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krek1111 [17]
3 years ago
5

When you lift a book of the ground to your desk, what kind of work do you do, negative or positive?

Physics
1 answer:
blondinia [14]3 years ago
8 0
It would be kinetic energy. Let's say the book is weighs 10 Newtons you need to use a force of 10 Newtons to lift the book. In other words it's positive. As you move the book you're giving it energy. Namely potential energy which will turn to kinetic energy if you let it go. So you're changing it's position and energy.
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Does adding electrons change the mass
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“Gaining or losing electrons has no impact on the number of protons and neutrons an atom has, which is another way of saying that the mass number is not affected by changes in the net charge.”
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Explain the difference in polarity between co2 and h2o by referring to the polarity of the bonds and the shape of the molecules
erik [133]
<span>CO2 is a linear molecule so despite of having to dipolar bonds they add up and exactly cancel each other rendering the molecule non-polar.</span>
4 0
3 years ago
Read 2 more answers
A cannon sends a projectile towards a target a distance 1420 m away. The initial velocity makes an angle 35◦ with the horizontal
ss7ja [257]

Answer:

v_{o}=141.51m/s

Explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

x=1420m\\g=-9.8m/s^{2}

From the equation of x-position we know that

x=v_{ox}t=v_{o}cos(35)t

Solving for v_{o}

v_{o}=\frac{x}{tcos(35)} =\frac{1420m}{tcos(35)} (1)

Now, if we analyze the equation of y-position we got

y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}

At the end of the motion y=0

0=v_{o}sin(35)t+\frac{1}{2}gt^{2}

Knowing the equation for v_{o} in (1)

0=\frac{1420}{tcos(35)}tsin(35)-\frac{1}{2}(9.8)t^{2}

\frac{1}{2}(9.8)t^{2}=1420tan(35)

Solving for t

t=\sqrt{\frac{2(1420tan(35))}{9.8} } =14.25s

Now, we can solve (1)

v_{o}=\frac{1420m}{(14.25s)cos(35)}=141.51m/s

6 0
3 years ago
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Which graph represents the relationship between the magnitude of the gravitational force exerted by earth on a spacecraft the di
Neko [114]

Answer:

B as distance increase force decrease, but it is not a linear relationship.

8 0
3 years ago
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

4 0
3 years ago
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