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3 years ago

A roller coaster car picks up speed as it rolls down the slope. It accelerates at a rate of 6 m/s. What is its

Physics

1 answer:

Rom4ik [11]3 years ago

5 0

Answer:

velocity = speed of an object Basically, as speed increases, velocity increases...

Explanation:

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A plane is heading south with a velocity of 150 kilometers/hour. It experiences a tallwind with a velocity of 20 kilometers/hour

pogonyaev

Answer: 170km/hr to the South

Explanation:

The resultant velocity of the plane ilwill be gotten by adding the value of the velocities of the plane with the tallwind. It should be noted that the direction of the wind is.takwn into consideration.

Therefore, the resultant velocity of the plane will be:

= 20km/hr + 150km/hr

= 170km/hr to the South

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3 years ago

How can wind put off fire?

kipiarov [429]

♥ If the wind is strong enough it can do so.
♥ By having a strong enough wind you can blow out the fire before the flame can consume any more vapor.
♥ If the wind is fast enough, like a birthday cake candle for example, the wind will burn out.

7 0

3 years ago

Read 2 more answers

A constant force of 5.00 N acts on a 2.50 kg object for 10.0 s. What are the changes in the object’s momentum and velocity?

dimulka [17.4K]

Hope this answer helps, cause Idk, I might be wrong, but I still, I used the correct formulas, so I might be correct

7 0

3 years ago

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$