In your egg drop you want to decrease.

A man pushes a shopping cart across a level floor. What force resists the effort force? A) gravity B) friction C) the normal for

Kipish [7]

Answer:

B) Friction

Explanation:

Friction is a force that acts when an object is sliding along a surface. Microscopically, this force is due to the fact that the two surfaces are not perfectly smooth, but they have "imperfections" that cause a force that opposes the motion of the object.

For an object sliding on a flat surface, the force of friction has magnitude:

$F_f = \mu_k mg$

where

$\mu_k$ is the coefficient of kinetic friction

m is the mass of the object

g is the acceleration of gravity

The direction of the force of friction is always opposite to the direction of motion of the object.

In reality, friction also acts if the object is at rest and it is pushed by a force; in this case, we talk about static friction, and its magnitude is

$F_f = \mu_s mg$

where $\mu_s$ is called coefficient of static friction, and it is generally larger than the coefficient of kinetic friction.

8 0

2 years ago

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

(c) $8.66in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):

$c^2=a^2+b^2-2abcos(\gamma)$

$c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in$

4 0

3 years ago

Identify and sketch all the external forces acting on the chair. because the chair can be represented as a point particle of mas

SOVA2 [1]

There are two external force acts on the chair.

1. The force due to earth gravity, acts in the downward direction.

2. Reaction force of the gravity, which acts in the Upward direction (Normal Force).

On every object, there is a force acts due to gravity of earth, which pulls the object towards the centre of earth, known as gravity force, always acts in the downward direction. Mathematically it's given as

F=mg

here, m is the mass of the object, and g is the acceleration of gravity.

To balance this gravity force, a counter force acts in the opposite direction, whose magnitude is equal to the force of gravity

8 0

2 years ago

Read 2 more answers

two ships leave a port at the same time. The first ship sails on a bearing of 40 degrees at 18 knots and the second at a bearing

Yuliya22 [10]

Answer:

The distance between the ships is 87.84 km.

Explanation:

Given that,

Angle of first ship= 40°

Speed of first ship = 18 knots

Angle of second ship= 130°

Speed of second ship = 26 knots

We need to calculate the resultant velocity

Using cosine rule

$v=\sqrt{v_{1}^2+v_{2}^2-2v_{1}v_{2}\cos\theta}$

Put the value into the formula

$v=\sqrt{18^2+26^2-2\times18\times26\times\cos90}$

$v=\sqrt{18^2+26^2}$

$v=\sqrt{324+676}$

$v=10\sqrt{10}$

We need to calculate the distance between the ships

$d =v\times t$

Put the value into the formula

$d=10\sqrt{10}\times1.5\times1.852$

$d=87.84\ km$

Hence, The distance between the ships is 87.84 km.

7 0

2 years ago

What is the distance from axis about which a uniform, balsa-wood sphere will have the same moment of inertia as does a thin-wall

andrey2020 [161]

Answer:

$D_{s}$ ≈ 2.1 R

Explanation:

The moment of inertia of the bodies can be calculated by the equation

I = ∫ r² dm

For bodies with symmetry this tabulated, the moment of inertia of the center of mass

Sphere $Is_{cm}$ = 2/5 M R²

Spherical shell $Ic_{cm}$ = 2/3 M R²

The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass

I = $I_{cm}$ + M D²

Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation

Let's start with the spherical shell, axis is along a diameter

D = 2R

Ic = $Ic_{cm}$ + M D²

Ic = 2/3 MR² + M (2R)²

Ic = M R² (2/3 + 4)

Ic = 14/3 M R²

The sphere

Is = $Is_{cm}$ + M [ $D_{s}$ ²

Is = Ic

2/5 MR² + M $D_{s}$ ² = 14/3 MR²

$D_{s}$ ² = R² (14/3 - 2/5)

$D_{s}$ = √ (R² (64/15)

$D_{s}$ = 2,066 R

3 0

3 years ago