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3 years ago

A forklift raises a crate weighing 8.35 × 102 newtons to a height of 6.0 meters. What amount of work does the forklift do?

Physics

2 answers:

vovangra [49]3 years ago

8 0

Answer:

You just need to use:W=Fd Where W is work done, F is force, d is distance. Units of work will be Nm

Explanation:

nydimaria [60]3 years ago

3 0

You are almost there, often you are given a mass. But, here you already have a force. You just need to use:W=FdWhere W is work done, F is force, d is distance. Units of work will be Nm

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Newtons Law of motion

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2 years ago

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The pressure of a box pushes down on the floor is 50 Pa if the box weighs 400 N what is the area of the base of the box

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Answer:8m^2

Explanation:

Area=force÷pressure

Area=400÷50

A=8m^2

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3 years ago

A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem

Viktor [21]

Answer:

The energy dissipated in the resistor ${U_k} = \frac{U}{k}$

The energy dissipated in the resistor ${U_k} = kU$

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{q ^2}{C}$

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation $q = CV$ is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

$U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2} kCV ^2$

In this equation, $Uk$ is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2$

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

$U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}$

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

$U_{k} = \frac{1}{2}kCV^2$

Here, V is voltage in the circuit.

Substitute $U =\frac{1}{2} CV^2$ in the equation of ${U_k}$

So,

$U_{k} = \frac{1}{2} kCV^2\\$

$= k(\frac{1}{2} CV^2)$

$U_{k} = kU$

4 0

2 years ago

The drawing shows a golf ball passing through a windmill at a miniature golf course. The windmill has 12 blades and rotates at a

tester [92]

Answer:

vmin = 0.23 m/s

Explanation:

The golf ball must travel a distance equal to its diameter in the time between blade arrivals to avoid being hit. If there are 12 blades and 12 blade openings and they have the same width, then each blade or opening is 1/24 of a circle of is 2π/24 = 0.26 radians across.

Therefore, the time between the edge of one blade moving out of the way and the next blade moving in the way is

time = angular distance/angular velocity

⇒ t = 0.26 rad / 1.35 rad/s = 0.194 s

The golf ball must get completely through the blade path in this time, so must move a distance equal to its diameter in 0.194 s, therefore the speed of the golf ball is

v =d/t

⇒ v = 0.045 m / 0.194 s = 0.23 m/s

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2 years ago

Why is it important to apply critical thinking to spam e-mail?

BARSIC [14]

Spam email can contain malicious computer code and viruses,it can be an attempt to commit fraud,and it can be an attempt to get personal and financial.

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2 years ago

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