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3 years ago

4

Physics

1 answer:

ollegr [7]3 years ago

7 0

Answer:

50 m/s

Explanation:

The motion of the roller coaster car parameters are;

The initial speed of the roller coaster, v = 2 m/s

The distance to reach the bottom, Δd = 200 meter

The time it takes to reach the bottom, Δt = 4 seconds

The speed at the bottom of the tracks = 22 m/s

The average speed, $v_{ave}$ = Δd/Δt

∴ $v_{ave}$ = (200 m)/(4 s) = 50 m/s

The average speed of the roller coaster ride over the 200 m drop, $v_{ave}$ = 50 m/s

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A very powerful vacuum cleaner which has a hose of circular cross section can lift a brick of mass 12 kg when the hose is placed

valkas [14]

Answer:

A). 1.9 cm

Explanation:

m = Mass of brick = 12 kg

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of hose

A = Area = $\pi r^2$

F = Force = $mg$

Let us assume that the pressure required to lift the brick would be atmospheric pressure

$P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{F}{\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{F}{\pi P}}\\\Rightarrow r=\sqrt{\dfrac{12\times 9.81}{\pi\times 101325}}\\\Rightarrow r=0.01923\ m=1.9\ cm$

The radius of the hose should be 1.9 cm

6 0

3 years ago

what is the value of the constant for a second order reaction if the reactant concentration drops from .657 M to ,0981 M in 17 s

yaroslaw [1]

Answer : The value of the constant for a second order reaction is, $0.51M^{-1}s^{-1}$

Explanation :

The expression used for second order kinetics is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = ?

t = time = 17s

$[A_t]$ = final concentration = 0.0981 M

$[A_o]$ = initial concentration = 0.657 M

Now put all the given values in the above expression, we get:

$k\times 17s=\frac{1}{0.0981M}-\frac{1}{0.657M}$

$k=0.51M^{-1}s^{-1}$

Therefore, the value of the constant for a second order reaction is, $0.51M^{-1}s^{-1}$

6 0

3 years ago

A vertical wire carries a current straight up in a region where the magnetic field vector points due north. What is the directio

Elanso [62]

Answer:

The direction of the resulting force on this current is due east.

Explanation:

Given;

direction of the magnetic field to be due north

Applying right hand rule which states that: to determine the direction of the magnetic force on a positive moving charge point the thumb of the right hand in the direction of velocity v, the fingers in the direction of magnetic field B, and a perpendicular to the palm points in the direction of magnetic force.

Since the magnetic force must be perpendicular to the magnetic field, and direction of the magnetic field is due north, then the magnetic force must be due East.

Therefore, the direction of the resulting force on this current is due east.

7 0

3 years ago

A manager at a local bank analyzed the relationship between monthly salary and three independent variables: length of service (m

Readme [11.4K]

Answer:

Explanation:

8 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

3 years ago