Find the force of gravity of 1000 kg car.
1 answer:
Answer:
The force of gravity acting on the car is <u>9800 N vertically downward.</u>
Explanation:
Given:
Mass of the car given is 1000 kg.
We know that the force of gravity is the force applied by the center of Earth on any body. The force of gravity is also called the weight of the body and always act towards the center of the Earth.
From Newton's second law, we know that the force acting on a body is equal to its mass and acceleration.
Here, the acceleration acting on the car is due to gravity and thus has a constant value of 9.8 m/s² on the surface of Earth.
Therefore, the force of gravity acting on the car is given using the Newton's second law as:
Force of gravity = Mass of car Acceleration due to gravity.
Force of gravity = (1000 kg) (9.8 m/s²)
Force of gravity = 9800 N [1 kg.m/s² = 1 N]
Therefore, the force of gravity acting on the car is 9800 N vertically downward.
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H has a positive 1 charge. This means that having 3H = +3<span>. This is a neutral compound so x= -3 because X+3H= 0
Y</span>
is also neutral so 2X+Y= 0
we know X=-3 So, 2(-3)+Y=0
-6+y=0
Y=+6 charge
Answer: The valency of X is -3. The valency of Y is 6
H has a positive 1 charge. This means that having 3H = +3<span>. This is a neutral compound so x= -3 because X+3H= 0
Y</span>
we know X=-3 So, 2(-3)+Y=0
-6+y=0
Y=+6 charge
Answer: The valency of X is -3. The valency of Y is 6
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6