A ball of 12 kg is attached to a string of 0.8 meter spun at 4

Joanna learned from customer satisfaction surveys that diners in her restaurant wanted to be able to make substitutions. She has

Dmitry [639]

Answer:

standards

Explanation:

Based on the information provided within the question in regards to the situation at hand it can be said that Joanna is demonstrating a standards gap. this is a gap caused by the difference between the customer service standards a company has created for itself and the expectations the company believes that the customers have for that company. Since Joanna did not tell all the servers of the customers expectations then the ones who do not know will not be able to provide this service to those customers, thus the restaurant will not be able to meet it's customer service standards.

3 0

3 years ago

Which quote best reveals why sowers was successful in achieving her goal?

BabaBlast [244]

Sowers has spent her career fighting for gender equality and it passionate about it .

5 0

2 years ago

A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exert

strojnjashka [21]

Answer:

$v=10m/sec$

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

$F_N+Fg$

where

$F_N =forceof the normal$

$Fg= force due to gravity$

Generally the net force is given to be $FC(force towards center)$

$F_C =F_N + Fg$

$F_N =F_C -Fg$

$F_N=F_C-F_g$

$F_N=\frac{mv^2}{R} -mg$

Mathematical we can now derive V

$m_g + \frac{8m}{4}= \frac{mv^2}{8}$

$\frac{5mg}{4} =\frac{mv^2}{8}$

$v^2 =\frac{40*10}{4}$

$v=10m/sec$

Therefore the speed of the roller coaster is given ton be $v=10m/sec$

5 0

2 years ago

Rearrange the formula for mechanical energy to solve for height:

docker41 [41]

Explanation:

Given formula:

ME= $\frac{1}{2}$ mv²+mgh

To make height the subject of the formula, follow the following procedures;

Subtract $\frac{1}{2}$ mv² from both side of equation

M.E - $\frac{1}{2}$ mv² = $\frac{1}{2}$ mv² - $\frac{1}{2}$ mv²+mgh

This gives:

M.E - $\frac{1}{2}$ mv² = mgh

Multiply both sides of the expression by $\frac{1}{mg}$

( M.E - $\frac{1}{2}$ mv² ) x $\frac{1}{mg}$ = $\frac{1}{mg}$ x mgh

h = ( M.E - $\frac{1}{2}$ mv² ) x $\frac{1}{mg}$

Learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

4 0

2 years ago

A circular loop with radius r is rotating with constant angular velocity ω in a uniform electric field with magnitude E. The axi

inn [45]

Answer:

$\Phi_{E} = E\pi r^2 \omega t$

Explanation:

The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

$\Phi_{E} = \vec{E}\vec{A}$

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.

The above formula can be rewritten as follows

$\Phi_{E} = EA\cos(\theta)$

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.

If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

$\theta = \omega t$

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.

Therefore the electric flux can be written as a function of time

$\Phi_{E} = E\pi r^2 \omega t$

3 0

2 years ago