When electromagnetic fields interact with charged particles ?

If an atom is neutral, then the number of electrons is equal to the number of

Aloiza [94]

Answer:

Neutral atom means the number of protons is equal to the number of electrons.

Explanation:

4 0

3 years ago

In a first order decomposition in which the rate constant is 0.204 sec-1, how long will it take (in seconds) until 0.399 mol/L o

DENIUS [597]

Answer:

2.902 to 3 decimal places

5 0

2 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

How would the attractive force between two spheres change if the mass of one sphere was doubled?

Dominik [7]

The force of attraction between two objects can be illustrated using Newton's Law of Universal Gravitation.
The relation between the different parameters is shown in the attached image.

Now, from the relation, we can deduce that the force between the two objects is directly proportional to the masses of the two objects.

This means that, if the mass of one object is doubled, then the force between the two objects will also be doubled.

8 0

2 years ago

An airplane travels at 300 mi/h south for 2.00 h and then at 250 mi/h north for 750 miles. what is the average speed for the tri

iren2701 [21]

<span>Th find the average speed of a trip we need to dived the total distance by the total time. Let's find the total distance d. d = (300 mi/h)(2.00 h) + 750 miles d = 600 miles + 750 miles d = 1350 miles The total distance is 1350 miles Let's find the total time t. t = 2.00 hours + (750 mi / 250 mi/h) t = 2.00 hours + 3.00 hours t = 5.00 hours The total time of the trip is 5.00 hours. We can find the average speed. d / t = 1350 miles / 5.00 hours d / t = 270 miles/ hour The average speed of the trip is 270 mi/h (Note that the direction does not matter when we find the average speed.)</span>

6 0

3 years ago