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4 years ago

Which types of magma has the highest viscosity?

Physics

1 answer:

MAXImum [283]4 years ago

5 0

Answer:

Acidic magma

Explanation:

Viscosity simply refers to the resistance to flow. The magma is comprised of different minerals intermixed with various gases.

There are three main factors that determine the viscosity of the magma, namely-

The temperature of the magma

Magma composition

Presence of dissolved gases in the magma

The acidic magma has the highest viscosity, as this type of magma is comprised of more than 65% of silica content and are mixed with the dissolved gases such as CO₂, SO₂, H₂S, N₂, CH₄ and so on.

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A ray of light strikes a mirror. The angle formed between the incident ray and the reflected ray measures 64 º. What are the mea

julia-pushkina [17]

Answer:

Draw line perpendicular to mirror

Angle of incidence = angle between incident ray and perpendicular

Angle of reflection = angle between reflected ray and perpendicular

32 + 32 = 64

Angle of incidence = angle of reflection = 32 deg

6 0

2 years ago

An object takes 5 seconds to move 2 meters upward. How fast did it go?

jonny [76]

Answer:

2.5

Explanation:

5/2=2.5

8 0

3 years ago

Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.30 kg and is initially moving to t

snow_lady [41]

Answer:

a) 3.632 m/s

b) 0.462 m/s

Explanation:

Using the law of conservation of momentum:

$m_{1} u_{1} + m_{2} u_{2}= m_{1} V_{1} + m_{2} V_{2}$ ..........(1)

$m_{1} = 0.30 kg\\u_{1} = 2.4 m/s\\m_{2} = 0.80 kg\\u_{2} = 0 m/s$

Substituting the above values into equation (1) and make V2 the subject of the formula:

$0.3(2.4) + 0.80(0)= 0.3 V_{1} + 0.8 V_{2}\\$

$V_{2} = \frac{0.72 - 0.3 V_{1}}{0.8}$ ..................(2)

Using the law of conservation of kinetic energy:

$0.5m_{1} u_{1} ^{2} + 1.2 = 0.5m_{1} V_{1} ^{2} + 0.5m_{2} V_{2} ^{2}\\0.5(0.3) (2.4) ^{2} + 1.2 = 0.5(0.3) V_{1} ^{2} + 0.5(0.8)V_{2} ^{2}\\$

$2.064 = 0.15 V_{1} ^{2} + 0.4V_{2} ^{2}$ .......(3)

Substitute equation (2) into equation (3)

$2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.72 - 0.3V_{1} }{0.8}) ^{2}\\2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.5184 - 0.432V_{1} + 0.09V_{1} ^{2} }{0.64}) \\1.32096 = 0.096 V_{1} ^{2} + 0.20736 - 0.1728V_{1} + 0.036V_{1} ^{2} \\0.132 V_{1} ^{2} - 0.1728V_{1} - 1.1136 = 0\\V_{1} = 3.632 m/s$

Substituting $V_{1}$ into equation(2)

$V_{2} = \frac{0.72 - 0.3 *3.632}{0.8}\\V_{2} = \frac{0.72 - 0.3 *(3.632)}{0.8}\\V_{2} = 0.462 m/s$

8 0

3 years ago

If the work function of a material is such that red light of wavelength 700 nm just barely initiates the photoelectric effect, w

marishachu [46]

Answer:

2.13 x 10^-19 J or 0.53 eV

Explanation:

cut off wavelength, λo = 700 nm = 700 x 10^-9 m

λ = 400 nm = 400 x 10^-9 m

Use the energy equation

$E = \frac{h c}{\lambda _{o}}+K$

Where, K be the work function

$\frac{h c}{\lambda} = \frac{h c}{\lambda _{o}}+K$

$K =hc\left ( \frac{1}{\lambda } -\frac{1}{\lambda _{0}}\right )$

$K =6.63\times 10^{-34}\times 3\times 10^{8}\left ( \frac{1}{4\times 10^{-7} } -\frac{1}{7\times 10^{-7}}\right )$

K = 2.13 x 10^-19 J

K = 0.53 eV

3 0

3 years ago

a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially

CaHeK987 [17]

Answer:

Velocity of the helium nuleus = 1.44x10⁴m/s

Velocity of the proton = 2.16x10⁴m/s

Explanation:

From the conservation of linear momentum of the proton collision with the He nucleus:

$P_{1i} + P_{2i} = P_{1f} + P_{2f]$ (1)

where $P_{1i}$ : is the proton linear momentum initial, $P_{2i}$ : is the helium nucleus linear momentum initial, $P_{1f}$ : is the proton linear momentum final, $P_{2f}$ : is the helium nucleus linear momentum final

From (1):

$m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f}$ (2)

where m₁ and m₂: are the proton and helium mass, respectively, $v_{1i}$ and $v_{2i}$ : are the proton and helium nucleus velocities, respectively, before the collision, and $v_{1f}$ and $v_{2f}$ : are the proton and helium nucleus velocities, respectively, after the collision

By conservation of energy, we have:

$K_{1i} + K_{2i} = K_{1f} + K_{2f}$ (3)

where $K_{1i}$ and $K_{2i}$ : are the kinetic energy for the proton and helium, respectively, before the colission, and $K_{1f}$ and $K_{2f}$ : are the kinetic energy for the proton and helium, respectively, after the colission

From (3):

$\frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2}$ (4)

Now we have two equations: (2) ad (4), and two incognits: $v_{1f}$ and $v_{2f}$ .

Solving equation (2) for $v_{1f}$ , we have:

$v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f}$ (5)

From getting (5) into (4) we can obtain the $v_{2f}$ :

$v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0$

$v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0$

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

$v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s}$ (6)

Now, by introducing (6) into (5) we get the proton velocity after the collision:

$v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4}$

$v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s}$

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!

7 0

4 years ago