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xeze [42]
3 years ago
10

two forces whose magnitude are in ratio of 3:5 gives a resultant of 35N.if the angle of inclination is 60degree.calculate the ma

gnitude of each force.​
Physics
1 answer:
nadya68 [22]3 years ago
7 0

Answer:

the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

Explanation:

The computation of the magnitude of the each force is shown below:

Provided that

Ratio of forces = 3: 5

Let us assume the common factor is x

Now

first force =  3x

And, the second force = 5x

Resultant force = 35 N

The Angle between the forces = 60 degrees  

Based on the above information

Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos\theta

35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]

 35 =√ 9x² + 25x² + 15x²    (cos 60° = 0.5)

35 = √49 x²

 x = 5

So, the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

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Reduce the friction at the surface

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2 years ago
The second-order decomposition of hi has a rate constant of 1.80 x 10−3 m−1 s−1. How much hi remains after 45.6 s if the initial
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2.9 M

Explanation:

The concentration-time equation for a second order reaction is:

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3 years ago
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4 years ago
In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the met
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B. The number of electrons emitted from the metal per second increases.

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3 years ago
A 500-Ω resistor, an uncharged 1.50-μF capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b)
Alexeev081 [22]

Answer:

a) 0.01232 A

b) 0.00075 s = 0.75 ms

c) 0.0045323 A = 4.532 mA

d) 3.894 V

Explanation:

R = 500 Ω

V = 6.16 V

C = 1.50 μF

Let Vs be the voltage of the emf source

Let Vc be the voltage across the capacitor at any time

a) Current flows as a result of potential difference between two points. So, the current flows according to difference in voltage between the emf source and the capacitor.

At time t = 0,

There is no voltage on the capacitor; Vc = 0 V

Current in the circuit is given by

I = (Vs - Vc)/R

I = (6.16 - 0)/500

I = 0.01232 A

b) Time constant for an RC circuit is given by τ

τ = RC = (500) (1.5 × 10⁻⁶) = 0.00075 s

c) The current decay in an RC circuit (called decay because the current in the circuit starts to fall as the capacitor's voltage rises as the capacitor charges) is given by

I = I₀ e⁻ᵏᵗ

where k = (1/τ)

I₀ = Current in the circuit at t = 0 s; I₀ = 0.01232 A

At t = τ = 0.00075 s, kt = (τ/τ) = 1

I = 0.01232 e⁻¹ = 0.0045323 A = 4.532 mA

d) The voltage for a charging capacitor is given by

Vc = Vs (1 - e⁻ᵏᵗ)

where k = (1/τ)

At t = τ = 0.00075 s, Vc = ?, Vs = 6.16 V, kt = 1

Vc = 6.16 (1 - e⁻¹) = 6.16 (0.6321)

Vc = 3.894 V

4 0
3 years ago
Read 2 more answers
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