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3 years ago

The shorter the wavelength the ____ the frequency

Physics

2 answers:

Yuri [45]3 years ago

8 0

Answer:

Yo, it's greater homies

Explanation:

I had just did it on USAtestprep

natali 33 [55]3 years ago

5 0

Higher the frequency and greater energy

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Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a second identical ball (Ball B) initially at rest. Afte

timurjin [86]

Answer:

Explanation:

We shall apply law of conservation of momentum during the collision of ball A and B .

Total momentum before collision of A and B = .35 x 10 = 3.5 kg m/s

Let the velocity of B after collision be v .

Total momentum after collision = .35 x 2 + .35v

According to law of conservation of momentum

.35 x 2 + .35v = 3.5

.35 v = 2.8

v = 8 m /s .

The direction of B will be same as direction of A .

3 0

3 years ago

IF YOU ANSWER THESE 4 QUESTIONS! I WILL GIVE YOU BRAINEST!!! 17 POINTS!!!

melisa1 [442]

Answer:

1.d

2.d

3.c

4.c

Okay, d & d, c & c. Iknow what you're thinking, I guessed. Nope! I just took a good look and chosed what I thought was the best answer

Explanation:

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3 years ago

Lake Preston's latitude is about 44.4 degrees north of the equator. Which global wind belt are

Serggg [28]

Answer:

Idk

Explanation:

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3 years ago

A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is

Soloha48 [4]

Answer: $V=5839.051m/s$

Explanation:

According to the Third Kepler’s Law of Planetary motion:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;:

$T=1.26(10)^{4}s$ is the period of the satellite

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=5.98(10)^{24}kg$ is the mass of the Earth

$a$ is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity $V$ is given by:

$V=\sqrt{\frac{GM}{a}}$ (2)

Now, from (1) we can find $a$ , in order to substitute this value in (2):

$a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}$ (3)

$a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}$ (4)

$a=11705845.57m$ (5)

Substituting (5) in (2):

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}$ (6)

$V=5839.051m/s$ (7) This is the speed at which the satellite travels

6 0

3 years ago

A 570 kg elevator accelerates downwards at 1.5 m/s2 for the first 13 m of its motion.

jeka94

Mass of the elevator (m) = 570 Kg
Acceleration = 1.5 m/s^2
Distance (s) = 13 m
Let the force be F.
We know, F = ma,
Therefore, F = (570 × 1.5) N = 855 N
Angle between distance and force (θ) = 0°
We know, work done = F s Cos θ
Therefore, work done by the cable during this part
= (855 × 13 × Cos 0°) J
= (855 × 13 × 1) J
= 11115 J

Answer:

11115 J

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0

3 years ago