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4 years ago

9

What causes the earths crust to move ?

2 answers:

tangare [24]4 years ago

8 0

Upwelling currents in the molten material beneath the crust.

The rocks that make up the crust are light, compared with the metal-rich material beneath. The crust floats on top like an iceberg. Slow-moving currents underneath propel the continents around the surface

Studentka2010 [4]4 years ago

7 0

The movement of magma underneath earth's crust.

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An object has a mass of 20 grams and a volume of 5 cubic centimeters what is the object density

Solnce55 [7]

20 g / 5 cm^3 = 4 g/cm^3 or 4000 kg / m^3.

3 0

3 years ago

which words are readable red word on white paper in red light or blue words on white paper in blue light

Angelina_Jolie [31]

Answer:

second

Explanation:

the first let's everything appear red, the send would produce black(-ish) words on paper that's appealing blue

6 0

4 years ago

Read 2 more answers

Consider a uniform sphere, which has a mass of 4.80 kg and a radius of 22.0 cm. A tangential force of 11.2 N is applied to the o

Tcecarenko [31]

Answer:

The moment of inertia of this sphere is $0.0929\ kg-m^2$ .

Explanation:

It is given that,

Mass of the sphere, m = 4.8 kg

Radius of the sphere, r = 22 cm = 0.22 m

Tangential force, F = 11.2 N

The moment of inertia of the uniform sphere is given by :

$I=\dfrac{2}{5}mr^2$

$I=\dfrac{2}{5}\times 4.8\ kg\times (0.22\ m)^2$

$I=0.0929\ kg-m^2$

So, the moment of inertia of this sphere is $0.0929\ kg-m^2$ . Hence, this is the required solution.

8 0

3 years ago

What angular velocity in revolution per second is needed for a centrifuge to produce an centripetal acceleration of a=9800m/s sq

max2010maxim [7]

Option (d) is correct.

Explanation:

centripetal acceleration = a = r ω²

r= radius= 10 cm= 0.1 m

ω= angular velocity

9800= (0.1) ω²

ω=313 rad/sec

ω= 313 rad/s * (1 rev/2π rad)

ω=49.8 rev/s

7 0

3 years ago

(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic fiel

sammy [17]

Answer:

Part a)

$R_1 = 0.072 m$

Part b)

$R_2 = 0.036 m$

Part c)

d = 0.072 m

Explanation:

Part a)

As we know that the radius of the charge particle in constant magnetic field is given by

$R = \frac{mv}{qB}$

now for single ionized we have

$R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}$

$R_1 = 0.072 m$

Part b)

Similarly for doubly ionized ion we will have the same equation

$R = \frac{mv}{qB}$

$R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}$

$R_2 = 0.036 m$

Part c)

The distance between the two particles are half of the loop will be given as

$d = 2(R_1 - R_2)$

$d = 2(0.072 - 0.036)$

$d = 0.072 m$

6 0

3 years ago