#2 Distance (d) = _______37.5cm___________

Select all of the statements that are true.

natta225 [31]

The correct statements are "Each orbit holds a fixed number of electrons" and "The n=1 orbit can only hold two electrons." According to the Bohr model, the maximum number of electrons that can occupy an orbit is given by $2n^2$ , where n is the number of the orbit. For instance, when n=1 it means $2n^2= 2(1)^2=2$ . This particular orbit can only hold up to two electrons. Even though the electrons can gain energy and move to higher orbits or electrons from higher orbits can lose energy and drop to the n=1 level, the energy level would not allow more electrons to enter the orbit once it is full. Again the octet rule, which states that atoms achieve stability by having 8 valence electrons, limits the maximum number of electrons that can be occupied by an orbit. The gain and loss of electrons is done to achieve the noble gas configuration and once that is reached no more electron can be added to an orbit

8 0

3 years ago

A car is driving 55 miles per hour east. The car changes direction to head north but still travels at 55 miles per hour. Which i

bezimeni [28]

Answer:

Just as distance and displacement have distinctly different meanings (despite their similarities), so do speed and velocity. Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. A fast-moving object has a high speed and covers a relatively large distance in a short amount of time. Contrast this to a slow-moving object that has a low speed; it covers a relatively small amount of distance in the same amount of time. An object with no movement at all has a zero speed.

7 0

3 years ago

Read 2 more answers

The motions of the toy robot shown above are driven by an electric motor. The power source for the toy's motor is a pair of batt

weeeeeb [17]

the C is correct one and surely I suggest you to go for it !

3 0

3 years ago

Read 2 more answers

Consider an elevator carrying Kermit the frog weighing 4000.0 N is held 5.00 m above a spring with a force constant of

tigry1 [53]

Answer:

The maximum compression distance of the spring is 0.375 m

Explanation:

The given parameters are;

The weight of the elevator and the frog = 4,000.0 N

The location of the elevator above the spring = 5.00 m

The force constant of the spring, k = 8,000.0 N/m

The frictional force of the brakes = 1,000.0 N

The net force, F, of the elevator on the spring is F = 4,000.0 N - 1,000.0 N = 3,000.0 N

F = 3,000.0 N

The maximum compression distance of the spring, x is given as follows;

F = k × x

Substituting the known values gives;

3,000.0 N = 8,000.0 N/m × x

∴ x = (3,000.0 N)/(8,000.0 N/m) = 0.375 m

x = 0.375 m = 37.5 cm

The maximum compression distance of the spring, x = 0.375 m.

4 0

3 years ago

A car that weighs 1.0 x 10^4 N is initially moving at a speed of 38 km/h when the brakes are applied and the car is brought to a

hram777 [196]

Answer:

Part a) Force on car = 2833.84 Newtons

Part b) Time to stop the car = 3.8 seconds

Part c) Factor for stopping distance is 4.

Part d) Factor for stopping time is 1.

Explanation:

The deceleration produced when the car is brought to rest in 20 meters can be found by third equation of kinematics as

$v^2=u^2+2as$

where

v = final speed of the car ( = 0 in our case since the car stops)

u = initial speed of the car = 38 km/hr = $\frac{38\times 1000}{3600}=10.56m/s$

a = deceleration produces

s = distance in which the car stops

Applying the given values we get

$0^2=10.56^2+2\times a\times 20\\\\a=\frac{0-10.56^2}{2\times 20}\\\\\therefore a=-2.78m/s^2$

Now the force can be obtained using newton's second law as

$Force=\frac{Weight}{g}\times a$

Applying values we get

$Force=\frac{1.0\times 10^4}{9.81}\times -2.78\\\\\therefore F=-2833.84Newtons$

The negative direction indicates that the force is opposite to the motion of the object.

Part b)

The time required to stop the car can be found using the first equation of kinematics as

$v=u+at$ with symbols having the same meanings

Applying values we get

$0=10.56-2.78\times t\\\\\therefore t=\frac{10.56}{2.78}=3.8seconds$

Part c)

From the developed relation of stopping distance we can see that the for same force( Same acceleration) the stopping distance is proportional to the square of the initial speed thus doubling the initial speed increases the stopping distance 4 times.

Part d)

From the relation of stopping time and the initial speed we can see that the stopping distance is proportional initial speed thus if we double the initial speed the stopping time also doubles.

8 0

3 years ago

#2 Distance (d) = _______37.5cm____________

#2 Distance (d) = _37.5cm______