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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?

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

Lostsunrise [7]

“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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3 years ago

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>

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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)

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

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3 years ago

Read 2 more answers

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