A block is on a frictionless table, on earth. The block accelerates at&nbsp;5.3&nbsp;ms2&nbsp;when a 10 N horizontal force is applied to it. The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is&nbsp;1.62&nbsp;ms2. A horizontal force of 5N is applied to the block when it is on the moon. The acceleration imparted to the block is closest to:&nbsp;a)&nbsp;2.9&nbsp;ms2&nbsp;b)&nbsp;3.2&nbsp;ms2&nbsp;c)&nbsp;3.4&nbsp;ms2&nbsp;d)&nbsp;2.7&nbsp;ms2&nbsp;e)&nbsp;2.4&nbsp;ms2

Question

A block is on a frictionless table, on earth. The block accelerates at&amp;nbsp;5.3&amp;nbsp;ms2&amp;nbsp;when a 10 N horizontal force is applied to it. The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is&amp;nbsp;1.62&amp;nbsp;ms2. A horizontal force of 5N is applied to the block when it is on the moon. The acceleration imparted to the block is closest to:&amp;nbsp;a)&amp;nbsp;2.9&amp;nbsp;ms2&amp;nbsp;b)&amp;nbsp;3.2&amp;nbsp;ms2&amp;nbsp;c)&amp;nbsp;3.4&amp;nbsp;ms2&amp;nbsp;d)&amp;nbsp;2.7&amp;nbsp;ms2&amp;nbsp;e)&amp;nbsp;2.4&amp;nbsp;ms2

coffentw · Accepted Answer

First, you must find the mass of the block.  m=Fa=10 N5.3 ms=1.887 kg  (gravity is not factored in because there is no friction onthe table, and it is only moving horizontal)  Now, find the acceleration on the moon:  a=Fm=5 N1.887 kg=2.65 ms  Answer: d.

Jeffrey Jordon · Accepted Answer

When horizontal force is applied on block then as per Newton&#039;s II law we will have  F=ma  10=m(5.3)  so here mass will be  m=1.89 kg  now on same block F = 5 N is applied on surface of Moon  so on surface on moon we will have  F=ma  5=1.89a  a=2.65m/s2  so above is acceleration on surface of moon

Nick Camelot · Accepted Answer

First, let&#039;s determine the mass of the block on Earth. We know that the block accelerates at 5.3ms2 when a 10 N force is applied. Using the formula above, we can rearrange it to solve for mass: m=Fneta. Substituting the given values, we have m=10N5.3ms2.On the moon, the acceleration due to gravity is 1.62ms2. We want to find the acceleration of the block when a 5 N force is applied. Using the same formula, we can solve for the new acceleration: a=Fnetm. Substituting the given values, we have a=5N10N5.3ms2.Simplifying the expression, we get a=5N·5.3ms210N. Canceling out the units, we have a=2.65ms2.The acceleration imparted to the block on the moon is closest to (a)&amp;nbsp;,&amp;nbsp;2.9ms2.

Mr Solver · Accepted Answer

Result:a)2.9ms2Solution:On Earth:The net force acting on the block can be calculated using Newton&#039;s second law:Net&amp;nbsp;force=Mass×AccelerationIn this case, the net force is given as 10 N, and the acceleration is given as 5.3 ms2. We need to find the mass of the block.10N=Mass×5.3ms2Solving for mass:Mass=10N5.3ms2=1.8868kgOn the Moon:The gravitational force acting on the block on the Moon can be calculated using the formula:Gravitational&amp;nbsp;force=Mass×Acceleration&amp;nbsp;due&amp;nbsp;to&amp;nbsp;gravityHere, the acceleration due to gravity on the Moon is given as 1.62 ms2. We need to find the gravitational force acting on the block.Gravitational&amp;nbsp;force=1.8868kg×1.62ms2=3.0533NSince the gravitational force on the Moon is 3.0533 N, and the applied force is 5 N, the net force can be calculated as:Net&amp;nbsp;force=Applied&amp;nbsp;force−Gravitational&amp;nbsp;force=5N−3.0533N=1.9467NNow, we can use Newton&#039;s second law again to find the acceleration on the Moon:Net&amp;nbsp;force=Mass×AccelerationSubstituting the known values:1.9467N=Mass×AccelerationSolving for acceleration:Acceleration=1.9467N1.8868kg=1.0311ms2The acceleration imparted to the block on the Moon is approximately 1.0311ms2.Among the given options, the closest value to this acceleration is a)2.9ms2.

A block is on a frictionless table, on earth. The block accelerates at 5.3\ \frac{m}{s^2} when a 10 N horizontal force is applied to it. The block and

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