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lina2011 [118]
3 years ago
11

How much force is needed to accelerate an object of mass 90 kg at a rate of 1.2 m/s2

Physics
2 answers:
algol133 years ago
7 0

Answer : The force needed is 108 N.

Explanation :

Force : It is defined as the product of mass of an object and acceleration of an object.

Formula used :

Force=Mass\times Acceleration

Given:

Acceleration = 1.2m/s^2

Mass = 90 kg

Now put all the given values in this formula, we get:

Force=90kg\times 1.2m/s^2

Force=108kg.m/s^2=108N

The unit of force is, N or kg.m/s^2

Therefore, the force needed is 108 N.

soldier1979 [14.2K]3 years ago
3 0
∑F = ma = (90 kg)(1.2 m/s²) = 108 N = 100 N (1 significant digit)

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The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t
dem82 [27]

Answer:

\frac{M_e}{M_s} = 3.07 \times 10^{-6}

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now for the time period of moon around the earth we can say

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T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

here we know that

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\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}

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Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa
zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, \rho _o = 926 kg/m³

density of the wood, \rho _{wood} = 974 kg/m³

density of water, \rho _w = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood}  -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood}  -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

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3 years ago
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