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4 years ago

Volcanism on Earth is always associated with plate tectonic activity.

1 answer:

hjlf4 years ago

7 0

Volcanism is associated with two of the plate boundary types: divergent and convergent margins. ... Volcanism can also occur at intraplate volcanoes. These volcanoes are believed to have sources deeper down in the Earth's mantle that remain in a relatively fixed location relative to the always migrating plate boundaries.

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A spaceship approaches the earth with a speed 0.50c. A passenger in the spaceship measures his heartbeat as 9- beats per minute.

JulijaS [17]

Answer:

D) 61 beats per minute

Explanation:

5 0

3 years ago

A vertical scale on a spring balance reads from 0 to 155 N . The scale has a length of 10.0 cm from the 0 to 155 N reading. A fi

Harrizon [31]

Answer:

mass of the fish is 8.11 kg

Explanation:

As we know that the frequency of oscillation of spring block system is given as

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

here we know that the reading of scale varies from 0 to 155 N from length varies from x = 0 to x = 10 cm

Now we have

$k = \frac{155}{0.10} N/m$

$k = 1550 N/m$

so now we have

$2.20 = \frac{1}{2\pi}\sqrt{\frac{1550}{m}}$

$m = 8.11 kg$

so mass of the fish is 8.11 kg

4 0

3 years ago

A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the

insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

$\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}$

Req =3.03Ω , R1 =12.18Ω

$\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}$

$\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}$

$\frac{1}{R2}=0.2479$

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0

4 years ago

Steam enters a one-inlet, two-exit control volume at location (1) at 360°C, 100 bar, with a mass flow rate of 2 kg/s. The inlet

yaroslaw [1]

Answer:

The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

Explanation:

Given that,

Mass flow rate = 2 kg/s

Diameter of inlet pipe = 5.2 cm

Fifteen percent of the flow leaves through location (2) and the remainder leaves at (3)

The mass flow rate is

$m_{2}=0.15\times2$

We need to calculate the mass flow rate at reach exit

Using formula of mass

$m_{3}=m_{1}-m_{2}$

$m_{3}=2-0.15\times2$

$m_{3}=1.7\ kg/s$

We need to calculate the inlet velocity

Using formula of velocity

$v=\dfrac{m}{\rho A}$

Put the value into the formula

$v=\dfrac{2}{42.868\times\dfrac{\pi}{4}\times(5.2\times10^{-2})^2}$

$v=21.9\ m/s$

Hence, The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

7 0

4 years ago

A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.50 107 m/s and experiences an acceleration

KIM [24]

Answer:

A) B = 0.009185 T

B) Drection is negative y-direction

Explanation:

A) We are given;

Speed(v) = 2.5 x 10^(7) m/s

Acceleration (a) = 2.2 x 10^(13) m/s²

We also know that charge of proton(q) = 1.6 x 10^(-19)

Mass of proton(m) = 1.67 x 10^(-27)

Now, Since the proton is moving by circular motion, this force is equal to the centripetal force which is given as;

F = qvBsinθ = ma

Since perpendicular, θ = 90°

And so, sinθ = sin 90 = 1

Thus, qvB = ma

Making B the subject gives;

B = ma/qv

B = (1.67 X 10^(-27) X 2.2 X 10^13)) / (1.6 X 10^(-19) X 2.5 X 10^(7))

= 0.009185 T

B) By use of Flemings right hand rule, we can see that the middle finger points toward negative y-direction, so the magnetic field is in the negative y-direction

4 0

4 years ago

Read 2 more answers