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3 years ago

The process by which metamorphic rock changes to igneous rock begins with?

Physics

1 answer:

Strike441 [17]3 years ago

5 0

Melting, as igneous rock is magma or lava that has cooled and hardened.

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In August 2011, the Juno spacecraft was launched from Earth with the mission of orbiting Jupiter in 2016. The closest distance b

Vikentia [17]

(a) 8927 mi/h

In order to calculate the average speed, we need to convert the time (t=5.0 y) into hours first. In 1 year, we have 365 days, each day consisting of 24 hours, so the time taken is:

$t=(5.0 y)(365 d/y)(24 h/d)=43,800 h$

The distance covered by the spacecraft is

$d=391 mil. mi = 391\cdot 10^6 mi$

Therefore, the average speed is just the ratio between the distance covered and the time taken:

$v=\frac{d}{t}=\frac{391\cdot 10^6 mi}{43,800 h}=8,927 mi/h$

(b) 35 minutes (2097 seconds)

The transmitted signals (which is a radio wave, which is an electromagnetic wave) travels back to the Earth at the speed of light:

$c=3.0\cdot 10^8 m/s$

Since 1 miles = 1609 metres, the distance covered by the signal is

$d=391\cdot 10^6 mi \cdot (1609 m/mi)=6.29\cdot 10^{11} m$

So, the time taken by the signal will be

$t=\frac{d}{v}=\frac{6.29\cdot 10^{11} m}{3.0\cdot 10^8 m/s}=2097 s$

And since 1 minute = 60 sec, the time taken is

$t=2097 s \cdot \frac{1}{60 s/min}\sim 35 min$

7 0

3 years ago

A spring loaded toy shoots straight upward with a velocity of 4.5 m/s. Determine the maximum height it reaches. Determine the ti

Angelina_Jolie [31]

Recall that

${v_y}^2-{v_{0y}}^2=2a_y(y-y_0)$

At its maximum height $y_{\mathrm{max}}$ , the toy will have 0 vertical velocity, so that

$-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}$

$\implies y_{\mathrm{max}}=1.0\,\mathrm m$

For the toy to reach this maximum height, it takes time $t$ such that

$\dfrac{0+v_0}2t=y_{\mathrm{max}}\implies t=0.46\,\mathrm s$

which means it takes twice this time, i.e. $t=0.92\,\mathrm s$ , for the toy to reach its original position.

The velocity of the toy when it falls 1.0 m below its starting point is

${v_y}^2-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(0-1.0\,\mathrm m)$

$\implies{v_y}^2=39.85\,\dfrac{\mathrm m^2}{\mathrm s^2}$

$\implies v_y=-6.4\,\dfrac{\mathrm m}{\mathrm s}$

where we took the negative square root because we expect the toy to be moving in the downward direction.

6 0

3 years ago

Suppose the voltage source for a series RL-circuit were given as V0sin(ωt) instead of V0cos(ωt). Write an expression for the cur

Gala2k [10]

Answer:

Explanation:

This is an RL circuit, therefore:

Impedance; z = $\mathbf{\sqrt{R^2+L^2}}$

$\mathbf{z = \sqrt{R^2+(Lw)^2}}$

Current amplitude

$\mathbf{I_o = \dfrac{V_o}{z}} \\ \\ \mathbf{I_o = \dfrac{V_o}{\sqrt{R^2+L^2\omega ^2}}}$

Given that:

$V_o = 1.9 \ V \\ \\ \omega= 51 \ rad/s\\\\ R = 21 \Omega \\ \\ L = 0.52 H$

∴

$I_o= \dfrac{1.9}{\sqrt{21^2+(0.52\times 51)^2}}$

$\mathbf{I_o= 0.0562} \\ \\ \mathbf{I_o = 56.2 \ mA}$

Phase constant :

$tan \ \phi = \dfrac{L \omega}{R } \\ \\ tan \ \phi = \dfrac{0.52 \times 51}{21} \\ \\ tan \phi = 1.263$

$\text{Phase constant : }\phi = tan^{-1} (1.263) \\ \\ \phi = 51.6^0\\ \\\text{To radians} \phi = 51.6 \times \dfrac{\pi}{180} \\ \\ \phi = 0.287 \pi \\ \\ \mathbf{\phi = 0.9 \ rad}$

3 0

3 years ago

8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott

nasty-shy [4]

Answer: The ball is travelling with a speed of 5.5 m/s after hitting the bottle.

Explanation:

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

$m_1,u_1\text{ and }v_1$ are the mass, initial velocity and final velocity of ball.

$m_2,u_2\text{ and }v_2$ are the mass, initial velocity and final velocity of bottle.

We are given:

$m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s$

Putting values in above equation, we get:

$(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s$

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0

3 years ago

How are contact forces and field forces different ?

Sedaia [141]

field is force at a distnce = falling under grav ... contact is force at point of contact

6 0

3 years ago