All mutations are favorable and increase the ability for the

vesna_86 [32]

A) the tension in the string.

4 0

3 years ago

When a machine is used to perform a task, work output is always more than work input. please select the best answer from the cho

Maurinko [17]

Properly input current above 40 voltage or 100 for example elecrical fan machine is used to perform a task, work output is always more than work

5 0

3 years ago

plz help me with hw A bus of mass 1000 kg moving with a speed of 90km/hr stops after 6 sec by applying brakes then calculate the

Lelechka [254]

Answer:

Mass, M = 1000 kg

Speed, v = 90 km/h = 25 m/s

time, t = 6 sec.

Distance:

${ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}$

Force:

${ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}$

5 0

2 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

2 years ago

If your weight is 120 pounds and your mass is 54 kilograms how would those values change if you were on the moon

BlackZzzverrR [31]

The gravitation acceleration on the moon is different than on Earth. It is 1.6 m/s^2. If you weigh 120 lbs, then you would multiply 120 pounds by the gravitational acceleration on the moon and then divide by the acceleration on Earth.

(120 lbs * 1.6) / 9.8 = 20 pounds.

The mass will always be the same no matter what planet you’re on, so it’s still 54 kg.

4 0

2 years ago