It took 500 newtons of force to push a car 4 meters. How much work was done?

The monkey experiment is an example of what?

Semenov [28]

Answer:

D.) Sensory adaptation

Explanation:

Assuming you are talking about the cloth and metal monkey experiment performed in the field of psychology (not physics), the monkey formed an attachment to the cloth mother because it felt closer to it, as it was more appealing to its senses.

3 0

3 years ago

Who was Georges Lemaitre and how did he contribute to the Big Bang theory?<br> (15 points)

Ivahew [28]

A Belgian Astronomer and Cosmotologist who formulated the modern Big Bang theory, which states that the universe began in a cataclysmic explosion of a small, primeval “super-atom”

7 0

3 years ago

A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much tim

Free_Kalibri [48]

Answer:

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

$y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\ where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]$

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

$(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]$

3 0

3 years ago

Two blocks, with masses M2>M1, are connected by ropes. You pull to the right on a second rope, with external force "T1".The b

Gre4nikov [31]

Answer:

$(M_1 + M_2) a > M_2 a$

Becuase $M_1 +M_2> M_2$

So then we can conclude that:

$T_1 > T_2$

And that makes sense since the force $T_1$ needs to accelerate the two masses and $T_2$ just need to accelerate $M_2$ .

So the best option for this case would be:

a. T1 > T2

See explanation below.

Explanation:

For this case we consider the system as shown on the figure attached.

Since the system is connected the acceleration for both masses are equal, that is $a_{M_1}= a_{M_2} = a$

From the second Law of Newthon we have that the force applied for the mass $M_2$ is $F_{M_2}= M_2 a$ and we know that the force acting on the x axis for the mass 2 is $F_{M_2}= T_2$ so then we have that $T_2= M_2 a$

Now when we consider the system of $M_1 +M_2$ as a whole mass, this system have the same acceleration $a$ and on this case we will see that the only force acting on the entire system would be $T_1$ and then by the second law of Newton we have that: