QUESTION 1 The speed of sound in air is 340 m/s. What is the wavelength of a soundwave that has a frequency of 968 Hz?

A 13,000 kg helicopter accelerates upward a 0.5 m/s^2 while lifting a 2000 pound car. to the nearest newton, what is the lift fo

Vinil7 [7]

Lift force exerted by the air on the rotors=143244 N

Explanation:

we use Newtons second law

F- (M+m)g=(M+m)a

F= lift force

m= mass of helicopter= 13000 Kg

M= mass of car= 2000 lb=907.2 kg

a= acceleration= 0.5 m/s²

g= acceleration due to gravity

F- (M+m)g=(M+m)a

F=(M+m)(a+g)

F=(13000+907.2)(0.5+9.8)

F=143244 N

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

how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?

ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

5 0

3 years ago

A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare

zvonat [6]

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

$F = \frac{\Delta P}{\Delta t}$

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

$F_{12} = F_{21}$

so we know that

$\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}$

so we have

$\Delta P_{12} = \Delta P_{21}$

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

6 0

3 years ago

2.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific Test

kow [346]

Answer:

0.15625 grams

Explanation:

Half life: It is related to the decay of radioactive material. The duration in which half of the material will be degraded/decayed. That means after half life 50% of the radioactive material will be left. Here the half life is 28 years.

Initial quantity of the sample: 2.5 grams.

After 28 years, the leftover quantity = 1.25 grams

After 56 years, the leftover quantity = 0.625 grams

After 84 Years, the leftover quantity = 0.3125 grams

After 112 years, the leftover quantity = 0.15625 grams

5 0

3 years ago

QUESTION 1 The speed of sound in air is 340 m/s. What is the wavelength of a soundwave that has a frequency of 968 Hz?​

QUESTION 1 The speed of sound in air is 340 m/s. What is the wavelength of a soundwave that has a frequency of 968 Hz?