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3 years ago

12

Physics problems you don’t need to do 13

1 answer:

galina1969 [7]3 years ago

5 0

Answer:

force of gravity

Explanation:

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The last is 80 meter per second,please in am in exam

inna [77]

Answer:

I have fond the answer

Explanation:

but my camera doesn't work

7 0

3 years ago

Read 2 more answers

An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. Its image is 16.0 cm in front of the mirro

svet-max [94.6K]

We can rearrange the mirror equation before plugging our values in.
1/p = 1/f - 1/q.
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'

So the height of the image produced by the mirror is 9.6cm.

6 0

3 years ago

the atomic number of cesium (Cs) is 55. If an atom of cesium has 78 neutrons, what is the atomic mass of cesium?

Slav-nsk [51]

Atomic mass= number of protons + number of neutrons
$55 + 78 = 133$
hope this helps

7 0

3 years ago

Read 2 more answers

As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o

STatiana [176]

Answer:

3335400 N/m² or 483.75889 lb/in²

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

A = Area = 1.5 cm²

m = Mass of woman = 51 kg

F = Force = mg

When we divide force by area we get pressure

$P=\frac{F}{A}\\\Rightarrow P=\frac{mg}{A}\\\Rightarrow P=\frac{51\times 9.81}{1.5\times 10^{-4}}\\\Rightarrow P=3335400\ N/m^2$

$1\ N/m^2=\frac{1}{6894.757}\ lb/in^2$

$3335400\ N/m^2=3335400\times \frac{1}{6894.757}\ lb/in^2=483.75889\ lb/in^2$

The pressure exerted on the floor is 3335400 N/m² or 483.75889 lb/in²

7 0

3 years ago

A long, solid rod 5.3 cm in radius carries a uniform volume charge density. Part A If the electric field strength at the surface

Blizzard [7]

Answer:

$\rho = 7.35\times \muC/m^3$

Explanation:

given,

Radius of the solid rod, R = 5.3 cm

Electric field strength,E = 22 kN/C

Let the volume charge density be ρ

From Gauss law

$E = \dfrac{Rl}{2\epsilon_0}$

ε₀ is the permitivity of free space

R is the radius of the rod

and also,

$\rho=\dfrac{2E\epsilon_0}{R}$

ρ is the volume charge density

$\rho=\dfrac{2\times 22\times 10^{3}\times 8.854\times 10^{-12}}{0.053}$

$\rho = 7.35\times 10^{-6}\ C/m^3$

$\rho = 7.35\times \muC/m^3$

Hence, the volume charge density is equal to $\rho = 7.35\times \muC/m^3$

6 0

3 years ago