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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
A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ
Elena L [17]

Answer:

The ocean is 6485.6m deep when measured from the vessel

Explanation:

v=1474m/s

t=8.88s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

velocity=\frac{distance}{time}\\ v=\frac{2d}{t} \\vt=2d\\d=\frac{vt}{2}

d=\frac{1474*8.8}{2}

d= 6485.6m

7 0
3 years ago
Is rose quartz igneous sedimentary or metamorphic
Dmitriy789 [7]
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7 0
3 years ago
A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s
Ede4ka [16]

When acceleration is constant, the average velocity is given by

\bar v=\dfrac{v+v_0}2

where v and v_0 are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}

where x,x_0 are the final/initial displacements, and t,t_0 are the final/initial times, respectively.

Take the car's starting position to be at t_0=0\,\mathrm s. Then

\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t

So we have

x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m

You also could have first found the acceleration using the equation

v=v_0+at

then solve for x via

x=x_0+v_0t+\dfrac12at^2

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of a anyway.

7 0
3 years ago
James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m
Over [174]

Answer:

4.1\cdot 10^8 N

Explanation:

First of all, we need to find the pressure exerted on the sphere, which is given by:

p=p_0 + \rho g h

where

p_0 =1.01\cdot 10^5 Pa is the atmospheric pressure

\rho = 1000 kg/m^3 is the water density

g=9.8 m/s^2 is the gravitational acceleration

h=11,000 m is the depth

Substituting,

p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m

So the total area of the sphere is

A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2

And so, the inward force exerted on it is

F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N

8 0
3 years ago
Read 2 more answers
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