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4 years ago

A car accelerates at a constant rate from 12 m/s to 27 m/s while it travels 125 m. How long does it take to achieve this speed?

Physics

1 answer:

MArishka [77]4 years ago

3 0

Answer:

6.41 s

Explanation:

Under constant acceleration we know that

average velocity × time taken = displacement

$s=\frac{u+v}{2}t$

$125=\frac{27+12}{2}t$

t = 6.41 s

The proof of used equation is given in the attachment.

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Serving at a speed of 164 km/h, a tennis player hits the ball at a height of 2.23 m and an angle θ below the horizontal. The ser

sweet-ann [11.9K]

Answer:

The angle θ is 6.1° below the horizontal.

Explanation:

Please, see the figure for a description of the situation.

The vector "r" gives the position of the ball and can be expressed as the sum of the vectors rx + ry (see figure).

We know the magnitude of these vectors:

magnitude rx = 11.6 m

magnitude ry = 2.23 m - 0.99 m = 1.24 m

Then:

rx = (11. 6 m, 0)

ry = (0, -1.24 m)

r = (11.6 m + 0 m, 0 m - 1.24 m) = (11.6 m, -1.24 m)

Using trigonometry of right triangles:

magnitude rx = r * cos θ = 11. 6 m

magnitude ry = r * sin θ = 2.23 m - 0.99 = 1.24 m

where r is the magnitude of the vector r

magnitude of vector r:

$r = \sqrt{(11.6m)^{2} + (1.24m)^{2}} = 11.667m$

Then:

cos θ = 11.6 m / 11.667 m

θ = 6.1°

Using ry, we should obtain the same value of θ:

sin θ = 1.24 m/ 11.667 m

θ = 6.1°

( the exact value is obtained if we do not round the module of r)

7 0

3 years ago

Which statement about real gases is true? A) Forces of attraction and repulsion exist between gas particles at close range. B) T

olga55 [171]

Answer: A) Forces of attraction and repulsion exist between gas particles at close range.

Explanation:

The <u>Ideal Gas equation</u> is:

$P.V=n.R.T$

Where:

$P$ is the pressure of the gas

$V$ is the volume of the gas

$n$ the number of moles of gas

$R$ is the gas constant

$T$ is the absolute temperature of the gas in Kelvin

According to this law, molecules in gaseous state do not exert any force among them (attraction or repulsion) and the volume of these molecules is small, therefore negligible in comparison with the volume of the container that contains them. In this sense, real gases can behave approximately to an ideal gas, under conditions of high temperature and low pressures.

However, at low temperatures or high pressures, real gases deviate significantly from ideal gas behavior. This is because at low temperatures molecules begin to move slower, allowing the repulsive and attractive forces among them to take effect. In fact, <u>the attraction forces are responsible for the condensation of the gas</u>. In addition, at high pressures the volume of molecules cannot be approximated to zero, hence the volume of these molecules is not negligible anymore.

7 0

3 years ago

The temperature at the surface of the Sun is approximately 5,300 K, and the temperature at the surface of the Earth is approxima

N76 [4]

Answer:

The entropy change of the Universe that occurs is 19.346 J/K

Explanation:

Given;

temperature of the sun, $T_s$ = 5,300 K

temperature of the Earth, $T_E$ = 293 K

radiation energy transferred by the sun to the earth, E = 6000 J

The sun loses Q of heat and therefore decreases its entropy by the amount

$\delta S_{sun} = \frac{-Q}{T_s}$

The earth gains Q of heat and therefore increases its entropy by the amount

$\delta S_{Earth} = \frac{-Q}{T_E}$

The total entropy change is:

$\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\ = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K$

Therefore, the entropy change of the Universe that occurs is 19.346 J/K

4 0

3 years ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)