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Vikki [24]
2 years ago
15

What kind of elements tend to have higher ionization energies.

Physics
1 answer:
algol132 years ago
5 0
Non metals tend to have higher ionization energies
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Lillian (mass 40.0 kg) standing at rest on slippery ice catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Wha
mestny [16]

Answer:

0.82 m/s

Explanation:

From the law of conservation of momentum,

Total momentum = Final momentum.

mu+m'u' = V(m+m')......................... equation 1

Where m = mass of Lillian, u = initial velocity of Lillian before she catches the dog, m' = mass of the dog, u' = initial velocity of the dog, V = velocity of Lillian and the dog.

make V the subject of the equation,

V = (mu+m'u')/(m+m')................ Equation 2

Given: m = 40 kg, m' = 15 kg, u = 0 m/s (Lillian was standing at rest), u' = 3.0 m/s.

Substitute into equation 2

V = (40×0+15×3)/(40+15)

V = (0+45)/55

V = 45/55

V = 0.82 m/s

3 0
3 years ago
Describe Newton's First Law of motion. Provide an example to support your explanation. (4 points
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An object will remain at rest or in uniform motion until acted upon by an external unbalanced source. Ex) a ball remaining still until kicked by a foot. 
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3 years ago
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When subjected to a force of compression, the length of a bone (compression Young's modulus 9.4 x 109 N/m2, tensile Young's modu
Anarel [89]

To solve this problem it is necessary to apply the definition of Young's Module which states that

Y_1 = \frac{\frac{F}{A}}{\frac{\Delta l_0}{l}}

Where,

F = Force

A = Cross sectional Area

L = Length

L_0 = Initial Length

We need to find the ratio between the two values when the another values are constant, that is

\frac{Y_1}{Y_2} = \frac{\frac{\frac{F}{A}}{\frac{\Delta l_1}{l}}}{\frac{\frac{F}{A}}{\frac{\Delta l_2}{l}}}

\frac{Y_1}{Y_2} = \frac{\Delta l_2}{\Delta l_1}

Re-arrange to find \Delta l_2,

\Delta l_2 = \frac{9.4*10^9}{1.6*10^{10}}*3.7*10^{-5}

\Delta l_2 = 2.17*10^{-5} m

Therefore the bone stretch around 2.17*10^{-5} m

7 0
3 years ago
How many photons per second are emitted by the antenna of a microwave oven if its power output is 1.00 kw at a frequency of 2515
Norma-Jean [14]
The power of the antenna is
P=1.00 kW=1000 W
and since the power is the emitted energy divided by the time, the energy emitted per second is
E=Pt = (1000 W)(1 s)=1000 J

The frequency of the emitted photons is
f=2515 MHz=2.515 \cdot 10^9 Hz
So the energy of a single photon is
E_1=hf=(6.6 \cdot 10^{-34} Js)(2.515 \cdot 10^9 Hz)=1.66 \cdot 10^{-24} J

Therefore, to find the number of emitted photons per second, we should divide the total energy emitted by the antenna in 1 second by the energy of a single photon:
N= \frac{E}{E_1}= \frac{1000 J}{1.66 \cdot 10^{-24} J}=  6.02 \cdot 10^{26}
8 0
3 years ago
A simple ideal Brayton cycle uses argon as the working fluid. At the beginning of the compression, P1 = 15 psia and T1 = 70°F, t
Shtirlitz [24]

B I thing hope this helps

7 0
3 years ago
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