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Lilit [14]
3 years ago
6

please what do i need to know when writing an alternative to practical exam on SIMPLE HARMONIC MOTION ​

Physics
1 answer:
Ivanshal [37]3 years ago
5 0

Answer:

had trouble

Explanation:

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1. What is technology?<br>techndegy​
Naya [18.7K]

Answer:

the application of scientific knowledge for practical purposes, especially in industry. Another answer:the sum of techniques,skills,methods and processes used in the production of goods or services or in the accomplishment of objectives, such as scientific investigation

6 0
4 years ago
How does the diameter of the disk of milky way galaxy compare to its thickness?
weqwewe [10]
The diameter is about 100 times as great as the thickness.
7 0
4 years ago
Calculate the pressure exerted on the floor by the boy standing on both feet if the weight of the boy is 40kg. Assume that the a
Jlenok [28]

Answer:

P = 1333.33 N

Explanation:

The pressure exerted by the boy on the floor can be calculated by the following equation:

P = \frac{F}{A}

where,

P = Pressure exerted by the boy = ?

F = Force Applied = Weight of Boy = 40 kg = 40 N (since 1 kg = 1N)

A = Area of application of force = 2(Area of one show) = 2(6 cm x 25 cm)

A = 2(0.06 m x 0.25 m) = 0.03 m²

Therefore,

P = \frac{40\ N}{0.03\ m^2}\\\\

<u>P = 1333.33 N</u>

4 0
3 years ago
An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above
AlekseyPX

Answer:

v0 = 25 m/s

vf = 0 m/s

a = -9.80 m/s^2

change in x = 31.89m

but that's only 1/3 of the hight, so i time it by 3 to get 96m

4 0
3 years ago
(a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivi
Mkey [24]

Answer:

a. 960 W b. One 1 kW room heater

Explanation:

a. The rate of heat conduction P = kA(T₂ - T₁)/d where k = 2 × 0.040 W/m-K =  0.080 W/m-K since the thermal conductivity of glass wool is 0.040 W/m-K and that of the material is twice the thermal conductivity of glass wool, A = area of walls = 120 m², T₁ = outside surface temperature = 5.0 °C, T₂ = inside surface temperature = 18.0 °C and d = thickness of wall = 13.0 cm = 0.13 m

P = kA(T₂ - T₁)/d

= 0.080 W/m-K × 120 m²(18.0 °C - 5.0 °C)/0.13 m

= 9.6 Wm/K × 13 K/0.13 m

= 124.8 Wm/0.13 m

= 960 W

b. The number of 1 kW room heater required will be

n = rate of heat conduction/power of one room heater = 960 W/ 1 kW = 960 W/1000 W = 0.96 ≅ 1

So we need only one 1 kW room heater.

6 0
4 years ago
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