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SOVA2 [1]
3 years ago
14

A NASCAR driver weighing 75 kg enters a corner with a radius of 155 meters. If his acceleration is 100 m/s2. What is tangential

velocity of the vehicle?
Physics
1 answer:
Ymorist [56]3 years ago
3 0

The tangential velocity of the car is 124.5 m/s

Explanation:

The centripetal acceleration of an object in uniform circular motion is given by:

a=\frac{v^2}{r}

where

a is the acceleration

v is the tangential velocity

r is the radius of the circle

For the car in this problem, we have:

a=100 m/s^2 is the centripetal acceleration

r = 155 m is the radius of the turn

Solving for v, we find the tangential velocity of the car:

v=\sqrt{ar}=\sqrt{(100)(155)}=124.5 m/s

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

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**100 points** PLEASE ANSWER IN 3 PARAGRAPHS
Deffense [45]

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.

hope it helps! stay safe and tell me if im wrong pls :D

(brainliest if you want, or if its right pls) :)

4 0
2 years ago
What is it that if you have, you want to share me, and if you share, you do not have?
Colt1911 [192]

Answer:

Secrets? Cuz if I share you the secrets then the secrets will no longer will be secrets.

4 0
3 years ago
Read 2 more answers
How does inertia affect someone who isn't wearing a seatbelt
Paul [167]
Basically, when someone is resting in an accelerated vehicle without restraint from a seatbelt, the force of stopping the vehicle will be when inertia occurs, and that force of the vehicle coming to a stop will affect the passenger (without a seatbelt/restraint from another force or object) greatly by throwing them.
For example;
If I were to be riding in a vehicle (without a seatbelt) that's accelerating at 40 m/s^2 and it suddenly gets slammed on the breaks, I will be thrown forward from inside the vehicle.

I hope this helps!
7 0
4 years ago
A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi
Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L)  ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L

Finally, using the Parallel Axis Theorem, we calculate I_B:

I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4}  ML^{2} =\frac{1}{3} ML^{2}

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More force needs to be applied
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