rightarrow v_m=0.025\times 355.71531\\\Rightarrow v_m=8.89288275\ m/s

The maximum velocity of a particle on the string is 8.89288275 m/s

5 0
2 years ago
Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of ea
dimaraw [331]

Answer:

stone A is diamond.

Explanation:

given,

Volume of the two stone =  0.15 cm³

Mass of stone A = 0.52 g

Mass of stone B = 0.42 g

Density of the diamond =  3.5 g/cm³

So, to find which stone is gold we have to calculate the density of both the stone.

We know,

density\density = \dfrac{mass}{volume}

density of stone A

\rho_A = \dfrac{0.52}{0.15}

\rho_A = 3.467\ g/cm^3

density of stone B.

\rho_B = \dfrac{0.42}{0.15}

\rho_B = 2.8\ g/cm^3

Hence, the density of the stone A is the equal to Diamond then stone A is diamond.

6 0
3 years ago
An isloated point charge produce an electric field with magnitude E at a point 2 m away. At a point 1 m from the charge magnitud
Nina [5.8K]

Answer:

the correct answer is C,      E’= 4E

Explanation:

In this exercise you are asked to calculate the electric field at a given point

         E = k \frac{q}{r^2}

indicates that the field is E for r = 2m

         E = \frac{ k q}{4}                  (1)

the field is requested for a distance r = 1 m

         E ’= k \frac{q}{r'^2}

         E ’= k q / 1

 

from equation 1

         4E = k q

       

we substitute

        E’= 4E

so the correct answer is C

8 0
2 years ago
Sanitation can be accomplished by all of the means except
Nonamiya [84]

Answer:

Air drying

Explanation:

Sanitation is the set of measures that aims to preserve or modify the conditions of the environment in order to prevent diseases and promote health, improve the quality of life of the population and the productivity of the individual and facilitate economic activity. Basic sanitation is a right guaranteed by the Constitution as the set of services, infrastructure and operational installations for water supply, sanitary sewage, urban cleaning, urban drainage, solid waste and rainwater management.

Sanitation can be carried out in several ways, except for dry air, as this would damage the quality of sanitation and the environment.

Sanitation is important in treated water, sewage collection and treatment services lead to an improvement in the quality of people's lives, especially in child health, with a reduction in child mortality, improvements in education, in the expansion of tourism, in the valuation of properties, in income of the worker, in the cleaning of rivers and preservation of water resources, etc.

5 0
3 years ago
Which of the following changes would cause the fusion rate in the Sun’s core to increase? Check all that apply. View Available H
fredd [130]

Answer:

1. An increase in the core temperature

2. A decrease in the core radius.

Explanation:

The sun is a Main Sequence star. A Main Sequence star is powered by fusing hydrogen into Helium within its core.

For this fusion to take place, a temperature of at least 10 million Kelvin is required, beyond this point, the fusion rate is directly related to the core temperature. If the temperature increases, the fusion rate will greatly increase.

Something similar happens if the core reduces its radius. This can happen at the end of the star's lifetime, shortly before it becomes a red giant. Once the hydrogen is depleted, the core will start to shrink because the force of gravity, and as it gets smaller, gets more compressed, and its temperature increases. The outer layers of remaining Hydrogen that were outside the core now begin to heat up, and as the core continues to shrink, the star gets hot enough to begin the fusion process again, and the fusion rate can even be higher than it was during the first phase of the star, as the star becomes a Red Giant.

7 0
2 years ago
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