C liquid at room temperature
Answer: AAAAAAAAGGGGGHHHHJJJGSSSUUUUUUUUYCCFVGBHNJM
Explanation: YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET
Explanation :
It is given that,
Mass of the car, m = 1000 kg
Force applied by the motor, 
The static and dynamic friction coefficient is, 
Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,




So, the acceleration of the car is
. Hence, this is the required solution.
Answer:
dt/dx = -0.373702
dt/dy = -1.121107
Explanation:
Given data
T(x, y) = 54/(7 + x² + y²)
to find out
rate of change of temperature with respect to distance
solution
we know function
T(x, y) = 54 /( 7 + x² + y²)
so derivative it x and y direction i.e
dt/dx = -54× 2x / (7 +x² + y²)² .........................1
dt/dy = -54× 2y / (7 + x² + y²)² .........................2
now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2
dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²
dt/dx = -0.373702
and
dt/dy = -54× 2(3) / (7 + (1)² + (3)²)²
dt/dy = -1.121107
Newton’s Law: F = MA
A = F/M (change equation)
12.6 N/ 2.4 kg = 5.25
Answer: acceleration is 5.25 m/s^2