Ask question

Ask question

All categories

2 years ago

5

An offshore oil platform houses a drill for drilling oil. The drill descends from the platform and enters the ocean. At the bott

om of the ocean, the drill starts to dig into the rock. It continues drilling into the rock for another 7,000 feet and then stops. Where does the drill stop?

2 answers:

bezimeni [28]2 years ago

4 0

Answer:

in the geosphere

Explanation:

butalik [34]2 years ago

3 0

Answer:

Geosphere

Explanation:

I did it and got the answer right.

You might be interested in

.Ryan boiled a liter of water and then stirred sugar into it, adding more sugar until no more would dissolve in the water, creat

kaheart [24]

Answer: The correct option is that all of the sugar will come out of solution, and pure water will float to the top

Explanation:

Solution in the field of Chemistry is usually made up of two or more substances which contains a solute that dissolves in a solvent.

A solution can either be:

-> Saturated

--> Unsaturated or

-> Supersaturated.

A saturated solution is a solution with solutes that dissolves until it is unable to dissolve anymore leaving the undissolved solute beneath.

When there is mixture of a solute and a solvent in a solution the reactions that occurs are called crystallization and dissolution. Crystallization causes solid solutes to remain undissolved while dissolution is simply the dissolving process of the solute.

When Ryan added more sugar after reaching the saturation point of the mixture, the process of crystallization set in which surpassed the process of dissolution of the sugar solute leading to precipitation of the solute of out the solution.

6 0

3 years ago

A war wolf is a device used during the middle ages to assault fortifications with large rocks. A simple trebuchet is constructed

Katarina [22]

Answer:v=41.23 m/s

Explanation:

Given

mass of heavy object $m_1=52 kg$

distance of $m_1$ from the axle $r_1=14 cm$

mass of rock $m_2=123 gm$

Length of rod $=4.1 m$

distance of $m_2$ from axle $r_2=4.1-0.14=3.96 m$

Net torque acting is

$T_{net}=m_1gr_1-m_2gr_2$

$T_{net}=52\times 0.14\times g-0.123\times 3.96\times g$

$T_{net}=6.793\times 9.8$

$T_{net}=66.57 N-m$

Work done by $T_{net}$ is converted to rock kinetic Energy

thus

$T_{net}\times \theta =\frac{mv^2}{2}$

Where $\theta =angle\ turned =\frac{\pi }{2}$

$v= velocity\ at\ launch$

$66.57\times \frac{\pi }{2}=\frac{0.123\times v^2}{2}$

$v^2=66.57\times \pi$

$v=\sqrt{1700.511}$

$v=41.23 m/s$

3 0

3 years ago

The electric motor of a model train accelerates the train from rest to 0.685 m/s in 21.5 ms. The total mass of the train is 875

Natali [406]

Answer:

P=9.58 W

Explanation:

According to Newton's second law, and assuming friction force as zero:

$F_m=m.a\\F_m=0.875kg*a$

The acceleration is given by:

$a=\frac{\Delta v}{t}\\a=\frac{0.685m/s}{21.5*10^{-3}s}\\\\a=31.9m/s^2$

So the force exerted by the motor is:

$F_m=0.875kg*31.9m/s^2\\F_m=27.9N$

The work done by the motor is given by:

$W_m=F_m*d\\\\d=\frac{1}{2}*a*t^2\\d=\frac{1}{2}*31.9m/s^2*(21.5*10^{-3}s)^2\\\\d=7.37*10^{-3}m$

$W_m=27.9N*7.37*10^{-3}m\\W_m=0.206J$

And finally, the power is given by:

$P=\frac{W_m}{t}\\P=\frac{0.206J}{21.5*10^{-3}s}\\\\P=9.58W$

8 0

3 years ago

Using the picture above, which ball has the greatest potential energy?

weqwewe [10]

Ball 4 because the higher the elevation is the greater the potential energy it has

4 0

2 years ago

A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k

Veronika [31]

Answer:

$14.98\ \text{kg m/s}$

$45.26^{\circ}$

Explanation:

$P_1$ = Initial momentum of the pin = 13 kg m/s

$P_i$ = Initial momentum of the ball = 18 kg m/s

$P_2$ = Momentum of the ball after hit

$55^{\circ}$ = Angle ball makes with the horizontal after hitting the pin

$\theta$ = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

$P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}$

Momentum in the y direction

$P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}$

$(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}$

The pin's resultant velocity is $14.98\ \text{kg m/s}$

$P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}$

The pin's resultant direction is $45.26^{\circ}$ below the horizontal or to the right.

4 0

2 years ago