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3 years ago

Average acceleration of an object is the:

Physics

1 answer:

Andreyy893 years ago

8 0

Answer:

Change in its velocity divided by the change in time

Explanation:

the equation for average acceleration (a) = (vf-vi)/t

a = (velocity final - velocity initial) / change in time

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If the change in velocity increases, what happens to the acceleration during the same time period?

vfiekz [6]

The answer should be “Acceleration increases.”

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3 years ago

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4. Three methods that people use are:

Mila [183]

The answers to question 4 d

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3 years ago

A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is t

GuDViN [60]

Answer:

The magnification is $m = 12$

Explanation:

From the question we are told that

The object distance is $u = 36.2 \ cm$

The focal length is $v = 39.5 \ cm$

From the lens equation we have that

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

=> $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

substituting values

$\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}$

$\frac{1}{v} = -0.0023$

=> $v = \frac{1}{0.0023}$

=> $v =-433.3 \ cm$

The magnification is mathematically represented as

$m =- \frac{v}{u}$

substituting values

$m =- \frac{-433.3}{36.2}$

$m = 12$

6 0

3 years ago

The lever allows Jeff to lift a much greater weight because

Tcecarenko [31]

The smaller force is applied over a longer distance.

I hope this helps. :)

7 0

3 years ago

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

The density of the flow at the exit is 2.2721 kg/m³

4 0

3 years ago