Phy 101- Assignment 1

PHY 101- Assignment 1
PHY 101- Assignment 1


Question 1

Answer:

No, it is not true to say that a ball will land at the same time whether you drop it straight down from the top of a tower or if you throw it out horizontally. The time it takes for the ball to reach the ground will be different in each scenario.

When you drop the ball straight down from the top of a tower, it falls vertically due to the force of gravity acting on it. The only force acting on the ball is the downward force of gravity, and it accelerates towards the ground at a constant rate (assuming negligible air resistance). In this case, the ball will accelerate solely in the vertical direction, and it will reach the ground in a certain amount of time.

On the other hand, when you throw the ball out horizontally, it will have an initial horizontal velocity in addition to the force of gravity acting on it. The horizontal velocity will keep the ball moving horizontally as it falls vertically due to gravity. This means that the ball will follow a curved path known as a projectile motion. Since the ball is moving horizontally as it falls, it will take longer to reach the ground compared to the scenario where it is dropped straight down.

The key point here is that the horizontal velocity of the ball adds to the time it takes to reach the ground when thrown horizontally. The vertical component of the ball's motion is still affected by gravity and follows the same path as in the vertical drop scenario, but the horizontal component adds extra distance traveled before reaching the ground.

Therefore, due to the additional horizontal motion when the ball is thrown horizontally, it will take longer to reach the ground compared to when it is dropped straight down from the top of a tower.

Question 2

Answer:

To calculate the x and y components of the car's average velocity we can use the formula: 

vx=Δx / t 
vy= Δy / t

Given data:

vx=5.0m
vy=4.0m
t=3.0s

Calculating the x-component of the average velocity (vx):
vx=Δx / t 
vx=5.0m/3.0s
vx≈ 1.67m/s

Calculating the x-component of the average velocity (vy):
vy= Δy / t
vy=4.0m/3.0s
vy≈ 1.33m/s

Average velocity in the x-direction = 1.67m/s
Average velocity in the y-direction = 1.33m/s

Question 3

Yes, there is a force involved when a body is moving while veering to the right at a constant speed.

When an object moves in a curved path or changes its direction, there must be a force acting towards the center of the curve, known as the centripetal force. In this case, as the body veers to the right, there must be a force acting towards the center of the circular path to keep it moving in that direction.

According to Newton's first law of motion (the law of inertia), an object at a constant velocity will continue to move in a straight line unless acted upon by an external force. Therefore, for the body to move in a curved path while veering to the right, there must be a force exerted on it, causing it to change its direction.

The force required to keep the body moving in a curved path is provided by a combination of factors. In the case of a vehicle, for example, the tires exert a frictional force against the road surface, which acts as the centripetal force to keep the vehicle on its curved trajectory while moving at a constant speed. This force allows the body to overcome the tendency to continue in a straight line and instead move in a circular or curved path.

In summary, when a body is moving while veering to the right at a constant speed, there is a centripetal force involved, which is necessary to maintain the curved path of motion.

Question 4

Solution:
From the given data:
F = 168 N (applied force)
m = 17 kg (mass of the suitcase)
mg = 166.6 N (weight of the suitcase due to gravity

Substituting the values into the formula: 
net force= F - mg 
net force = 168 N - 166.6 N 
net force ≈ 1.4 N

Rearranging the equation, we solve for acceleration (a):

a = net force / m 
a ≈ 1.4 N / 17 kg 
a ≈ 0.0824 m/s²

The magnitude of the suitcase's acceleration is approximately 0.0824 m/s².

To determine the direction of the acceleration, we need to consider the direction of the net force. 

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