Answer:
A) Cold object will start getting hot
B) Heat exchange will stop as the two object acquire the same temperature.
Explanation:
A) When one hot object and one cold object are kept in contact then the heat is transferred from the hot object to the cold object via different modes of heat transmission. Hence, the cold object starts getting hot
B) The transmission of heat from the hot object to the cold object will stop as the temperature of the two object becomes equal to each other.
Answer:
hgt tdyrghgcderhgfe
Explanation:
rghttyhhhggggftttgf d's weyhvfswertfcdsettyhgftgdfvhgg fee e4y
- Angle (θ) = 60°
- Force (F) = 20 N
- Distance (s) = 200 m
- Therefore, work done
- = Fs Cos θ
- = (20 × 200 × Cos 60°) J
- = (20 × 200 × 1/2) J
- = (20 × 100) J
- = 2000 J
<u>Answer</u><u>:</u>
<u>2</u><u>0</u><u>0</u><u>0</u><u> </u><u>J</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer:
Speed =0.283m/ s
Direction = 47.86°
Explanation:
Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane
MU1 =MU2cos38 + MV2cos y ...x plane
0 = MU2sin38 - MV2sin y .....y plane
Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2
Substitute into equation above
.46 = .34cos38 + V2cos y ...equ1
.34sin38 = V2sin y...equ2
.19=V2cos Y...x
.21=V2sin Y ...y
From x
V2 =0.19/cost
Sub V2 into y
0.21 = 0.19(Sin y/cos y)
1.1052 = tan y
y = 47.86°
Sub Y in to x plane equ
.19 = V2 cos 47.86°
V2=0.283m/s
