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3 years ago

9

A. How much work is done by a man standing with a load of 100 kg?ANS=0

1 answer:

damaskus [11]3 years ago

7 0

Explanation:

W = Fd

W for work

F for force

D for displacement

Since the man did not move nor did he move anything, the displacement is 0, thus work is 0 no matter what the force is

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#3: a container has the dimensions of 30 cm x 50 mm x 0.2 m. the density of its contents is 2.5 g/cm3. what is the mass of the s

WARRIOR [948]

Good afternoon!

We calculate the volume of the container in cm³. To do that, we must put the units in cm:

30 cm → 30 cm
50 mm → 5 cm
0.2 m → 20 cm

The volume is:

V = 30 . 5 . 20

V = 3000 cm³

Now, we calculate the mas with the formula:

m = dV

m = 2.5 · 3000

m = 7500 g

Dividing by 1000, we have the mass in kg:

m = 7.5 kg

4 0

4 years ago

Read 2 more answers

g a stone with mass m=1.60 kg IS thrown vertically upward into the air with an initial kinetic energy of 470 J. the drag force a

otez555 [7]

Answer:

Height reached will be 28.35 m

Explanation:

Here we can use the work energy theorem to find the maximum height

As we know by work energy theorem

Work done by gravity + work done by friction = change in kinetic energy

$-mgh - F_f h = 0 - \frac{1}{2}mv_i^2$

now we will have

$-1.60(9.8)(h) - 0.900(h) = - 470$

$-16.58 h = -470$

$h = 28.35 m$

so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction

5 0

3 years ago

Determine the potential difference between two charged parallel plates that are 0.50 cm apart and have an electric field strengt

fiasKO [112]

$E = \frac{V}{r} \\ V = \frac{E}{r} \\ V = \frac{9.0V/cm}{0.5cm} \\ V = 18V$

4 0

4 years ago

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

7 0

4 years ago

A can, containing only air, has its lid tightly screwed on and is left in strong sunlight

allsm [11]

Answer:

Because everyone knows that when you increase temperature activity withn molecules increases they will collide more making the can probably explode

3 0

3 years ago