All categories

2 years ago

Which type of electromagnetic radiation cannot be focused?

Physics

2 answers:

Irina18 [472]2 years ago

7 0

The answer is A. Hope this helps. :)

dalvyx [7]2 years ago

5 0

(A) Gamma rays

Focussing is a process in which a beam of light is passed and concentrated on the particular point.

Gamma rays are the type of elecromagnetic radiation that cannot of focused. Gamma rays has high frequency and also are quite energetic due to which when the beam of light is passed through it, it becomes too difficult to focus on a particular point as they interacts strongly with the matter and destroys itself. Hence, Gamma rays are not easily focused.

You might be interested in

If an object placed in a fluid experiences a buoyant force equal to the weight of the fluid it displaces, the object will float.

enyata [817]

Answer: The answer is False

Explanation: This is for the one's in apex <>

6 0

2 years ago

If the ankylosaurs starts running and accelerates at 1.3m/s^2 for 3 seconds how far does he make it before the velociraptor catc

sveta [45]

Acceleration is the change of velocity, and velocity is the change of distance. The opposite of finding change, or differentiation, is integration.

Acceleration = 1.3 m/s²

Velocity: ∫ 1.3 dx = 1.3x + c m/s

Distance: ∫ 1.3x dx = 1.3x²/2 + c m

Distance run: 1.3*3²/2 = 5.85 m

What bad thing happened?

7 0

2 years ago

A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f

hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

$h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2$ .... (2)

By equation the equation (1) and (2), we get

$41.7=1.12 u +4.9 \times 1.12^{2}$

u = 31.75 m/s

5 0

2 years ago

What is a measure between the difference in start and end positions?

crimeas [40]

Answer:

Displacement

General Formulas and Concepts:

Kinematics

Displacement vs Total Distance

Explanation:

Displacement is the difference between the start position and end position.

Total Distance is the entire distance traveled between the start and end position.

Topic: AP Physics 1 Algebra-Based

Unit: Kinematics

5 0

2 years ago

A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t

rusak2 [61]

Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

$v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt$

v_ave = v0+a(T/2)

We can esaily note that when t=T/2

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

$x(t) = x_A + v_0 t + \frac{1}{2}at^2$

Where x_A is the position of point A. Therefore, the car will be at:

x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2

8 0

2 years ago