All categories

3 years ago

What are the units for the spring constant, k?

Physics

2 answers:

vitfil [10]3 years ago

5 0

Answer:

C. newtons/meter

Explanation:

To find the units of spring constant you can use the hook's law, which says,

F = k × x where

F = force (unit Newton)

k = spring constant

x = expansion or contraction ( unit m)

So by this equation you can get that

k = F / X

So you get the units as N/m

Tanya [424]3 years ago

4 0

The units for the spring constant, k, are Newtons per meter (N/m).)

You might be interested in

Examine the cross-sectional slice of a stem of a mint plant. Which statement most specifically describes mint?

Maslowich

The available options are:

Mint is a dicot.

Mint is a monocot.

Mint is an angiosperm.

Mint is a bulb plant.

Answer:

Mint is a dicot.

Explanation:

Given the fact that Mint is considered to be a member of Lamiaceae, an angiosperm plant which is characterized by typically having leaves that consist of reticulate vacation and appears like veins in structure. It also has a seed that contains two cotyledons.

Hence, it is considered a DICOT PLANT due to these characteristics. The botanical name of Mint is referred to as Mentha arvensis.

7 0

2 years ago

The current theory of the structure of the

myrzilka [38]

Answer:

hiiiiiiiiiiii

Explanation:

4 0

2 years ago

What is the speed of a wave that has a wavelength of 0.4 m and a frequency of 10 Hz

Shkiper50 [21]

<span>speed = wavelength x frequency
speed = 0.4m X 10 Hz
speed = 4 m/s</span>

4 0

2 years ago

Scott is walking 11 m/s due south along a road when he spots Karla, who is walking toward him at 1.5 m/s. On her way, Karla sees

emmasim [6.3K]

Answer: A. Scott and Karla

Explanation: Velocity includes speed AND direction.

8 0

2 years ago

An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t

mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a) the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F $_L$ = C $_L$ $\frac{1}{2}$ ρV²A = W

we substitute

0.45 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F $_D$ = T = C $_D$ $\frac{1}{2}$ ρV²A

we substitute

= 0.065 × $\frac{1}{2}$ × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

8 0

3 years ago