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3 years ago

Why is it so important to know how the parts of a can opener work

Physics

1 answer:

Dafna1 [17]3 years ago

3 0

Answer:

Hi there!

There are many possible answers!

My best guesses are:

1) Knowing how they work will prevent injury! For example, knowing that the gears twist will stop you from putting your finger in it!

2) Allowing you to use it! Knowing how to line up the edge of the can with the tool will let you use it properly!

Hope this helps

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A person kicks a 4.0-kilogram door with a 48-newton force causing the door to accelerate at 12 meters per second squared. What i

inna [77]

Answer:

-48 N

Explanation:

mass of door (m) = 4 kg

acceleration of the door = 12 m/s^{2}

force exerted by the person = 48 N

From Newton's third law of motion, action and reaction are equal but opposite. Therefore the force exerted on the door by the person which is 48 N will be the same as the force exerted on the person by the door but opposite in its direction, and this would be - 48 N

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3 years ago

How many dots belong in the electron dot diagram of a boron(B) atom

sergij07 [2.7K]

three dots belong in the electron dot diagram of a boron(B) atom.

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3 years ago

Read 2 more answers

Using the formula F = M* A. What is the acceleration of a .5 kg

Katyanochek1 [597]

Answer:32 m/s/s

Explanation: since F=M*A, F=16N, M=0.5kg, A= F/M

A=16/0.5

A=32 m/s/s

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3 years ago

Define inner core <br> How thick is inner core <br> Temperature of inner core

Alex777 [14]

Answer:

The solid material found in the centre of some planets at extremely high temperature and pressure, distinct from the liquid outer core.

about 1250 km

approximately 5700 K (5430 °C or 9806 °F)

Explanation:

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3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

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3 years ago