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Norma-Jean [14]
3 years ago
10

Suppose you have a large boulder in your yard that you’d like to move. How could you break it into smaller pieces without using

a sledgehammer?
Physics
2 answers:
Scrat [10]3 years ago
8 0

Is the question that is trying to be answered here how to move the boulder, or how to break up a boulder? If the purpose of the question is how to move the boulder from A to B, then you roll it.

laiz [17]3 years ago
8 0

-- Build a long, skinny campfire all around the rock, as close as possible to the rock, so the hot coals are right around the underside of the rock, and the flames lick up and around it.

-- Let it burn there for a few hours, so that a lot of the outside of the rock gets really hot.

-- Throw a big bunch of cold water on top of the rock ... enough so that it runs down the whole outside of the rock.

The thermal stresses that suddenly propagate through the rock, on the surface and for some distance into it, crack a layer of some thickness into several pieces, and the pieces fall off onto the ground.

If the cracking and crumbling doesn't extend far enough in, then just pick up the pieces of the layer that DID crumble off, and then go through the  whole same process again, with the inside smaller rock that remains.

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What medical technique uses sound wave pulses and their reflections to map structures and organs inside the body?
garri49 [273]

The correct answer is “C” ultrasound. Hope this helps!

7 0
3 years ago
Read 2 more answers
a 1210 kg roller coaster car is moving 6.33 m/s. as it approaches the station, brakes slow it down to 2.38 m/s over a distance o
Stels [109]

Answer:

4960 N

Explanation:

First, find the acceleration.

Given:

v₀ = 6.33 m/s

v = 2.38 m/s

Δx = 4.20 m

Find: a

v² = v₀² + 2aΔx

(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)

a = -4.10 m/s²

Next, find the force.

F = ma

F = (1210 kg) (-4.10 m/s²)

F = -4960 N

The magnitude of the force is 4960 N.

4 0
3 years ago
Read 2 more answers
What is the slope of the line plotted below? (-2,-1) (4,2) A. 1 B. 2 C. -0.5 D. 0.5
Dmitry [639]

Answer: D. 0.5

Explanation:

The slope formula is y2-y1/x2-x1.

2-(-1)/4-(-2) = 2+1/4+2

2+1/4+2 = 3/6 = 1/2

1/2 = 0.5

The slope is 1/2, or 0.5.

8 0
3 years ago
What is the angle of reflection of the incident angle is 35°?<br> 35<br> 25°<br> 55<br> 90
Anestetic [448]

Answer:

i think its the third one or second one

4 0
2 years ago
Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th
RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2} (1)

Where:

P_{1}, P_{2} - Pressures of the water on the first and second floors, measured in pascals.

\rho - Density of water, measured in kilograms per cubic meter.

v_{1}, v_{2} - Speed of the water on the first and second floors, measured in meters per second.

z_{1}, z_{2} - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]

\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]

v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})

v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) } (2)

If we know that P_{1}-P_{2} = 0\,Pa, \rho=1000\,\frac{kg}{m^{3}}, v_{1} = 15.5\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and z_{1}-z_{2} = -3.5\,m, then the speed of the water flow in the pipe on the second floor is:

v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}

v_{2} \approx 13.100\,\frac{m}{s}

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

4 0
2 years ago
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