What kind of elements tend to have higher ionization energies.

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

serious [3.7K]

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.

7 0

3 years ago

Read 2 more answers

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