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2 years ago

The strongest ernergy wave in the EM Spectrum is 'microwave' true false

Physics

1 answer:

Ratling [72]2 years ago

6 0

Pretty sure it’s false....................

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A planet orbits a start with the path shown below.

kvv77 [185]

PART a)

As we know that gravitational potential energy is given by the formula

$U = -\frac{Gm_1m_2}{r}$

here we can see that gravitational potential energy inversely varies with the distance

so here when distance from the sun is minimum then magnitude of gravitational potential energy is maximum while since it is given with negative sign so its overall value is minimum at that position

So gravitational potential energy is minimum at the nearest point and maximum at the farthest point

PART b)

Since we know that sum of kinetic energy and potential energy is constant here

so the points of minimum potential energy is the point where kinetic energy is maximum which means speed is maximum

So here speed is maximum at the nearest point

Part C)

since gravitational potential energy inversely varies with distance so it's graph will be like hyperbolic graph with distance

4 0

3 years ago

what is the mass of a pure platinum disk with a volume of 113 cm3? the density of platinum is 21.4 g/cm3

Oduvanchick [21]

V=m×d
m=v/d
m=113/21.4
m=5.28g

6 0

3 years ago

Read 2 more answers

While bats emit a wide variety of sounds, one type emits pulses of sound at a frequency between 39 kHz and 78 kHz. What is the r

sdas [7]

Answer:

The range of wavelengths of the sound is 7692.30 m and 3846.15 m

Explanation:

A bat emits pulses of sound at a frequency between 39 kHz and 78 kHz. It is required to find the range of wavelengths of this sound.

Bat uses ultrasonic waves. It moves with the speed of light.

If f = 39 kHz,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{39\times 10^3}\\\\\lambda=7692.30\ m$

If f = 78 kHz,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{78\times 10^3}\\\\\lambda=3846.15\ m$

So, the range of wavelengths of the sound is 7692.30 m and 3846.15 m.

6 0

3 years ago

A projectile is fired horizontally from a height of 78.4 m at a speed of 300 m/sec. How far did it travel horizontally before hi

Mice21 [21]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g

U is the speed = 300m/s

H is the maximum height = 78.4m

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

Substitute into the fromula;

R = 300√2(78.4)/9.8

R = 300 √(16)

R = 300*4

R = 1200m

Hence the projectile travelled 1200m before hitting the ground

3 0

2 years ago

The angular speed of the hour hand of a clock, in rad/min, is:___________

lapo4ka [179]

The angular speed is defined as:

where

$T=12*60=720min$

$\omega=\frac{2\pi}{720}$

$\omega=4.4**10^{-3} rad/min$

4 0

3 years ago