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3 years ago

Which of the following terms refers to the gas that escapes a comet?

Physics

1 answer:

Gnoma [55]3 years ago

7 0

Answer:

The correct option is "a plasma tail"

Explanation:

Gases <u>escaping from a comet</u> are ionized by ultraviolent rays from the sun. These ionized gases are then carried away from the sun by the solar wind to form the plasma tail. While the plasma tail is made up of gases, the dust tail is made up of tiny solid particles (dust) that escapes from the comet and are also pushed away from the sun.

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A basketball player spins the ball with an angular acceleration of 10 rad/s2. What is the ball’s final angular velocity if the b

Lena [83]

Explanation:

It is given that,

The angular acceleration of the basketball, $\alpha=10\ rad/s^2$

Time taken, t = 3 seconds

We need to find the ball’s final angular velocity if the ball starts from rest. It can be calculated using definition of angular acceleration i.e.

$\alpha=\dfrac{\omega_f-\omega_i}{t}$

$\omega_i=0\ (rest)$

$\omega_f=\alpha t$

$\omega_f=10\ rad/s^2\times 3\ s$

$\omega=30\ rad/s$

So, the ball's final angular velocity is 30 rad/s. Hence, this is the required solution.

8 0

3 years ago

The height of the mercury column in a barometer is 756 mm Hg on

notsponge [240]

Answer:

p = 1.0076 10⁵ Pa

Explanation:

Atmospheric pressure is given by the relation

P = rho g h

In this case they indicate that the height of the column of mercury is h = 756 mm Hg

let's reduce the height to the SI system

h = 756 mm (1m / 1000 mm)

h = 0.756 m

let's calculate

P = 13600 9.8 0.756

p = 1.0076 10⁵ Pa

7 0

2 years ago

Two charges are sitting 1.5 m apart with a force of 3 N between them. They are now moved farther apart to 2.25 m and one of the

ANTONII [103]

Answer: 2.37N

Explanation:

According to coulombs law which states that the force of attraction (F) between two charges (q1 and q2) is directly proportional to the product of their charges and inversely proportional to the square of the distance (r) between them. Mathematically,

F = kq1q2/r²

For the first two charges that are sitting 1.5 m apart with a force of 3 N between them, we have

3 = kq1q2/1.5²

3 = kq1q2/2.25

Kq1q2= 6.75... (1)

If the charges are now moved farther apart 2.25 m and one of the charges is increased by a factor of 4. The formula becomes

F2 = k(4q1)q2/2.25² (q1 has been increased by factor of 4)

k(4q1)q2 = 5.06F2 ... (2)

Dividing 2 by 1 we have

k(4q1)q2/kq1q2 = 5.06F2/3

4 = 5.06F2/3

5.06F2 = 12

F2= 12/5.06

F2 = 2.37N

Therefore the magnitude of the new force between the two charges is 2.37N

5 0

3 years ago

A 30 ton rail car and a 90 ton rail car, initially at rest,are connected together with a giant but massless compressed springbet

Liula [17]

Answer:

V2 = 1.33m/s

Explanation:

M1 = 30 ton

M2 = 90 ton

V1 = 4 m/s

V2 = ?

Assumption: momentum conserved, no friction

initial momentum = final momentum = 0

momentum : p = MV

(M1*V1) - (M2*V2) = 0

V2 = (30*4)/90 = 1.33m/s

6 0

3 years ago

A ball is thrown upward with an initial vertical velocity of v0 to a maximum height of hmax and falls back toward earth. on the

Vlad1618 [11]

Answer: $v_f=g\frac{t}{2}$

The ball was thrown at the speed of $v_o$ .

Maximum height achieved is $h_{max}$

Time of flight is t.

Now, the time the ball takes to achieve maximum height = the time taken by ball to fall back = $\frac{t}{2}$

let us just consider the second half of the flight. At $h_{max}$ , the velocity would be zero. let us consider as the initial velocity for the second half of the flight i.e. $v_i=0$

Using the equation of motion:

$v_f=v_i+at$

where, $v_f$ is the final velocity, a is the acceleration, t is the time taken.

Because the ball would fall under gravity, hence a=g and time of flight would be t/2

$\Rightarrow v_f=0+g\frac {t}{2}$

5 0

3 years ago