What is the eccentricity of an ellipse with a foci distance of 50,000,000 km and

Air can be humidified by passing through a long tube whose inside diameter is lined with a wick saturated with liquid water. The

Oduvanchick [21]

Answer:

IV because the process of water is equal to 5,8 to 783253.23 to the hendroxagram of 4.

Explanation:

4 0

3 years ago

A 150kg person stands on a compression spring with spring constant 10000n/m and nominal length of 0.50.what is the total length

Ivahew [28]

Answer:

The total length of the spring would be 0.65 m

Explanation:

The Concept

Hooke's law evaluates the increment of spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;

F = k x ..................1

where F is the force applied to the spring

k is the spring constant

x is the spring stretch or extension

Step by Step Calculations

We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;

x = F/k

but F = m x g

so, x = (m x g)/k

given that, the mass of the person m =150 kg

g is the acceleration due to gravity = 9.81 m/ $s^{2}$

k is the spring constant = 10000 N/m

then x = (9.81 m/ $s^{2}$ x 150 kg)/10000 N/m

x = 0.14715 m

the extension experienced by the spring after the compression is 0.14715 m

The total length of the spring would be;

L = 0.14715 m + 0.5 m = 0.64715

L ≈ 0.65 m

Therefore the total length of the spring would be 0.65 m

4 0

3 years ago

A net force of 24 N is acting on a 4.0-kg object. Find the acceleration in m/s2.

Inessa [10]

Hi there!

We can use Newton's Second Law:
$\Sigma F = ma$

ΣF = Net force (N)
m = mass (kg)
a = acceleration (m/s²)

We can rearrange the equation to solve for the acceleration.

$a = \frac{\Sigma F}{m}\\\\a = \frac{24}{4} = \boxed{6 \frac{m}{s^2}}$

6 0

2 years ago

Ramesh announced in class: ''Yesterday I had fever and my body temperature was 100 degrees.'' Ravi said: ''We learnt in the last

pogonyaev

Answer:

D. Ramesh and Ravi are correct, but they are using different measurement scales.

\Huge{\underline{\textrm{Explanation}}}Explanation

Here, Ravi says that his body temperature is 100 degrees, but does not mention that whether it is 100 degrees Celsius or 100 degrees Fahrenheit. When the temperature of a human body is more than 100.4 degree Fahrenheit (38°C), or near to it, the person is considered to have fever.

The boiling point of water is 100 degrees Celsius and not 100 degrees Fahrenheit.

Thus, they both are using different measurement scales.

7 0

2 years ago

A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re

slamgirl [31]

Answer:

tension: 19.3 N
acceleration: 3.36 m/s^2

Explanation:

Given

mass A = 2.0 kg

mass B = 3.0 kg

θ = 40°

Find

The tension in the string

The acceleration of the masses

Solution

Mass A is being pulled down the inclined plane by a force due to gravity of ...

F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

F/m = a = F/m

(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0

3 years ago