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2 years ago

PWEESE HELPPPP I'LL GIVE BRAINLIEST, (NO LINKS OR I'LL REPORT)

Physics

2 answers:

miss Akunina [59]2 years ago

8 0

Answer:

extensions.

Explanation:

Hope this helps

Sever21 [200]2 years ago

7 0

Answer:

extensions i think because sound waves travel through the air and to our ears so its extended

Explanation:

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Why are infrared waves located between microwaves and visible light on the electromagnetic spectrum?

dangina [55]

Answer: The infra red waves is located between microwave and visible light based on their WAVELENGTH and FREQUENCY of occurrence.

Explanation:

Electromagnetic waves are those waves that do not require or need a material medium for its propagation, but they are able to travel through a vacuum. They exhibit or show all properties associated or connected with light. They are undeflected in electric and magnetic fields. These electromagnetic waves are arranged in order of their FREQUENCY and WAVELENGTHS which is known as ELECTROMAGNETIC SPECTRUM.

FREQUENCY is defined as the number of cycles which the wave completes in one second and is measured in Hertz(Hz). While WAVELENGTH is defined as the distance between two successive crests or troughs of waves which is measured in meter (m).

The electromagnetic spectrum is made up of the following rays which is arranged from the biggest wavelengths to the smallest:

--> Radiowaves

--> microwave :

--> infrared rays:

--> visible light:

--> ultraviolet rays

--> x-rays and

--> Gamma rays.

According to the arrangement of the spectrum above, the microwave has a higher wavelength and frequency than the infrared rays, while the visible light has a lower wavelength and frequency than the infrared rays.

3 0

2 years ago

A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t

levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

$F=m\omega^2 r$

where

m is the mass of the object

$\omega$ is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

$\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s$ is the angular velocity

r = 30 cm = 0.30 m is the radius of the circle

Substituting, we find the force:

$F=(0.040)(6.28)^2(0.30)=0.47 N$

3 0

2 years ago

Two pans of water are on different burners of a stove. One pan of water is boiling vigorously, while the other is boiling gently

vichka [17]

Answer:

The correct answer is C.

Explanation:

3 0

3 years ago

Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many

dimulka [17.4K]

Explanation:

It is given that,

Velocity of the particle moving in straight line is :

$v(t)=t^2e^{-2t}\ m/s$

We need to find the distance (x) traveled by the particle during the first t seconds. It is given by :

$x=\int\limits {v.dt}$

$x=\int\limits {t^2e^{-2t}dt}$

Using by parts integration, we get the value of x as :

$x=\dfrac{-(2t^2+2t+1)e^{-2t}}{4}\ meters$

Hence, this is the required solution.

6 0

3 years ago

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh

Karolina [17]

Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

We are told that the salmon started downstream, 3.18 m away from a waterfall.

Thus, range = 3.18 m

Since the horizontal velocity component is constant, then;

Range = vcosθ × t

Thus,

vcosθ × t = 3.18 - - - (eq 1)

We are told the salmon reached a height of 0.294 m

Thus, using distance equation;

s = v_y•t + ½gt²

g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

0.294 = v_y•t - ½gt²

v_y = vsinθ

Thus;

0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

We are given g = 9.81 m/s² and θ = 45°

0.295 = (3.18 × tan 45) - ½(9.81) × t²

0.295 = 3.18 - 4.905t²

3.18 - 0.295 = 4.905t²

4.905t² = 2.885

t = √2.885/4.905

t = 0.5882 s

Thus;

v = 3.18/(0.5882 × cos45)

v = 7.65 m/s

8 0

3 years ago