Ask question

Ask question

All categories

1 year ago

14

In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric

al, in this problem we'll model it as a sphere with a mass of 1.0 × 1013 kg and a radius of 1.6 km. If a rock were dropped from a height of 1.0 m above the comet's surface, how long would it take to hit the surface?

1 answer:

Viktor [21]1 year ago

8 0

To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are

Comet mass $M = 1.0*10^{13} kg$

Radius $r = 1.6km = 1600 m$

Rock was dropped from a height 'h' from surface = 1m

The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is

$g = \frac{GM}{R^2}$

Where G means gravitational universal constant and M the mass of the planet

$g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}$

$g = 2.607*10^{-4} m/s^2$

Now calculate the value of the time

$h = \frac{1}{2} gt^2$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}$

$t = 87.58s$

The time taken for the rock to reach the surface is t = 87.58s

You might be interested in

A metal disk weighing 1 N is resting on an index card that is balanced on top of a glass. When the index card is quickly pulled

Burka [1]

Answer: D

Step by step explanation:

8 0

2 years ago

Read 2 more answers

Write a hypothesis about how the height of the cylinder affects the temperature of the water. Use the "if . . . then . . . becau

IgorLugansk [536]

If the mass of the cylinder increases, the temperature of the water increases, because a greater mass means the cylinder has more potential energy that can be converted to thermal energy, increasing the temperature of the water.

4 0

2 years ago

Consider four different oscillating systems, indexed using i = 1 , 2 , 3 , 4 . Each system consists of a block of mass mi moving

Rzqust [24]

Answer:

The order is 2>4>3>1 (TE)

Explanation:

Look up attached file

4 0

2 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

6 0

2 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.240 rev/s. The magnitude

bazaltina [42]

Explanation:

Given that,

Angular velocity = 0.240 rev/s

Angular acceleration = 0.917 rev/s²

Diameter = 0.720 m

(a). We need to calculate the angular velocity after time 0.203 s

Using equation of angular motion

$\omega_{f}=\omega_{i}+\alpha t$

Put the value in the equation

$\omega_{f}=0.240+0.917\times0.203$

$\omega_{f}=0.426\ rev/s$

The angular velocity is 0.426 rev/s.

(b). We need to calculate the tangential speed of the blade

Using formula of tangential speed

$v= r\omega$

Put the value into the formula

$v = \dfrac{0.720 }{2}\times0.426\times2\pi$

$v=0.963\ m/s$

The tangential speed of the blade is 0.963 m/s.

(c). We need to calculate the magnitude at of the tangential acceleration

Using formula of tangential acceleration

$a_{t}=r\alpha$

Put the value into the formula

$a_{t}=0.36\times0.917\times2\pi$

$a_{c}=2.074\ m/s^2$

The tangential acceleration is 2.074 m/s².

Hence, This is required solution.

4 0

2 years ago