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3 years ago

Some cells divide rapidly. Give an example.

Physics

1 answer:

Travka [436]3 years ago

4 0

Answer:

hey I hope you're having a great day! An example of a cell that divides rapidly are cells in hair follicles!

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An adult human is 60 percent water; a third of this water is in extracellular fluid, and 20 percent of extracellular fluid is in

Yanka [14]

Answer:

32 pounds

Explanation:

The amount of water in the 200 pound person is

200 * 60% = 200*0.6 = 120 pounds

Of the 120 pounds, a third of this is extracellular fluid, the amount of extracellular fluid is

120 / 3 = 40 pounds

20 % of this is in the blood, which amounts to

40 * 20% = 40 * 0.2 = 8 pounds

The rest is interstitial fluid, which is

40 - 8 = 32 pounds

5 0

3 years ago

A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS

Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;

transferred energy + generated energy = 0

(radiation + convection) + generated energy = 0

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0$

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s$

(a) the electrical power generated for still summer day

$T_s = T_{\infty} = 35 ^oC = 308 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W$

(b)the electrical power generated for a breezy winter day

$T_s = T_{\infty} = -15 ^oC = 258 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W$

3 0

2 years ago

A record turntable rotates through 5.0 rad in 2.8 s as it is accelerated uniformly from rest. What is the angular velocity at th

Usimov [2.4K]

Answer:

$\omega_f = 3.584\ rad/s$

Explanation:

given,

turntable rotate to, θ = 5 rad

time, t = 2.8 s

initial angular speed = 0 rad/s

final angular speed = ?

now, using equation of rotational motion

$\theta = \omega_i t + \dfrac{1}{2}\alpha t^2$

$5 = 0+ \dfrac{1}{2}\alpha\times 2.8^2$

$\alpha= \dfrac{10}{2.8^2}$

α = 1.28 rad/s²

now, calculation of angular velocity

$\omega_f = \omega_i + \alpha t$

$\omega_f =0 +1.28\times 2.8$

$\omega_f = 3.584\ rad/s$

hence, the angular velocity at the end is equal to 3.584 rad/s

4 0

3 years ago

Question 18 according to newton's second law of motion, the force acting on an object varies directly with the object's accelera

leonid [27]

Solve for "x"
X=force
18/6=x/9
cross multiply
162=6x
x=27
Hope this helps

3 0

3 years ago

The format specifier ________ is a placeholder for an int value. %d %n %int %s

Grace [21]

%d is a format specifier that is a placeholder for an int value. It tells the compiler that we want to print an integer value that is present in variable a. In this way there are several format specifiers in c.

8 0

3 years ago