According to motion in straight line t1≠t2
A biker travels d1 meters in t1 seconds at v1 m/s for the first leg and d2 meters in t2 seconds at v2 m/s for the second leg. It's possible that t1t2 if his average speed is equal to the average of v1 and v2.
An object is said to be in motion if its position in relation to its surroundings changes over time. It is a shift in an object's position over time. The only type of motion that exists is motion in a straight line.
A reference system is constantly used to describe a particle's motion. An arbitrary origin point is used to create a reference system, and a coordinate system is imagined to be connected to it. The reference system for a specific problem is the coordinate system that has been selected for it. For the majority of the problems, we typically select an earth-based coordinate system as the reference system.
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Some substances look alike and the density can help you determine which is which. It also helps to tell which is heavier than the other.
Because the gravitational force, which points downward, is perfectly balanced by the normal reaction of the floor of the bowling lane, which points upward. The two forces are equal in magnitude, so the net force acting vertically on the bowling ball is zero, therefore there is no acceleration along this direction. Moreover, since the ball is moving in the horizontal direction, the gravitational force has no component along this direction, so it does not change the velocity of the ball.
Explanation:
Given that,
The box of oranges cannot exceed a mass of 10.222 Kg if we are sending to a friend by mail.
The mass of each orange is 198 g
We know that,
1 kg = 1000 g
10.222 kg = 10.222×1000 g
Let there are n number of oranges. So,

It means she can send 52 oranges and it is maximum quantity.
The negative sign on the acceleration is only a vector quantity that means the object is accelerating to the left. Hence, we can only focus on it magnitude which is 4 m/s^2. Acceleration is the change in velocity over time. The change in velocity must be 24 m/s - 0 m/s, if you want the object to stop. Therefore,
a = (v2 - v1)/t
4 = (24 - 0)t
t = 6 seconds
The object will stop after 6 seconds.