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ludmilkaskok [199]

3 years ago

11

One way in which elements differ from each other is the structure of the electron cloud in each element’s atoms. In an electron

cloud, an electron that is farther away from the nucleus has

2 answers:

dedylja [7]3 years ago

5 0

Electrons that are further away from the nucleus have more energy. As they enter an "excited" state, they jump up orbits.

worty [1.4K]3 years ago

4 0

Answer:

Explanation:

Electron that is farther away from the nucleus have high level of energy because they are in excited state. When the electrons are farther from nucleus then there is decreased interaction between the nucleus and electrons, and thus these electrons can be removed by spending less amount of energy. Hence we say that they are in excited state.

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g How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for

lbvjy [14]

Answer:

2

Explanation:

We know that in the Fraunhofer single-slit pattern,

maxima is given by

$a\text{sin}\theta=\frac{2N+1}{2}\lambda$

Given values

θ=2.12°

slit width a= 0.110 mm.

wavelength λ= 582 nm

Now plugging values to calculate N we get

$0.110\times10^{-3}\text{sin}2.12=(\frac{2N+1}{2})582\times10^{-9}$

Solving the above equation we get

we N= 2.313≅ 2

4 0

3 years ago

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

vaieri [72.5K]

Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

$R = 106.4 m$

so the distance from the fence is given as

$d = 106.4 - 97.5$

$d = 8.86 m$

7 0

3 years ago

A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.

$d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s$

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0

2 years ago

Two spacemen are floating together with zero speed in a gravity-free region of space. The mass of spaceman A is 120 kg and that

maw [93]

Answer:

your answer is A

Explanation:

4 0

2 years ago

An airplane starts from rest and accelerates at 10.8 m/s2 . what is its speed at the end of a 400 m long runway?

Cloud [144]

The final speed of an airplane is v = 92.95 m/s

The rate of change of position of an object in any direction is known as speed i.e. in other word, Speed is measured as the ratio of distance to the time in which the distance was covered.

Solution-

Here given,

Acceleration a= 10.8 m/s2 .

Displacement (s)= 400m

Then to find final speed of airplane v=?

Therefore from equation of motion can be written as,

v²=u²+ 2as

where, u is initial speed, v is final speed ,a is acceleration and s is displacement of the airplane. Therefore by putting the value of a & s in above equation and (u =0) i.e. the initial speed of airplane is zero.

v²= 2×10.8 m/s²×400m

v²=8640m/s

v=92.95m/s

hence the final speed of airplane v =92.95m/s

To know more about speed

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5 0

1 year ago