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3 years ago

If someone travels 3000 feet in one second, how far will it travel in 18 minutes

Physics

2 answers:

MaRussiya [10]3 years ago

8 0

Answer:3240000 feet's

Explanation:

Since 60seconds equal 1minute therefore 1080 seconds equal 18minutes

Therefore the total number of feet traveled for 18minutes(1080seconds)=3000×1080=3240000 feet

34kurt3 years ago

3 0

Answer:

3240000 ft

Explanation:

There are 60 seconds in a minute so, multiply 60 seconds by 18 minutes. That equals 1080 seconds.

Since there are 3000 ft in each second, just multiply 1080 seconds by 3000 ft.

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Where in the motion is the magnitude of the force from the spring on the object zero? Where in the motion is the magnitude of th

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Answer:

1. The magnitude of the force from the spring on the object is zero on Equilibrium.

2. The magnitude of the force from the spring on the object is a maximum on The top and bottom.

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4 years ago

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

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moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

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