State Newtons first law of motion.

1. What are the units of area under the line/curve?<br> 2. Does the area have any meaning?

TiliK225 [7]

The area under the velocity time graph is 125 m and the meaning of the area is displacement.

<h3>What is area under velocity - time graph?</h3>

The area under a velocity time graph represents the displacement of the object.

total area of the graph = A1 + A2

total area of the graph = ¹/₂ (base₁)(height₁) + ¹/₂ (base₂)(height₂)

total area of the graph = ¹/₂(4)(40) + ¹/₂(3)(30)

total area of the graph = 125 m

Thus, the area under the velocity time graph is 125 m and the meaning of the area is displacement.

Learn more about velocity time graph here: brainly.com/question/4710544

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6 0

1 year ago

1. What is the name given to the total possibility of sound frequencies

exis [7]

Explanation:

1. The entire span of possible sound waves is called the acoustic spectrum. It is subdivided into infrasonic sounds, audible sounds, and ultrasonic sounds.

2. The difference between a musical note and another note at twice the frequency is called an octave.

3. Sound intensity varies with the inverse square of distance.

4 0

3 years ago

A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas

Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

(c) the maximum acceleration of the object, a_max = 15.4 m/s²

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

$V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s$

(c) the maximum acceleration of the object

$a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2$

6 0

3 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$