Here mass of the iron pan is given as 1 kg
now let say its specific heat capacity is given as "s"
also its temperature rise is given from 20 degree C to 250 degree C
so heat required to change its temperature will be given as



now if we give same amount of heat to another pan of greater specific heat
so let say the specific heat of another pan is s'
now the increase in temperature of another pan will be given as


now we have

now as we know that s' is more than s so the ratio of s and s' will be less than 1
And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature
So correct answer is
<u>A) The second pan would reach a lower temperature.</u>
Answer:
734.215N
Explanation:
First we calculate the angle that corresponds to a 5% slope using the Tan-1 function

then we use the component that corresponds to the direction parallel to the road, additionally we must multiply by the gravity value to find the weight(g=9.81m/s^2)
Wx=M*g*sen(2.86)=1500kg*9.81*sen(2.86)=734.215N
Answer:
a) 12.8212 N
b) 12.642 N
Explanation:
Mass of bucket = m = 0.54 kg
Rate of filling with sand = 56.0 g/ sec = 0.056 kg/s
Speed of sand = 3.2 m/s
g= 9.8 m/sec2
<u>Condition (a);</u>
Mass of sand = Ms = 0.75 kg
So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg
== > total weight = 1.29 × 9.8 = 12.642 N
Now impact of sand = rate of filling × velocity = 0.056 × 3.2 = 0.1792 kg. m /sec2=0.1792 N
Scale reading is sum of impact of sand and weight force ;
i-e
scale reading = 12.642 N+0.1792 N = 12.8212 N
<u>Codition (b);</u>
bucket mass + sand mass = 0.54 +0.75=1.29 kg
==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)
Answer:
ACTION REACTION FORCES
Explanation:
When there is an action frce there will be a reaction force