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3 years ago

Carrying Capacity is...?

Physics

1 answer:

Elodia [21]3 years ago

3 0

Carrying capacity is the maximum amount of individuals a space can hold.

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A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m an

Ber [7]

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is $X = 38.3 \ m$

Explanation:

From the question we are told that

The acceleration along the x axis is $a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)$

The position of the object at t = 0 is x = -14.0 m

The velocity at t = 0 s is $v_{0}x = 7.10 m/s$

Generally from the equation for acceleration along x axis we have that

$a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)$

=> $\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt$

=> $V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

At t =0 s and $v_{0}x = 7.10 m/s$

=> $7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1$

=> $K_1 = 7.10$

$\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

=> $\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}$

=> $X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2$

At t =0 s and x = -14.0 m

$-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2$

=> $K_2 = -14$

$X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14$

At t = 10.0 s

$X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14$

=> $X = 38.3 \ m$

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3 years ago

A certain ideal gas has molar heat capacity at constant volume CV. A sample of this gas initially occupies a volume V0 at pressu

ANTONII [103]

Answer:

Explanation:

The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:

$P_{2} \cdot V_{2}^{\gamma} = P_{3} \cdot V_{3}^{\gamma}$

Where:

$\gamma = \frac{c_{p}}{c_{v}}$

$\gamma = 1 + \frac{R}{c_{v}}$

Final pressure is:

$P_{3} = P_{2}\cdot \left(\frac{V_{2}}{V_{3}} \right)^{\gamma}$

$P_{3} = P_{o}\cdot \left(\frac{1}{2}\right)^{\gamma}$

$P_{3} = \frac{P_{o}}{2^{\gamma}}$

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3 years ago

Ason lives near a state park, and he likes hiking through the grassy trails that lead to a nearby lake. He appreciates nature an

ddd [48]

Answer:.

Explanation:

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3 years ago

A virtual image larger than the object can be produced by

ss7ja [257]

Answer:

concave mirror

Explanation:

Why This is Correct Because, concave mirrors only form a virtual image when a object is larger than the other. when the object is produced between the focus, object and the image object, it becomes a virtual image.

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3 years ago

Read 2 more answers

Urgent! please help!

kap26 [50]

$_{94}^{239}Pu+ _{0}^{1}n--->_{40}^{100}Zr+_{x}^{y}Element+2_{0}^{1}n\\ \\ 94+0=40+x+0, x= 54,\\ \\ 239+1 = 100+y+2*1, y=138\\ \\_{54}^{138}Element= _{54}^{138}Xe$

Answer is A.

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3 years ago