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3 years ago

Nick raises a 4-kg mass vertically 0.05 meters above his head. Determine the work done on the mass.

1 answer:

8 0

The work done by Nick as he raises the given mass over the given distance is approximately 2 Joules.

Given the data in the question:

Mass raised; $m = 4kg$

Distance covered; $d = 0.05m$

Work done; $W = \ ?$

First we calculate the force used to raise the mass to the given distance.

Since the movement was vertical, the force must over come the mass and the pull of gravity ( $g = 9.8m/s^2$ ).

Using Newton's second law of Motion;

$F = m * g$

We substitute our values into the equation

$F = 4kg * 9.8m/s^2\\\\F = 39.2 kgm/s^2$

Now, we determine the work done by Nick.

Work done is simply the energy transferred from one store to another.

It is expressed as;

$Work\ done = force * distance\\\\W = F * d$

So we substitute our values into the equation

$W = 39.2kgm/s^2 * 0.05m\\\\W = 1.96kgm^2/s^2\\\\W = 2 J$

Therefore, the work done by Nick as he raises the given mass over the given distance is approximately 2 Joules.

Learn more about Newton's law of motion: brainly.com/question/20066791

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Chromosomes that carry the same set of genes is called a ________ Chromosome.

Leokris [45]

It is called a homologous chromosome meaning it carries the same gene

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3 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

6 0

3 years ago

The North American Plate is moving west at a rate of approximately 20 mm/yr. How long will it take for New York to move 10° long

scZoUnD [109]

Answer:

55000000 years

Explanation:

Rate of North American Plate moving (Velocity) = 20 mm/yr

Degrees New York has to move = 10° west

New York's longitude = 110 km/°

Distance of New York = 110×10 = 1100 km

= 1100×10⁶ mm

$\text {Time taken by the continental plate}=\frac {\text {Distance of New York}}{\text {Rate of North American Plate moving (Velocity)}}\\\Rightarrow \text {Time taken by the continental plate}=\frac{1100\times 10^6}{20}\\\Rightarrow \text {Time taken by the continental plate}=55\times 10^6\ years$