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4 years ago

Which of the following is a conductor for electricity?

Physics

1 answer:

zhannawk [14.2K]4 years ago

4 0

It would be wool. (: Like rubbing your feet against carpet ? same affect. (:
Good luck, rockstar, and I hope you pass!

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A rock weighs 110 N in air and has a volume of 0.00337 m3 . What is its apparent weight when submerged in water? The acceleratio

pickupchik [31]

Explanation:

It is given that,

Weight of the rock in air, W = 110 N

Since, W = mg

$m=\dfrac{W}{g}$

$m=\dfrac{110\ N}{9.8\ m/s^2}$

m = 11.22 kg

We need to find the apparent weight of the rock when it is submerged in water. Apparent weight is equal to the weight of liquid displaced i.e.

$M=d\times V$

d is the density of water, $d=1000\ kg/m^3$

V is the volume of rock, $V=0.00337\ m^3$

$M=1000\ kg/m^3\times 0.00337\ m^3$

M = 3.37 kg

The apparent weight in water, W = m - M

$W=7.85\ kg\times 9.8\ m/s^2$

W = 76.93 N

So, the apparent weight of the rock is 76.93 N. Hence, this is the required solution.

4 0

3 years ago

Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum

Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

$\frac{1}{2}mv^2 = mgh$

where

m is the mass of the man

v is the speed after jumping

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s$

In the acceleration phase, we know that the initial velocity is

$u=0$

And the force exerted on the floor is 2.3 times the gravitational force, so

$F=2.3 mg$

This means the net force on you is

$F_{net} = F-mg=2.3mg-mg=1.3 mg$

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

$a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g$

Now we can use the following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

$v^2-u^2=2as$