Answer:
Explanation:
When we accelerate in a car on a straight path we tend to lean backward because our lower body part which is directly in contact with the seat of the car gets accelerated along with it but the upper the upper body experiences this force later on due to its own inertia. This force is accordance with Newton's second law of motion and is proportional to the rate of change of momentum of the upper body part.
Conversely we lean forward while the speed decreases and the same phenomenon happens in the opposite direction.
While changing direction in car the upper body remains in its position due to inertia but the lower body being firmly in contact with the car gets along in the direction of the car, seems that it makes the upper body lean in the opposite direction of the turn.
On abrupt change in the state of motion the force experienced is also intense in accordance with the Newton's second law of motion.
<span>Density is a value for
mass, such as kg, divided by a value for volume, such as m^3. Density is a
physical property of a substance that represents the mass of that substance per
unit volume. The mass of the salty water is calculated as follows:
Mass = 1.166 g/mL ( 2000 mL ) = 2332 g of salty water</span>
Complete question :
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts. The supply storage area of the lunar outpost, where gravity is 1.63 m/s2, can only support 1 x 10^5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost
Answer:
601,220N
Explanation:
Given that:
Gravity at lunar outpost = 1.63m/s²
Acceleration due to gravity on earth = 9.8m/s²
Supported weight = 1 * 10^5 N
Maximum weight of supplies as measured on earth;
(Ratio of the gravities) * weight of supplies
(9.8m/s² / 1.63m/s²) * (1 * 10^5 N)
6.0122 * (1 * 10^5)
6.0122 * 10^5 N
= 601,220 N
Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
