The object D is made up of material Lead. The correct option is D.
<h3>What is specific heat?</h3>
The specific heat is the amount of heat required to change the temperature by 1°C. It is denoted by C.
Two 1-kg objects, C and D, increase in temperature by the same amount, but the thermal energy transfer of object C is greater than the thermal energy transfer of object D. The object C has a specific heat of 235 J/kg-K.
Q = m C ΔT
Qc > Qd
The energy transfer is proportional to specific heat.
Specific heat of D must be less. The possible material with specific heat less than the given value is for Lead material.
Thus, the correct option is D.
Learn more about specific heat,
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Answer: (a) The magnitude of its temperature change in degrees Celsius is
.
(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is
.
Explanation:
(a) Expression for change in temperature is as follows.

= 15.1 K
= 
= 
= 
Therefore, the magnitude of its temperature change in degrees Celsius is
.
(b) Change in temperature from Celsius to Fahrenheit is as follows.
F = 1.8C + 32
C = 
Since, K = C + 273
or, 

= 1.8 (15.1)
= 
or, = 
Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is
.
Explanation:
since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.
We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.
range can be calculated by the formula :-

u is the velocity during its take off and
is the angle at which its thrown
Given that
- u = 8m/ s
= 40°
calculating range using the above formula


value of sin 80 = 0. 985



Hence,
