A block of stone measures 15cm x 15cm x 20 cm.What is the total surface area of the stone

When you speak, your voice is produced by your vibrating vocal chords. These vibrations produce the ________ which carry _______

Len [333]

Produce waves that carry sound

7 0

2 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

$\rho(r) = ar^2 - br^3$

r is the radius of the spherical shell

dr is the thickness

volume of shell

$dV = 4 \pi r^2 dr$

mass of shell

$dM = \rho(r)dV$

$\rho = \rho_0 - br$

now,

$dM = (\rho_0 - br)(4 \pi r^2)dr$

integrating both side

$M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr$

$M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})$

$M = \pi R^3(\dfrac{\rho_0}{3}+\rho)$

we know,

$a = \dfrac{GM}{R^2}$

$a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}$

$a =\pi RG(\dfrac{\rho_0}{3}+\rho)$

$a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)$

a = 9.94 m/s²

7 0

2 years ago

A box (m = 20 kg) is sliding on a horizontal surface. it is connected to a massless hook by a light string passing over a massle

Varvara68 [4.7K]

mass of the box = 20 kg

force of friction on the box due to surface

$F_s = \mu_s N$

$F_s = 0.80 * 20 * 9.8$

$F_s = 156.8 N$

similarly kinetic friction on it

$F_k = \mu_k N$

$F_k = 0.30* 20 * 9.8$

$F_k = 58.8 N$

now the weight of the suspended block will be

$W = mg = 15*9.8 = 147 N$

so here the weight of the suspended block is less than the limiting friction on it

So here we will say that friction will counter balance the weight of the suspended block and it will not move at all

So acceleration of the box will be zero

5 0

2 years ago

A block of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. A force F of magnitude 5.

Marina86 [1]

Answer:

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

Coefficient of static friction = 11.71/18.44= 0.64

Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)

b) coefficient of kinetic friction = frictional force/ normal force

Fr = 0.4* mgcos 20o = 7.375N

F due to motion = ma = total force - frictional force

Ma = 11.71 - 7.375 = 4.335

a= 4.335/2(mass of the block) = 2.17m/s^2

C) work done = net force *distance = 4.335*2= 8.67Joules

8 0

2 years ago

Which is NOT an adaptation made by mammals during the Cenozoic era?

Anettt [7]

Becoming Cold Blooded im pretty sure

3 0

2 years ago

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