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Mandarinka [93]
3 years ago
6

David lifts a book 1.2 meters from the floor to the top of his desk. If the book weighed 0.50 kg, what is the gravitational pote

ntial energy gained by the book? Estimate g as 9.81.
Physics
1 answer:
Anika [276]3 years ago
7 0

Answer:

5.886 J

Explanation:

Given:

The mass of the book is, m=0.5 kg

Height of lift is, \Delta h=1.2 m

Acceleration due to gravity is, g=9.81 m/s²

Now, gain in gravitational potential energy is a function of change in position and is given as:

\Delta PE=mg\Delta h

Here, \Delta PE is the change in gravitational potential energy.

Plug in 0.5 kg for m, 9.81 for g and 1.2 for \Delta h. Solve for \Delta PE

\Delta PE=0.50\times 9.81\times 1.2=5.886\ J

Therefore, the gain in gravitational energy of the book is 5.886 J.

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how much energy must be transferred to a normal incandescent bulb (efficiency 0.05) for it to transfer 9J of energy by light?
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Revise energy transfers and use sankey diagrams to calculate the efficiency of these conversions with BBC ... Efficiency is a measure of how much useful energy is converted. Part of ... This is the Sankey diagram for a typical filament lamp: 100 joules of electrical energy is converted to 10 joules of light energy and 90 joules.

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3 years ago
An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode
sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell v_{0} in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

                v_{x}=v_{0} \times \cos \theta

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of v_{x} and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

           v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}

           v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}

For motion with constant acceleration, we know

            s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}

Along the horizontal, x-axis, we might write this as

            x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}

Measuring distances relative to the firing point means

               x_{0}=0

we know that,

              a_{x}=0

or,

             v_{x}=v_{x 0}=\text { constant }

By applying the values, we get,

           x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}

The acceleration of gravity is vertically downward and is g=-9.8 \mathrm{m} / \mathrm{s}^{2} , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

           y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}

we know, y_{0}=0 and a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}, so,

          y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)

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3 years ago
A 7.00-kg object undergoes an acceleration given by ax=3.00 and ay= 9.00 m/s2. Find (a) the components of the force acting on th
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Answer:

66.4 N

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Because the object has acceleration in two directions and the mass is constant, the force will be in two directions. The component of the forces are:

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The magnitude of the resultant force is given by

F = \sqrt{F_x^2+F_y^2}

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