First, we write the SI prefixed. The SI unit for distance is meters.
Kilo = 10³
Mega = 10⁶
Giga = 10⁹
Terra = 10¹²
Because our value has ten to the power of 11, we will use the closest and lowest power prefix, which is giga.
1.5 x 10¹¹ / 10⁹
= 1.5 x 10² Gm or 150 Gm
Writing in kilometers, we simply repeat the procedure except we divide by 10³ this time.
1.5 x 10¹¹ / 10³
= 1.5 x 10⁸ km
To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.
Where
A = Cross-sectional Area
Y = Young's modulus
= Coefficient of linear expansion for steel
= Temperature Raise
Our values are given as,
Replacing we have,
Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N
Answer:
option (E) 1,000,000 J
Explanation:
Given:
Mass of the suspension cable, m = 1,000 kg
Distance, h = 100 m
Now,
from the work energy theorem
Work done by the gravity = Work done by brake
or
mgh = Work done by brake
where, g is the acceleration due to the gravity = 10 m/s²
or
Work done by brake = 1000 × 10 × 100
or
Work done by brake = 1,000,000 J
this work done is the release of heat in the brakes
Hence, the correct answer is option (E) 1,000,000 J
Answer:
.c. −160°C
Explanation:
In the whole process one kg of water at 0°C loses heat to form one kg of ice and heat lost by them is taken up by ice at −160°C . Now see whether heat lost is equal to heat gained or not.
heat lost by 1 kg of water at 0°C
= mass x latent heat
= 1 x 80000 cals
= 80000 cals
heat gained by ice at −160°C to form ice at 0°C
= mass x specific heat of ice x rise in temperature
= 1 x .5 x 1000 x 160
= 80000 cals
so , heat lost = heat gained.
Answer: a) 456.66 s ; b) 564.3 m
Explanation: The time spend to cover any distance a constant velocity is given by:
v= distance/time so t=distance/v
The slower student time is: t=780m/0.9 m/s= 866.66 s
For the faster students t=780 m/1,9 m/s= 410.52 s
Therefore the time difference is 866.66-410.52= 456.14 s
In order to calculate the distance that faster student should walk
to arrive 5,5 m before that slower student, we consider the follow expressions:
distance =vslower*time1
distance= vfaster*time 2
The time difference is 5.5 m that is equal to 330 s
replacing in the above expression we have
time 1= 627 s
time2 = 297 s
The distance traveled is 564,3 m
