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3 years ago

9

Where would you find an inclined plane in a fan

2 answers:

ivanzaharov [21]3 years ago

8 0

The blades that spin around in the fan, because they are flat and produce work

Licemer1 [7]3 years ago

6 0

I don't think there's none cause the blades would be wedges

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Choose only one correct option. Explanation needed.

Mama L [17]

Answer:

$\large \boxed{\mathrm{C. \ \ \frac{500}{7 \times 15 \times 8} \ g/cm^3 }}$

Explanation:

$\displaystyle \sf Density = \frac{mass}{volume}$

$\displaystyle \rho = \frac{m}{V}$

$\sf mass=500 \ g$

$\sf volume \ of \ a \ cuboid=width \times length \times height=( 7 \times 15 \times 8) \ cm^3$

$\displaystyle \rho = \frac{500}{7 \times 15 \times 8}$

7 0

3 years ago

A snowball is launched horizontally from the top of a building at v = 16.9 m/s. If it lands d = 44 meters from the bottom, how h

Aloiza [94]

$v_x=16,9\frac{m}{s}\\
s=44m\\
g=10\frac{m}{s^2}\\\\
v_x=\frac{s}{t}\Rightarrow t=\frac{s}{v_x}\\\\
t=\frac{44}{16,9}\\\\
t=2,6s\\\\
h=\frac{gt^2}{2}\\\\
h=\frac{10*2,6^2}{2}\\\\
h=5*6,76\\\\
h=33,8m$

6 0

3 years ago

Two satellites, A and B are in different circular orbits

jek_recluse [69]

Answer:

The ratio of the orbital time periods of A and B is $\frac{1}{2}$

Solution:

As per the question:

The orbit of the two satellites is circular

Also,

Orbital speed of A is 2 times the orbital speed of B

$v_{oA} = 2v_{oB}$ (1)

Now, we know that the orbital speed of a satellite for circular orbits is given by:

$v_{o} = \farc{2\piR}{T}$

where

R = Radius of the orbit

Now,

For satellite A:

$v_{oA} = \farc{2\piR}{T_{a}}$

Using eqn (1):

$2v_{oB} = \farc{2\piR}{T_{a}}$ (2)

For satellite B:

$v_{oB} = \farc{2\piR}{T_{b}}$ (3)

Now, comparing eqn (2) and eqn (3):

$\frac{T_{a}}{T_{b}} = \farc{1}{2}$

6 0

2 years ago

The magnitude of the electrostatic force between two identical ions that are separated by a distance of 5.0A is 3.7×10^-9N.a) wh

jeyben [28]

Explanation:

Given that,

Electrostatic force, $F=3.7\times 10^{-9}\ N$

Distance, $r=5\ A=5\times 10^{-10}\ m$

(a) $F=\dfrac{kq^2}{r^2}$ , q is the charge on the ion

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}$

$q=3.2\times 10^{-19}\ C$

(b) Let n is the number of electrons are missing from each ion. It can be calculated as :

$n=\dfrac{q}{e}$

$n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}$

n = 2

Hence, this is the required solution.

8 0

3 years ago

Is the process of introducing a non blood fluid into the blood

vladimir1956 [14]

The lymphatic system

4 0

3 years ago