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mina [271]
3 years ago
6

A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseba

ll travels a horizontal distance of 16.5 m and rotates through an angle of 49.0 rad. The baseball has a radius of 3.67 cm and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?
Physics
1 answer:
Marina CMI [18]3 years ago
5 0

Answer:

4.5m/s

Explanation:

Linear speed (v) = 42.5m/s

Distance(x) = 16.5m

θ= 49.0 rad

radius (r) = 3.67 cm

= 0.0367m

The time taken to travel = t

Recall that speed = distance / time

Time = distance / speed

t = x/v

t = 16.5/42.5

t = 0.4 secs

tangential velocity is proportional to the radius and angular velocity ω

Vt = rω

Angular velocity (ω) = θ/t

ω = 49/0.4

ω = 122.5 rad/s

Vt = rω

Vt = 0.0367 * 122.5

Vt =4.5 m/s

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3 years ago
In the example given below, Aaron applies a force of 300N and Bob applies a force of 450N :
garri49 [273]

Answer:

Explanation:

This problem is all about torque. The "rules" are that in order for a system to be in rotational equilibrium, the sum of the torques on the system have to equal 0 (in other words, they have to equal each other {cancel each other out}). The equation for torque is

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The torques have to be the same on both sides of the fulcrum; mathematically, that looks like this:

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