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3 years ago

There are 10 deciliters in a liter how many liters are in 82 deciliters?

Physics

1 answer:

Masteriza [31]3 years ago

4 0

A deciliter is 1/10 of a liter, so 82 deciliters is 8.2 liters.

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Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

2 years ago

In an arcade game, a 0.126 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and relea

bearhunter [10]

Answer:

4.156 m/s

Explanation:

See attachment

4 0

2 years ago

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f

worty [1.4K]

Answer:

$v_{min} \approx 17.153\,\frac{m}{s}$

Explanation:

The roller coaster begins with maximum kinetic energy and no gravitational potential energy. The gravitational potential energy reaches its maximum when roller coaster is upside down at the top of the circle. The physical model for the roller coaster is constructed by means of the Principle of Energy Conservation:

$\frac{1}{2}\cdot m \cdot v_{min}^{2} = m\cdot g \cdot h$

The minimum velocity is:

$v_{min} = \sqrt{2\cdot g \cdot h}$

Let assume that radio of curvature is measured in meters. Hence:

$v_{min} = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot(15\,m)}$

$v_{min} \approx 17.153\,\frac{m}{s}$

8 0

2 years ago

What side of the worm faces up when placed in the tray when dissection?

Lelechka [254]

The side of the worm that faces up when placed in the tray for dissection is the smoother side.

You can observe the organs of these tiny creatures by dissecting a preserved earthworm. In dissecting a worm, lay the worm on your dissecting tray with its dorsal side facing up.

8 0

3 years ago

An inelastic collision of two objects is characterized by the following.

serious [3.7K]

Options:

(a) Total kinetic energy of the system remains constant.

(b) Total momentum of the system is conserved.

(d) Neither A nor B are true.

Answer:

(b) Total momentum of the system is conserved.

Explanation:

An inelastic collision is a type of collision in which momentum is conserved and kinetic energy is not conserved. That is, there is loss of kinetic energy.

In an inelastic collision:

Total momentum before collision = Total momentum after collision

An example of inelastic collision is seen in the ballistic pendulum, The ballistic pendulum is a device in which a projectile such as a bullet is fired into a suspended heavy wooden stationary block.

8 0

3 years ago