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oee [108]
3 years ago
5

Which of the following is not a dwarf planet?

Physics
1 answer:
Fantom [35]3 years ago
8 0
The correct answer is (a.) Hydra. Hydra is not a dwarf planet, instead, it is the moon of the dwarf planet, Pluto. There are only four accepted dwarf planets by the International Astronomical Union which were the Haumea, Pluto, Eris, and Makemake. 
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The results of a recent television survey of American TV households revealed that 86 out of every 100 TV households have at leas
Lelechka [254]

Answer:

q = \dfrac{14}{100}

Explanation:

given,

86 out of every 100 TV households have at least one remote control

Probability of any material is calculated by.

Probability = \dfrac{favourable\ outcome}{Total\ outcome}

P(at\ least\ one\ remote\ control)=\dfrac{86}{100}

P(at\ least\ one\ remote\ control)=0.86

now, Calculating probability that at least one remote control

   q = 1 - 0.86

   q = 0.14

P(doesnot\ have\ at\ least\ one\ remote\ control)=\dfrac{14}{100}

q = \dfrac{14}{100}

7 0
3 years ago
The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
3 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
3 years ago
a car initially at 65.00 m/s accelerates at 22.39 m/s^2. How far has the car traveled before it comes to a stop
Gennadij [26K]

Answer:

94.35 meter

Step by step explanation

7 0
2 years ago
Vector a has a magnitude of 12.3 units and points due west. vector b points due north. what is the magnitude of b if a - b has a
algol [13]

The magnitude of vector b is 8.58 Unit.

Since both the vectors a and b are perpendicular to each other, so we can apply the Pythagoras theorem to calculate the magnitude of the vector b.

Applying the Pythagoras theorem

(a-b)^2=a^2+b^2

15^2=12.3^2-b^2

b=8.58 unit

Therefor the magnitude of the vector b is 8.58 unit.

8 0
2 years ago
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