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grigory [225]
4 years ago
9

a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second

than there are seconds in a year? (b) Humans have existed for about 106 years, whereas the universe is about 1010 years old. If the age of the universe is defined as 1 "universe day," where a universe day consists of "universe seconds" as a normal day consists of normal seconds, how many universe seconds have humans existe
Physics
1 answer:
lawyer [7]4 years ago
3 0

Answer:

a) Yes, there are 10^8 shakes in a second (1 s \frac{1 shake}{10^{-8}s}=10^8 shake) in a year there are 31,536,000 seconds... that is 3.1536x10^7 (1year\frac{365 day}{1 year} \frac{24 h}{1 day} \frac{60 min}{1 h}\frac{60s}{1min}=3.1536 *10^7)

b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s

That is about 2 and half hours.

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A beam of protons moves in a circle of radius 0.20 m. The protons move perpendicular to a 0.36-T magnetic field. (a) What is the
gregori [183]

Answer:

a) v=6.898\times 10^{6}\ m.s^{-1} is the speed of each proton

b) F_c=3.97\times 10^{-13}\ N

Explanation:

Given:

radius of path of motion, r=0.2\ m

we know charge on protons, q=1.6\times 10^{-19}\ C

magnetic field strength, B=0.36\ T

we've mass of proton, m=1.67\times 10^{-27}\ kg

a)

From the equivalence of magnetic force and the centripetal force on the proton:

F_B=F_C

q.v.B=\frac{m.v^2}{r}

q.B=\frac{m.v}{r}

where:

v = speed of the proton

(1.6\times 10^{-19})\times 0.36=\frac{1.67\times 10^{-27}\times v}{0.2}

v=6.898\times 10^{6}\ m.s^{-1} is the speed of each proton

b)

Now the centripetal force on each proton:

F_c=m.\frac{v^2}{r}

F_c=1.67\times 10^{-27}\times \frac{(6.898\times 10^6)^2}{0.2}

F_c=3.97\times 10^{-13}\ N

6 0
4 years ago
Determine the MA based on the diagram below.
artcher [175]
There are two things that you should remember while dealing with the "Lever Mechanical Advantage" problems:

1) The Effort Arm;
2) The Resistance Arm.

Some books label the Effort Arm as in-lever arm and the Resistance Arm as out-lever arm. (Physics Jargon that you need to remember in order to solve problems)

The Effort Arm is that "part" of the lever where the force can be applied. The Resistance Arm is where some mass is placed. In the diagram, as you can see, the mass is placed on one arm of the lever. Therefore, it is the Resistance Arm.

Now the formula for the "Mechanical Advantage(MA)" is:
MA =  \frac{L_{e} }{ L_{r} }

Where L_{e} is the length of the Effort Arm(the subscript "e" stands for Effort), and L_{r} stands for the length of the Resistance Arm(here "r" stands for Resistance).

Plug in the values:

L_{e} = 15m.
L_{r} = 7m.

Therefore, L_{e} / L_{r} = 15/7 = 2.143 = MA

The correct answer is option C(2.14).

-i

3 0
3 years ago
Read 2 more answers
An object (mass 12.5 kg) slides on a frictionless surface at 1.32 m/s. How much work must be done to bring the object to rest
mote1985 [20]

Answer:

10.89 J.

Explanation:

The following data were obtained from the question:

Mass (m) = 12.5 kg

Velocity (v) = 1.32 m/s

Work done =?

To obtain the workdone, we shall determine the kinetic energy of the object since work and energy has the same unit of measurement. This is illustrated below:

Mass (m) = 12.5 kg

Velocity (v) = 1.32 m/s

Kinetic energy (K.E) =?

K.E = ½mv²

K.E = ½ × 12.5 × 1.32²

K.E = 6.25 × 1.7424

K.E = 10.89 J

The kinetic energy of the object is 10.89 J. Hence, the workdone in bringing the object to rest is 10.89 J.

8 0
3 years ago
What would be the distance moved if we had a 70 n force and work done is 8j
Lostsunrise [7]

Answer:

0.1143m

Explanation:

W=f×s

8=70s

make s the subject of the formula

s=8/70

=0.1143m

3 0
3 years ago
An object, accelerating from rest at a constant rate, travels over 28 m in 11 s. What is its final velocity?
Paul [167]

Answer:

answer is 3

Explanation:

by using s= [(v+ u)/2] x t

28= (v+ 0)/2 x 11

v= 5.09 ms^-1

4 0
3 years ago
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