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3 years ago

Which is the length of the route between two points

Physics

1 answer:

pychu [463]3 years ago

8 0

Answer:

Can you show us the route and then I would be happy to help :)

Explanation:

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A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s . The coefficient o

kobusy [5.1K]

2.04 meters distance is traveled by the sled before stopping.

Mass of the sled = m

The initial speed of the sled = 2 m/s

Coefficient of kinetic friction between sled and ice = 0.100

Let the distance the sled moves before it stops be d.

Gravity = 9.8 m/ s²

Let the initial kinetic energy sled be

$= K _{i}$

$K_{i} = \frac{1}{2} mv ^{2}$

The work done by the frictional force is,

$Work \: done \: by \: frictional \: force =W_{f}$

$W _{f} = μ_{k}mgd$

Work done by frictional force= Initial kinetic energy of the sled

$W_{f} = K_{i}$

$μ_{k}mgd= \frac{1}{2} mv ^{2}$

So, the distance traveled by the sled before stopping is

$d= \frac{1mv ^{2} }{2 \:μ_{k}mg}$

$d= \frac{1v ^{2} }{2 \:μ_{k}g}$

$d= \frac{2^{2} }{2 \times \:0.100 \times 9.8}$

$d= 2.04 \: m$

Therefore, the distance traveled by the sled before stopping is 2.04 meters.

To know more about work done, refer to the below link:

brainly.com/question/13662169

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5 0

2 years ago

What is the net worth of a car moving 30 miles per hour

miss Akunina [59]

Do you know the mass of the car?

6 0

3 years ago

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. Initially open and at re

xxTIMURxx [149]

Answer:

The final angular speed is 0.223 rad/s

Explanation:

By the conservation of angular moment:

ΔL=0

L₁=L₂

L₁ is the initial angular moment

L₂ is the final angular moment

L₁ is given by:

$L_1=L_{door} + L_{mud}$

As the door is at rest its angular moment is zero and the angular moment of mud can be considered as a point object, then:

$L_1= L_{mud}= mvr$

where

r is the distance from the support point to the axis of rotation (the mud hits at the center of the door; r=0.5 m)

v is the speed

m is the mass of the mud

L₂ is given by:

$L_2= (I_{door} + I_{mud}) \omega_f$

ωf is the final angular speed

The moment of inertia of the door can be considered as a rectangular plate:

$I_{door}=\frac{1}{3}MW^2$

M is the mass of the door

W is the width of the door

The moment of inertia of the mud is:

$I_{mud}=mr^2$

Hence,

$L_1=L_2\\mvr= (I_{door} + I_{mud}) \omega_f\\\omega_f=\frac{mvr}{I_{door} + I_{mud}} \\\omega_f=\frac{mvr}{I_{door} + I_{mud}}$

$\omega_f=\frac{0.5kg \times 12m/s \times 0.5m}{\frac{1}{3}40kg(1m)^2+0.5kg \times (0.5m)^2}$

$\omega_f=0.223 \frac{rad}{s}$

6 0

4 years ago

Need some help please

11Alexandr11 [23.1K]

Answer:

$x^2\neq -\frac{5}{13}$

First option

Explanation:

<u>Operations with functions </u>

Given two functions f, g, we can perform a number of operations with them including addition, subtraction, product, division, composition, and many others .

We have

$f(x)=20x^3-7x^2+3x-7$

$g(x)=-13x^2-5$

We are required to find

$\displaystyle \frac{f}{g}(x)$

We simply divide f by g as follows

$\displaystyle \frac{f}{g}(x)=\frac{20x^3-7x^2+3x-7}{-13x^2-5}$

We know rational functions may have problems if the denominator can be zero for some values of x. We must find out if there are such values and exclude them from the domain of the new-found function. We must ensure

$-13x^2-5\neq 0$

or equivalently

$x^2\neq -\frac{5}{13}$

Thus the first option is correct

Note: Since $x^2$ is always a positive number (for x real), our function does not really have any restriction in its domain

8 0

3 years ago

What is the nature of force between two charges if q1+q2=0

MAXImum [283]

Answer:
The force will be attractive.

Explanation:
Since we the sum of the charges is 0 they must have opposite charges and equal magnitudes.

And Opposite charges are attaractive towards each other.

3 0

4 years ago