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il63 [147K]
2 years ago
12

An 80 N rightward force is applied to a 10 kg object to accelerate it to the right.

Physics
1 answer:
ella [17]2 years ago
3 0

net force = 30 N

mass = 8.16 kg

acceleration = 3.68 m/s²

<h3>Further explanation</h3>

Given

80 N force applied

mass of object = 10 kg

Friction force = 50 N

Required

Net force

mass

acceleration

Solution

  • net force

Net force = force applied(to the right) - friction force(to the left)

Net force = 80 - 50 = 30 N

  • mass

Gravitational force(downward) : F = mg

m = F : g

m = 80 : 9.8

m = 8.16 kg

  • acceleration

a = F net / m

a = 30 / 8.16

a = 3.68 m/s²

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\lambda=6.83\times 10^{-5}\ m

Explanation:

Given that,

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5 0
2 years ago
Two coulombs of charge pass through a wire in 5 seconds. Calculate the current in the wire. 1​
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0.4A.

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7 0
2 years ago
Glycerin is poured into an open U-shaped tube until the height in both sides is 20cm. Ethyl alcohol is then poured into one arm
Liula [17]

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining \rho_A: Ethyl alcohol density and \rho_G: Glycerin density , we can write:

\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3

Simplifying:

\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)

On the other hand:

h_1 + h_2 = \Delta h + h_3

Rearranging:

h_1 - \Delta h = h_3 - h_2 (2)

Replacing  (2) in (1):

\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)

Rearranging:

\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h

Data:

h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3}

\frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}}  = \Delta h

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7 0
3 years ago
Below is an oscilloscope trace from an AC supply. Calculate the frequency of the supply if each horizontal division represents 0
GuDViN [60]

Answer:

20 hertz of frequency produced.

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Here we will find frequency and period should be in second, here given: 0.05 seconds

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8 0
2 years ago
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This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

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Hence we sense it as 4 times fainter than the nearer star.

5 0
2 years ago
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