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3 years ago

Start with a puddle on a sunny day. How might water move through the water cycle and eventually fall as rain

1 answer:

8 0

The puddle would warm up then begin to evaporate and then the gas could create a cloud the condensate and then make rain again to create another puddle then it happens all over again

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The variable that the experimenter decides to change to see if there is or is not an effect is the Independent Variable.

EastWind [94]

The correct answer is: [A]: "TRUE" .
_________________________________________________

6 0

3 years ago

Does a can opener make work easier by increasing force, increasing distance, or changing direction?

ivann1987 [24]

A can opener makes work easier by increasing force.

Explanation

Technically a can opener is a simple lever that increases the force with which we remove the cap of the can. Levers are rod shaped objects with fulcrum at one end and the length of the lever has a positive impact on the force.

considering the equation

work=force*distance

indicates that increasing the length of the lever/can opener increases the work done by it.

6 0

4 years ago

Need help with both questions!

xenn [34]

#14 isn't really a Physics problem. It's more of just reading a graph.

A). When speed changes, acceleration is

(change in speed) / (time for the change) .

To be correct about it, acceleration can be positive ... when speed
is increasing ... or it can be negative ... when speed is decreasing.
So, on this graph, there are two periods of acceleration:

From zero to 2 seconds, acceleration = (8 m/s) / (4 sec) = 2 m/s² .

From 10 to 12 seconds, acceleration = (-4 m/s) / (2 sec) = -2 m/s² .

B). From 12 to16 seconds, you can read the speed right from
the graph. It's 4 m/s .

C). From 2 to 10 seconds, the objects speed is a steady 8 m/s.
Covering 8 m/s every second for 8 seconds, it covers 64 meters.
Do you remember that distance is the area under the speed/time
graph? You can see that plainly on this graph. From 2 to 10 sec,
there are 16 blocks. Each block is (2 m/s) high and (2 sec) wide,
so its area is (2 m/s) x (2 sec) = 4 meters. The area of 16 blocks
is (16) x (4 meters) = 64 meters.
====================================

#15.

a). constant velocity on a distance graph is a line that slopes up;
constant velocity on a velocity graph is a horizontal line;

b). positive constant acceleration on a distance graph is a
line that curves up;
positive constant acceleration on a velocity graph is a
straight line that slopes up;

c). "uniformly slowing down to a stop" on a distance graph
is a line that's less and less curved as time goes on, and
eventually reaches the x-axis.
"uniformly slowing down to a stop" on a velocity graph is
a straight line that slopes down, and stops when it reaches
the x-axis.

7 0

3 years ago

The radius RH of a black hole, also known as the event horizon, marks the location where not even light itself can escape from t

olga nikolaevna [1]

Solution:

a) We know acceleration due to gravity, g = GM/r²

Differential change, dg/dr = -2GM/r³

Here, r = 50*Rh = 50*2GM/c² = 100GM/c ²

My height, h=dr = 1.7 m

Difference in gravitational acceleration between my head and my feet, dg = -10 m/s²

or, dg/dr = -10/1.7 = -2GM/(100GM/c²)³

or, 5.9*100³*G²*M² = 2c⁶

or, M = 0.59*c³/(1000G) = 2.39*1032 kg = [(2.39*1032)/(1.99*1030 )]Ms = 120*Ms

Mass of black hole which we can tolerate at the given distance is 120 time the mass of Sun.

b) This limit an upper limit ,we can tolerate smaller masses only.

4 0

3 years ago

Escribe verdadero o falso :

Oduvanchick [21]

By defining kinematics and derivative relations we can find which statements are true or false:

a) False. The derivative is the instantaneous velocity.

b) True. The second drift is the instantaneous acceleration.

c) False the derivative is

d) False the derivative is

a) The velocity is defined with the variation of the position with respect to time.

$v= \frac{dx}{dt}$

Wher x is the position and t the time.

In the change of the average velocity is the average value of the velocity in an interval

$v = \frac{v_f - v_o}{t}$

We can see that the derivative is the speed in a very small timet, that is, the speed instantaneous. Therefore the statement is False.

The prime derivative of the position is the instantaneous velocity, not the average velocity.

b) Acceleration is defined as the change in position with respect to time..

$a = \frac{dv}{dt}$

Let's use the chain rule.

a = $\frac{d}{dt} \frac{dx}{dt}$

a = $\frac{d^2 x}{dt^2}$

Therefore the second derivative of the position is the instantaneous acceleration.

The statement is True.

Questions c and d ask the derivative of a function

f (x) = 2 x aⁿ

c) derivative with respect of x

$\frac{df}{dx} = 2 a^n$

The answer is False.

d) derivative with respect to a.

$\frac{df}{da} = 2n \ x a^{n-1}$

Answer d is false

In conclusion using the definition of kinematics and derivative relations we can find which statements are true.

a) False. The derivative is the instantaneous velocity.

b) True. The second drift is the instantaneous acceleration.

c) False the derivative is: $\frac{df}{dx} = 2 a^n$

d) False the derivative is: $\frac{df}{da} = 2n \ x a^{n-1}$

Learn more about the relationship between derivatives and kinematics here: brainly.com/question/15344251

8 0

3 years ago