Answer:
Explanation:
The equation that relates heat Q with the temperature change 
of a substance of mass <em>m </em>and specific heat <em>c </em>is 
.
We want to calculate the final temperature <em>T, </em>so we have:
Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add 
to our temperature in 
to have it in 
as it must be):
Answer:
The magnitude of the acceleration of the car is 35.53 m/s²
Explanation:
Given;
acceleration of the truck, 
= 12.7 m/s²
mass of the truck, 
= 2490 kg
mass of the car, 
= 890 kg
let the acceleration of the car at the moment they collided = 
Apply Newton's third law of motion;
Magnitude of force exerted by the truck = Magnitude of force exerted by the car.
The force exerted by the car occurs in the opposite direction.
Therefore, the magnitude of the acceleration of the car is 35.53 m/s²
Answer:
<u><em>Rate of dissolving compounds:</em></u>
If we increase the temperature of the solution, then the dissolving compound would dissolve more easily.
<u><em>Boiling Point of Compounds:</em></u>
If the inter-molecular forces of any compound is really strong, then the boiling point of the compound would be really high.
Answer:
1.) Magnitude = 5596 N
2.) Direction = 60 degrees
Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N
Let us resolve the two forces into X and Y component
Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N
Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )
= 2000 + 2828.43
= 4828.43 N
The resultant force R will be
R = sqrt ( X^2 + Y^2 )
Substitutes the forces at X component and Y component into the formula
R = sqrt ( 2828.43^2 + 4828.43^2 )
R = sqrt ( 31313752.53 )
R = 5595.87 N
The direction will be
Tan Ø = Y/X
Substitute Y and X into the formula
Tan Ø = 4828.43 / 2828.43
Tan Ø = 1.707106
Ø = tan^-1( 1.707106 )
Ø = 59.64 degree
Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.
