All categories

3 years ago

For the hypothetical salaries in the following

Physics

1 answer:

FrozenT [24]3 years ago

3 0

Answer: $43400 per year

Explanation:

The Data set of hypothetical salaries is given as:

$46,000 per year,

$20,000 per year,

$33,000 per year,

$27,000 per year,

$91,000 per year.

The average salary per year will be the addition of the salaries per year above which will then be divided by 5. This will be:

= ($46,000 + $20,000 + $33,000 + $27,000 + $91,000) / 5

= $217000/5

= $43400 per year

The average salary per year will be $43400

You might be interested in

A 0.49-kg cord is stretched between two supports,7.3m apart. When one support is struck by a hammer, a transverse wave travels d

Wewaii [24]

Answer:

T= 5.18N

Explanation:

u = mass of chord / length of chord

u = 0.49/ 7.3

u = 0.067 kg/m

Velocity of sound waves (v) =length of chord / time taken for wave to travel

v = 7.3 / 0.83 = 8.795m/s

Tension is calculated below using the formula

T = v² * u

T = (8.795)² x 0.067

T= 5.18N

3 0

3 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago

How do I know if i’m doing number 2 right?

Ksju [112]

We are given an object that is speeding up on a level ground.

Let's remember that the gravitational energy depends on the change in height, therefore, if the object is not changing its height it means that the gravitational energy remains constant.

The kinetic energy depends on the velocity. If the velocity is increasing this means that the kinetic energy is also increasing.

Now, every change in velocity requires acceleration and acceleration requires a force. The force and the distance that the object moves are equivalent to the work that is transferred to the object and therefore, the change in kinetic energy. This means that the total energy of the system increases as work is transferred to the mass.

We have that the total energy of the system increases in the form of kinetic energy and that the gravitational potential energy remains constant. Therefore, the diagrams should look like pie charts that grow but the area of the segment of the potential energy stays the same. It should look similar to the following.

8 0

1 year ago

Types of mechanical waves include

USPshnik [31]

I think surface waves

4 0

3 years ago

Consider a highway composed of concrete. If the road is 4.5 km long, 30 m wide, and 0.70 m thick, what is its mass?

Aloiza [94]

Answer:

Explanation:

We need to assume that the density of the concrete is about 2350 Kg/m^3. And using the dimensions of the highway we can calculate the volume of the highway.

$vh=4500 * 30 * 0,70\\vh=94500 m^3\\den=m/v\\m=den*v\\m=2350*94500\\\\m=222075 ton$

5 0

3 years ago