A current relationship is one that

1) Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.

Anton [14]

Answer:

The order of increasing energy is as follows

"microwave < infrared < visible < ultraviolet"

Option (A) is correct.

Explanation:

Given:

Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.

From the formula of energy in terms of frequency.

$E = hf$

Where $h =$ planck constant, $f =$ frequency of light.

From above formula we can conclude that higher frequency means higher energy.

In our case ultraviolet has higher frequency and microwave has lower frequency.

So ultraviolet has higher energy and microwave has lower energy.

microwave < infrared < visible < ultraviolet

Therefore, the order of increasing energy is as follows

"microwave < infrared < visible < ultraviolet"

7 0

3 years ago

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

$I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2$

$I = 0.394kg\cdot m^2$

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$ , therefore

$W = \tau \theta$

$W = 11.5* \frac{\pi}{3}$

$W = 12.09J$

4 0

3 years ago

A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.

Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

= 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = $0.5mv^{2}$

= $0.5 x 25 x 5.6^{2}$ = 392 J

(C) average frictional force = $\frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}$

change in KE (ΔKE) = initial KE - final KE
ΔKE = $0.5mv^{2}$ - $0.5mvf^{2}$
when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE = $0.5x25x5.6^{2}$ - 0 = 392 J

\frac{-(ΔKE+ΔU)}{h}[/tex] = $\frac{-(392 + (-2940))}{12}$

= $\frac{(-392 + 2940)}{12}$ = 212.33 N

5 0

3 years ago

What is the change in momentum of a 50-kg rock that falls freely for 3 seconds?

Elan Coil [88]

Answer:

1470kgm/s

Explanation:

Given parameters:

Mass of the rock = 50kg

Time taken for the free fall = 3s

Unknown:

Change in momentum = ?

Solution:

The change in momentum will be difference between the ending momentum and finishing momentum.

Momentum is the product of mass and velocity

Momentum = mass x velocity

Initial momentum = 0, the velocity is 0

Final momentum = mass x final velocity

let us find the final velocity;

V = U + gt

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity = 9.8m/s²

t is the time

V = 0 + 9.8x3 = 29.4m/s

So;

Change in momentum = 50 x 29,4 = 1470kgm/s

6 0

3 years ago

A student, who weighs 720N, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is

jasenka [17]

Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

Explanation:

The given parameters are;

The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

The acceleration due to gravity is a constant = 9.81 m/s²

Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, $F_{net}$ = The weight of the student + The inertia force of the slowing elevator

∴ The net force, $F_{net}$ = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N

3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg

The value the scale will read = 96.59 kg.

3 0

3 years ago