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NNADVOKAT [17]
3 years ago
14

What rock type would most likely be found in the location marked by the black rectangle in the diagram below?

Physics
2 answers:
pychu [463]3 years ago
5 0

There is no diagram below so I can't answer the question

Brums [2.3K]3 years ago
5 0

Just took the assessment its Metamorphic.

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If an object accelerates at 40 m/s^2 in four minutes ("careful this is in minutes),
yaroslaw [1]

Change in speed = (acceleration) x (time)

4 minutes = 240 seconds

Change in speed = (40 m/s²) x (240 seconds)

Change in speed = <em>9,600 m/s</em>

What you're actually describing here is a car pulling 4 G's for 4 minutes, and ending up going 21,475 miles per hour.

The driver would definitely NOT get a speeding ticket, because nobody could catch him.

Also, his car would heat up and shoot flames from atmospheric friction.

(He could avoid this with some fancy steering, leave the atmosphere, and end up in low-Earth-orbit.)

Actually, I hope there's nobody in the car.  His experience wouldn't be pretty.

3 0
3 years ago
What affect does doubling the net force have on the acceleration of the object (when
Triss [41]
<h3>Answer: The acceleration doubles</h3>

===========================================================

Explanation:

Consider a mass of 10 kg, so m = 10

Let's say we apply a net force of 20 newtons, so F = 20

The acceleration 'a' is...

F = ma

20 = 10a

20/10 = a

2 = a

a = 2

The acceleration is 2 m/s^2. Every second, the velocity increases by 10 m/s.

---------------

Now let's double the net force on the object

F = 20 goes to F = 40

m = 10 stays the same

F = ma

40 = 10a

10a = 40

a = 40/10

a = 4

The acceleration has also doubled since earlier it was a = 2, but now it's a = 4.

---------------

In summary, if you double the net force applied to the object, then the acceleration doubles as well.

8 0
3 years ago
Read 2 more answers
Why do i lose at corn hole
aivan3 [116]

Answer:

Maybe because you not throwing it stringer or lighter. And you need to control yourself prob. maybe you did line it correctly or maybe your just bad to be honest. -.-

Explanation:

8 0
3 years ago
Read 2 more answers
You throw a 20 N rock into the air from ground level and observe that, when it is 15.0 m high, it is traveling upward at 25.0 m/
siniylev [52]

Answer:

a) 30.3 m/s b) 46.8 m

Explanation:

a)

  • Assuming no friction present, the work-energy theorem says that the work done on the rock by the only force acting on it (gravity force, aiming downward), is equal to the change of kinetic energy of the rock, at any point.
  • When it's at 15.0 m high, the work done by gravity on the rock can be written as follows:

        W = m*g* h* cos 180\º = -20 N* 15.0 m = -300 J (1)

  • The change in kinetic energy of the rock can be expressed in this way:

        \Delta K = K_{f} - K_{0}  = \frac{1}{2} * m* (v_{f}^{2} -v_{0} ^{2} ) (2)

  • where m= 2.04 kg, vf = 25.0 m/s.
  • As (1) and (2) are equal each other, we can solve for v₀, as follows:

        = -300 J = 0.5*2.04 kg *(( 25.0 m/s)^{2} -v_{0}^{2} )\\ \\ v_{0}^{2} = \frac{300*2 J}{2.04kg} + 625 (m/s)2 = 894. 1 (m/s)2 \\ \\  v_{0} = \sqrt{894.1 (m/s)2} = 30.3 m/s

b)

  • When the rock arrives to its highest point, it will momentarily come to a stop, before changing direction to start falling.
  • So, the final kinetic energy, will be zero.
  • Applying the work --energy theorem again, we can write the following equation:

        \Delta K = K_{f} - K_{0}  =( 0- \frac{1}{2} *m*v_{0}^{2}) = -m*g*h_{max}

  • Replacing by v₀ = 30.3 m/s, rearranging and simplifying common terms, we can solve for hmax, as follows:

        h_{max} = \frac{v_{0} ^{2} }{2*g} =\frac{(30.3 m/s)^{2}}{2*9.8 m/s2} = 46.8 m

7 0
3 years ago
Why is it that when you throw a Frisbee, it flies better when it is normally thrown and not upside down?
Katena32 [7]
Because the air gets stuck in the curves of the frissbee
6 0
3 years ago
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