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1 year ago

Consider the position vs. time graph below for a woman's movement in a hallway. What is the woman's velocity from 4 to 5 s?

Physics

1 answer:

Ksenya-84 [330]1 year ago

7 0

Answer:

The answer is " $6\ \frac{m}{s}$ "

Explanation:

The formula for velocity:

$\to \overline{v}={\frac{\Delta x}{\Delta t}}$

$=\frac{6}{1}\\\\=6\ \frac{m}{s}$

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Some car manufacturers claim that their vehicles could climb a slope of 42 ∘. For this to be possible, what must be the minimum

Paladinen [302]

Answer:

D. 0.9

Explanation:

Calculating minimum coefficient of static friction, we first resolve the forces (normal and frictional) acting on the vehicle at an angle to the horizontal into their x and y components. After this, we can now substitute the values of x and y components into equation of static friction. Diagrammatic illustration is attached.

Resolving into x component:

∑ $F_{x} = F_{s} - mgsin\alpha =0$

$F_{s} = mgsin\alpha$ ------(1)

Resolving into y component:

∑ $F_{y} = F_{n} - mgcos\alpha =0$

$F_{n} = mgcos\alpha$ ------(2)

Static frictional force, $F_{s} \leq$ μ $F_{n}$ ------(3)

substituting $F_{s}$ from equation (1) and $F_{n}$ from equation (2) into equation (3)

$mgsin\alpha \leq$ μ $mgcos\alpha$

$sin\alpha \leq$ μ $cos\alpha$

μ $\geq \frac {sin\alpha}{cos\alpha}$

μ $\geq tan\alpha$

The angle the vehicles make with the horizontal α = 42°

μ ≥ tan 42°

μ ≥ 0.9

6 0

1 year ago

A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Bro

aleksley [76]

Answer:

Potential difference though which the electron was accelerated is $2.67\times 10^{-6}\ uV\ .$

Explanation:

Given :

De Broglie wavelength , $\lambda=750\ nm.$

Plank's constant , $h=6.626\times 10^{-34}\ J.s \ .$

Charge of electron , $e=-1.6\times 10^{-19}\ C.$

Mass of electron , m=9.11\times 10^{-31}\ kg. $m=9.11\times 10^{-31}\ kg.$

We know , according to de broglie equation :

$\lambda=\dfrac{h}{mv}\\\\ v=\dfrac{h}{m\lambda}\\\\v=\dfrac{6.626\times 10^{-34}\ J.s \ }{9.11\times 10^{-31}\ kg\times 750\times 10^{-9}\ m }= 969.78\ m/s .$

Now , we know potential energy applied on electron will be equal to its kinetic energy .

Therefore ,

$qV=\dfrac{mv^2}{2}\\\\ V=\dfrac{mv^2}{2q}$

Putting all values in above equation we get ,

$V=2.67\times 10^{-6}\ uV .$

Hence , this is the required solution.

5 0

2 years ago

De Vico Comet orbits the Sun every 74.0 years and has an orbital eccentricity of 0.96. Find the comet's average distance from th

Debora [2.8K]

Answer: The comet's average distance from the sun is 17.6AU

Explanation:

From Kepler's 3rd Law, P^2=a^3

Where P is period in years

and a is length of semi-major axis or the average distance of the comet to the sun.

Given the orbital period to be 74 years

74^2 =a^3

5476 = a^3

Cube root of 5476 =a

17.626 = a

Approximately a= 17.6 AU

5 0

2 years ago

Julius competes in the hammer throw event. The hammer has a mass of 7.26 kg and is 1.215 m long. What is the centripetal force o

nevsk [136]

In the circular motion of the hammer, the centripetal force is given by
$F=m \frac{v^2}{r}$
where m is the mass of the hammer, v its tangential speed and r is the distance from the center of the motion, i.e. the length of the hammer.
Using the data of the problem, we find:
$F=m \frac{v^2}{r}=(7.26 kg) \frac{(31.95 m/s)^2}{1.215 m}=6100 N$

4 0

2 years ago

Read 2 more answers

In Paul Hewitt's book, he poses this question: "If the forces that act on a bullet and the recoiling gun from which it is fired

Sauron [17]

They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.

Hope this helps :)

4 0

2 years ago