In the waves lab, you changed ____ to see how it would affect wave speed.

When an object like a tree is illuminated by the sun, and you are looking toward the tree, light rays leave the object _________

Dima020 [189]

Objects absorb and reflect light differently depending on their physical characteristics, such as their shape or composition. Thanks to the reflection we can see the objects. Reflection can be defined as the change of direction of a wave, which, when in contact with the separation surface between two changing means, returns to the point where it originated. When the light illuminates the object, such as the tree, the rays of light will disperse in all directions allowing observation.

The correct answer is A. From every point on the surface of the tree, and in every direction

6 0

4 years ago

Students will consider the following items: ice, glass, copper, and aluminium. what happens to each item when hammered? what pro

vladimir2022 [97]

All metals except potassium and sodium, have a property known as malleability. Malleability is the quality of something that can be shaped into something else without breaking. So when aluminium and copper are hammered they will not break. Rather they will change shape and become thin or flat at the area where its hammered.

All Non- metals except diamond are brittle in nature, so when we hammer it , they will break down into pieces. So when ice and glass will be hammered they will shatter into pieces.

4 0

3 years ago

A stone is thrown, run a velocity of 15m/s is projected of an elevation of 30° to the horizontal calculate the time rate of flig

frozen [14]

Answer:

1.53 seconds

Explanation:

Applying,

T = 2usin∅/g................ Equation 1

Where, T = time of flight, u = initial velocity, ∅ = angle of projectile to the horizontal, g = acceleration due to gravity

From the question,

Given: u = 15 m/s, ∅ = 30°

Constant: g = 9.8 m/s²

Substitute these values in equation 1

T = 2(15)(sin30°)/9.8

T = 15/9.8

T = 1.53 seconds

Hence the time rate of flight is 1.53 seconds

3 0

3 years ago

Which tagline from a car advertisement has the BEST connotative appeal to the parents of a new teenage driver?

Nonamiya [84]

A. Advanced new safety features that get everyone home at night.

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

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