Answer:
In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.
You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes.
Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine Δs becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration ac because centripetal means center seeking.
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Answer:
Work done, W = 0.0219 J
Explanation:
Given that,
Force constant of the spring, k = 290 N/m
Compression in the spring, x = 12.3 mm = 0.0123 m
We need to find the work done to compress a spring. The work done in this way is given by :
W = 0.0219 J
So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.
<span>Frequency x Wavelength = Speed of light
Now, speed of light = 3 x 10^5 km/s = 3 x 10^8 m/s = 3 x 10^10 cm/s
Frequency = speed/Wavelength
= (3 x 10^10)/(4.257 x 10^-7)
=7 x 10^16 Hz</span>
Answer:
a) 
b) 
c) 
Explanation:
From the exercise we know the initial velocity of the projectile and its initial height
To find what time does it take to reach maximum height we need to find how high will it go
b) We can calculate its initial height using the following formula
Knowing that its velocity is zero at its maximum height
So, the projectile goes 1024 ft high
a) From the equation of height we calculate how long does it take to reach maximum point
Solving the quadratic equation
So, the projectile reach maximum point at t=2s
c) We can calculate the final velocity by using the following formula:
Since the projectile is going down the velocity at the instant it reaches the ground is:
Answer:
F=ma
Explanation:
Force = mass * acceleration
