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2 years ago

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

Physics

2 answers:

MaRussiya [10]2 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

34kurt2 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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A block of mass m=1kg sliding along a rough horizontal surface is traveling at a speed v0=2m/s when it strikes a massless spring

topjm [15]

Here we can use the work energy theorem

$W_f + W_s = K_f - K_i$

here we know that

$K_f = 0$

as it come to rest finally

$K_i = \frac{1}{2}mv_i^2$

$K_i = \frac{1}{2}\times 1\times 2^2$

$K_i = 2 J$

now work done by friction force will be given as

$W_f = - F_f \times d = -\mu mg d$

$W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu$

Work done by spring force is given as

$W_s = \frac{1}{2}k(x_i^2 - x_f^2)$

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$- 0.05 - \mu(0.98) = 0 - 2$

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3 years ago

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b

Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy= $\frac{1}{2} kx^{2} +PE1$

Energy at the point where the block will stop consists of only gravitational potential energy= $PE2$

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒ $\frac{1}{2} kx^{2} +PE1=PE2$

⇒ $PE2-PE1=\frac{1}{2} kx^{2}$

Also $PE2-PE2=mgh$

where $m$ is the mass of block

$g$ is acceleration due to gravity=9.8 m/s

$h$ is the difference in height between two positions

⇒ $mgh=\frac{1}{2} kx^{2}$

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x=11cm=0.11 m

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⇒ $20580*h=13.31$

⇒ $h=\frac{13.31}{20580}$

⇒h=0.0006467m= $6.5e-4$

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Answer:

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