Current= voltage divided by resistance
120/30=4
Given :
Mass of water, m = 2 grams.
The temperature of water drops from 31 °C to 29 °C .
The specific heat of water is 4.184 J/(g • °C).
To Find :
Amount of heat lost in this process.
Solution :
We know, heat lost is given by :

Therefore, amount of heat lost in this process is 16.736 J.
Fk = μK N
N = m a
N = 4 × a
N = 4a
Fk = μK N
17 = 5 × 4a
17 = 20 a
a = ¹⁷/₂₀ = 0.85 m/s²
Answer:
The magnitude is "3.8 m/s²", in the upward direction.
Explanation:
The given values are:
Mass,
m = 88 kg
Scale reads,
T = 900 N
As we know,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
Now,
⇒ 
On substituting the given values in the above equation, we get
⇒ 
On subtracting "862.4" from both sides, we get
⇒ 
⇒ 
⇒ 
⇒
(upward direction)