Which statements describe the factors affecting the strength of an electrical force? Check all that apply.

In what fundamental way did the work of Galileo differ from his predecessors who had thought about the sky?

kompoz [17]

Answer: The correct answer is:

Galileo used instruments and experiments to show him what nature was doing, instead of relying on pure logic

Explanation:

Galileo can be considered as one of the precursors of experimentation and the scientific method. A method that doesn't rely on "common sense" and rationalization and logic, but instead is fuelled by a disposition of skepticism and rather makes claims about reality based on experimentation and empirical data shows.

Galileo differed from his predecessors because he actually used and developed instruments and method to reliable measure and observe what nature was doing, instead of relying on pure logic.

7 0

2 years ago

Use the slit example to explain why you can hear a noise in another room through an open door.

defon

Because the wall reflects sound waves to your ears bouncing off of the walls, even if it's in another room.

8 0

3 years ago

If a baseball pitch leaves the pitcher's hand horizontally at a velocity of 150 km/h by what percent will the pull of gravity ch

Slav-nsk [51]

<span>0.52% First, let's convert that speed into m/s. 150 km/h * 1000 m/km / 3600 s/h = 41.667 m/s Now let's see how much time gravity has to work on the ball. Divide the distance by the speed. 18 m / 41.667 m/s = 0.431996544 s Now multiply that time by the gravitational acceleration to see what the vertical component to the ball's speed that gravity adds. 0.431996544 s * 9.8 m/s^2 = 4.233566131 m/s Use the pythagorean theorem to get the new velocity of the ball. sqrt(41.667^2 + 4.234^2) = 41.882 m/s Finally, let's see what the difference is (41.882 - 41.667)/41.667 = 0.005159959 = 0.5159959% Rounding to 2 figures, gives 0.52%</span>

8 0

3 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

2 years ago

A less massive moving object has an elastic collision with a more massive object that is not moving. Compare the initial velocit

stepladder [879]

Assume that the small-massed particle is $m$ and the heavier mass particle is $M$ .

Now, by momentum conservation and energy conservation:

$mv = mv_{m} + Mv_{M}$

$mv^{2} = mv^{2}_{m} + Mv^{2}_{M}$

Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

$v_{m} = v\frac{m - M}{m + M}\\\\\\v_{M} = v\frac{m + M}{m + M}\\$

So now, we see that $v_{m} < 0$ and $v_{M} > 0$ . So therefore, the smaller mass recoils out.

Hope this helps you!

Bye!

7 0

2 years ago