Answer:
a = 6.97 × 10⁻²⁵ m/s²
Explanation:
given,
mass = 0.42 kg
g = 9.8 m/s²
mass of the earth = 5.98 × 10²⁴ kg
magnitude of earth acceleration = ?
F = m g
F = 0.42 × 9.8
F = 4.17 N
according to newtons third law Force exerted by the earth will be equal to force exerted by apple
m a = 4.17 N
[tex]a = \dfrac{4.17}{5.98 \times 10^{24}}[/tex]
a = 0.697 × 10⁻²⁴ m/s²
a = 6.97 × 10⁻²⁵ m/s²
Hence, acceleration of earth toward apple is equal to a = 6.97 × 10⁻²⁵ m/s²
The magnitude of the earth’s acceleration toward the apple is [tex]6.889 \times 10^{-25}m/s^2[/tex]
According to Newton's second law, the formula for calculating the force is expressed as:
F = mg
m is the mass of the object
g is the acceleration due to gravity
Given the following
mass of apple = 0.42 kg
g = 9.81m/s²
F = 0.42 × 9.81
F = 4.1202 N
Next is to get the magnitude of the earth’s acceleration toward the apple using the formula:
F = ma
[tex]a=\frac{F}{m} \\a =\frac{4.1202}{5.98\times 10^{24}}\\a= 0.6889\times 10^{-24}\\a=6.889 \times 10^{-25}m/s^2[/tex]
Hence the magnitude of the earth’s acceleration toward the apple is [tex]6.889 \times 10^{-25}m/s^2[/tex]
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