All categories

3 years ago

How do you identify short and long term comets

1 answer:

8 0

A second important difference between comets coming from the Kuiper Belt and from the Oort cloud is represented by their different characteristic periods.
In fact, short period comets are thought to generate in the Kuiper belt and have rather predictable orbits with short periods (up to 200 years). There are two major families of short period comets: the Jupiter family with periods of less than 20 years and the Halley family with periods form 20 to 200 years. That's short

You might be interested in

Find the work done "by" the electric field on a positively charged point particle with a charge of 2.1x10^-6 C as it is moved fr

Alexxx [7]

Answer

Work done will be $14.7\times 10^{-6}J$ and it will be positive

Explanation:

We have given charge $2.1\times 10^{-6}C$

We have to find work done in moving the charge from 15 volt to 8 volt

Let $V_1=15V\ and\ V_2=8volt$

So potential difference $V=V_1-V_2=15-8=7volt$

We know that work done $W=QV$ , here Q is charge and V is potential difference

So work done $W=QV=2.1\times 10^{-6}\times 7=14.7\times 10^{-6}J$

It will be positive work done because work is done in moving charge from higher potential to lower potential

5 0

4 years ago

Water is diffrent from other substances because:

sergeinik [125]

Its B.hope this helps

8 0

3 years ago

Read 2 more answers

Here's an interesting challenge you can give to a friend. Hold a $1 bill by the upper corner. Have a friend prepare to pinch the

grin007 [14]

I would say that it would take her 35 * 2 cashing Bill properly because I multiply 0.25 times 16 which gave me 1.50 + 2.50 equals 3.50

4 0

3 years ago

Read 2 more answers

Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and

scZoUnD [109]

Answer:

20 cm

Explanation:

Te electric potential enery U = kq₁q₂/r were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U = -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².

So r = kq₁q₂/U

x - 2 = kq₁q₂/U

x = 0.02 + kq₁q₂/U m

x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

x = 0.02 + 0.18 = 0.2 m = 20 cm

7 0

3 years ago

At 30°c an almunium spher is 30cm³ long

kicyunya [14]

Answer:

Change volume / volume = .5 / 30 = .01666

coefficient of volume expansion = 3 * 24E-6 = 72E-6

Change in temp = .0167 / 24E-6 = 232 deg C

Final temp = 232 + 30 = 262 deg C

Change in volume / volume = 3 * coeff linear expansion * change in deg C

8 0

3 years ago