Describe what it means to view something from a frame of reference.Give an example to illustrate your explanation.

A 2000 kg car moves along a horizontal road at speed vo

cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

$\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a$ (Eq. 1)

$\Sigma F_{y} = N-m\cdot g = 0$ (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

$a = -\mu_{k}\cdot g$ (Eq. 3)

Where:

$a$ - Deceleration of the car, measured in meters per square second.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $\mu_{k} = 0.0735$ and $g = 9.807\,\frac{m}{s^{2}}$ , then deceleration of the car is:

$a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )$

$a = -0.721\,\frac{m}{s^{2}}$

The stopping distance of the car ( $\Delta s$ ), measured in meters, is determined from the following kinematic expression:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$ (Eq. 4)

Where:

$v_{o}$ - Initial speed of the car, measured in meters per second.

$v$ - Final speed of the car, measured in meters per second.

If we know that $v_{o} = 15.9\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $a = -0.721\,\frac{m}{s^{2}}$ , stopping distance of the car is:

$\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}$

$\Delta s = 175.319\,m$

The shortest possible stopping distance of the car is 175.319 meters.

8 0

3 years ago

This is my first question don’t really know how to use this app yet lol but somebody answer it for me pls!! Seded the corredare

miskamm [114]

3x6

(This just for extras )

7 0

3 years ago

A force of 12 N changes the momentum of a toy car from 3kgm/s t0 10kgm/s. Calculate the time the force took to produce this chan

yan [13]

Answer:

Time = 0.58 seconds

Explanation:

Given the following data;

Initial momentum = 3 kgm/s

Final momentum = 10 kgm/s

Force = 12 N

To find the time required for the change in momentum;

First of all, we would determine the change in momentum.

$Change \; in \; momentum = final \; momentum - initial \; momentum$

$Change \; in \; momentum = 10 - 3$

Change in momentum = 7 kgm/s

Now, we can find the time required;

Note: the impulse of an object is equal to the change in momentum experienced by the object.

Mathematically, impulse (change in momentum) is given by the formula;

$Impulse = force * time$

Making "time" the subject of formula, we have;

$Time = \frac {impulse}{force}$

Substituting into the formula, we have;

$Time = \frac {7}{12}$

Time = 0.58 seconds

6 0

3 years ago

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.23 g/cm3. The area of ea

motikmotik

Explanation:

Work done by gravity is given by the formula,

W = $\rho A (h_{1} - h)g (h - h_{2})$ ......... (1)

It is known that when levels are same then height of the liquid is as follows.

h = $\frac{h_{1} + h_{2}}{2}$ ......... (2)

Putting value of equation (2) in equation (1) the overall formula will be as follows.

W = $\frac{1}{4} \rho gA(h_{1} - h_{2})^{2})$

= $\frac{1}{4} \times 1.23 g/cm^{3} \times 9.80 m/s^{2} \times 3.89 \times 10^{-4} m^{2}(1.76 m - 0.993 m)^{2})$

= 0.689 J

Thus, we can conclude that the work done by the gravitational force in equalizing the levels when the two vessels are connected is 0.689 J.

3 0

3 years ago

Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of oneâs birth. The on

Anton [14]

Answer:

Explanation:

Gravitational force between two objects having mass m₁ and m₂ at a distance R

F = G m₁ m₂ / R²

Force between baby and father F₁ = 6.67x10⁻¹¹ x 4.1 x 120 / .18²

= 1.01 x 10⁻⁶ N

b )

Force between baby and Jupiter

F₂ = 6.67x10⁻¹¹ x 1.9x 10²⁷ x 4.1 / ( 6.29 x 10¹¹ )²

= 1.31 x 10⁻⁶ N

c )

Ratio = 1.01 / 1.31

= .77

4 0

3 years ago