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2 years ago

In which direction does gravitational force act

Physics

1 answer:

Sunny_sXe [5.5K]2 years ago

4 0

Gravitational force pushes down on us to keep us on the ground. So it pushes in a downward direction.

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A 30° incline permanently sits on a 1.1 meter high table. Starting from rest a ball rolls off the incline with a velocity of 2m/

exis [7]

It would be a good game for you but if I get a pic I don’t want you can you come to my crib I just

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3 years ago

A 28-kg beginning roller skater is standing in front of a wall. By pushing against the wall, she is propelled backward with a ve

Scorpion4ik [409]

Answer:

33.6 Ns backward.

Explanation:

Impulse: This can be defined as the product of force and time. The S.I unit of impulse is Ns.

From Newton's second law of motion,

Impulse = change in momentum

I = mΔv................................. Equation 1

Where I = impulse, m = mass of the skater, Δv = change in velocity = final velocity - initial velocity.

Given: m = 28 kg, t = 0.8 s, Δv = -1.2-0 = -1.2 m/s (Note: the initial velocity of the skater = 0 m/s)

Substituting into equation 1

I = 28(-1.2)

I = -33.6 Ns

Thus the impulse = 33.6 Ns backward.

3 0

3 years ago

What is parallelogram law of vector addition ???

Vladimir [108]

Answer:

According to the parallelogram law of vector addition if two vectors act along two adjacent sides of a parallelogram(having magnitude equal to the length of the sides) both pointing away from the common vertex, then the resultant is represented by the diagonal of the parallelogram passing through the same common vertex

Explanation:

6 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

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