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3 years ago

6

The actions of an employee are not attributable to the employer if the employer has not directly or indirectly encouraged the em

ployee to violate a law and if:

1 answer:

zepelin [54]3 years ago

4 0

Answer: the answer is d

Explanation: there are not more than 10 violations within a twelve month period hope this helps

You might be interested in

What is its maximum altitude above the ground? The answer is the maximum height above the ground

Kisachek [45]

Answer:

Maximum altitude above the ground = 1,540,224 m = 1540.2 km

Explanation:

Using the equations of motion

u = initial velocity of the projectile = 5.5 km/s = 5500 m/s

v = final velocity of the projectile at maximum height reached = 0 m/s

g = acceleration due to gravity = (GM/R²) (from the gravitational law)

g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)

g = -9.82 m/s² (minus because of the direction in which it is directed)

y = vertical distance covered by the projectile = ?

v² = u² + 2gy

0² = 5500² + 2(-9.82)(y)

19.64y = 5500²

y = 1,540,224 m = 1540.2 km

Hope this Helps!!!

3 0

3 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

6 0

3 years ago

Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim

AnnZ [28]

Answer:

$\boxed{v=\frac {V}{\sqrt {2}}}$

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that $f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}$

Now, the speed of the father will be $V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})$ while for the child the speed will be $v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})$

The ratio of the father’s speed to the child’s speed will be

$\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}$

8 0

2 years ago

A family took a trip in a car traveling East from Greensboro to Wilmington, NC. Use the Graph to answer the questions below.

kotykmax [81]

Answer:

1). Average speed = 1.5 m per second

2). Average velocity = 1.5 m per second

Explanation:

1). Since, speed is a scalar quantity

Therefore, average speed of the trip = $\frac{\text{Total distance covered}}{\text{Total time taken}}$

From the graph attached,

Total distance covered = 10 + 10 + 20 + 0 + 20 + 30

= 90 meters

Total time taken = 60 seconds

Average speed = $\frac{90}{60}$

= 1.5 meter per second

2). Velocity is a vector quantity.

Therefore, average velocity = $\frac{\triangle d}{\triangle t}$

= $\frac{d_{60}-d_0}{60-0}$

= $\frac{90-0}{60-0}$

= 1.5 meter per second

7 0

3 years ago

A 5kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N opposing the motion. Draw a fo

Taya2010 [7]

Answer:

Net force on the block is 32 N.

Acceleration of the object is 6.4 m/s².

Explanation:

Let the acceleration of the object be $a$ m/s².

Given:

Mass of the block is, $m=5\ kg$

Force of pull is, $F=40\ N$

Frictional force on the block is, $f=8\ N$

The free body diagram of the object is shown below.

From the figure, the net force in the forward direction is given as:

$F_{net}=F-f=40-8=32\ N$

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

$F_{net}=ma\\32=5a\\a=\frac{32}{5}=6.4\ m/s^2$

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².

3 0

3 years ago