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andrew-mc [135]
3 years ago
14

Why do we break up angled forces into components? How does that help us solve force problems?

Physics
1 answer:
galina1969 [7]3 years ago
7 0

Answer:

As the angle is increased, the acceleration of the object is increased. The explanation of this relates to the components that we have been drawing. As the angle increases, the component of force parallel to the incline increases and the component of force perpendicular to the incline decreases.

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Explain two ways of magnetising an object​
Nina [5.8K]

Answer:

There are two methods generally used to magnetize permanent magnets: static magnetization and pulse magnetization.

5 0
3 years ago
The Henry's law constant for CO2 is 3.6 ✕ 10−2 M/atm at 25°C. What pressure (in atm) of carbon dioxide is needed to maintain a C
xeze [42]

To solve this problem we will use Henry's law. This law states that at a constant temperature, the amount of gas dissolved in a liquid is directly proportional to the partial pressure exerted by that gas on the liquid. Mathematically it is formulated as follows:

C = K_H*P

Where,

K_H = Henry's constant for C02 at 25°C is equal to 3.6*10^{-2}M/atm

C = Gas concentration is 0.19M

Replacing we have,

0.19 M = (3.6*10^-2 M/atm)*P

P = 5.277 atm

Therefore the pressure of carbon dioxide is 5.277 atm

6 0
3 years ago
Help what is the answer
Ainat [17]

Answer:

C

Explanation:

F=k\dfrac{Q_1Q_2}{r^2}= \\\\(8.99 \times 10^9)\dfrac{(-2\times 10^{-4})(8\times 10^{-4})}{0.3^2}\approx 1.6\times 10^4 N

Hope this helps!

5 0
3 years ago
Lillie is running. She increases her initial speed of 30 km/h to 40 km/h so she can win the race. If she takes 0.05 hours to com
Ganezh [65]

Explanation:

First convert the speed into m/s and time into seconds

5 0
3 years ago
A motion sensor emits sound, and detects an echo 0.0115 s after. A short time later, it again emits a sound, and hears an echo a
Mekhanik [1.2K]

Answer:

1.17 m

Explanation:

From the question,

s₁ = vt₁/2................ Equation 1

Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.

Given: v = 343 m/s, t = 0.0115 s

Substitute into equation 1

s₁ = (343×0.0115)/2

s₁ = 1.97 m.

Similarly,

s₂ = vt₂/2.................. Equation 2

Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo

Given: v = 343 m/s, t₂ = 0.0183 s

Substitute into equation 2

s₂ = (343×0.0183)/2

s₂ = 3.14 m

The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁

s₂-s₁ =  (3.14-1.97) m = 1.17 m

7 0
3 years ago
Read 2 more answers
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