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3 years ago

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A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationary seal of m

ass 200 kg. The seal grabs onto the whale and holds fast. What is the momentum of these two sea creatures just after their collision

1 answer:

anzhelika [568]3 years ago

5 0

Answer:

Momentum after collision will be 6000 kgm/sec

Explanation:

We have given mass of the whale = 1000

Initial velocity v = 6 m/sec

It collides with other mass of 200 kg which is at stationary

Initial momentum of the whale = 1000×6 = 6000 kgm/sec

We have to find the momentum after collision

From conservation of momentum

Initial momentum = final momentum

So final momentum = 6000 kgm/sec

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A block with mass 0.5 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2 m. W

lianna [129]

Answer:

So coefficient of kinetic friction will be equal to 0.4081

Explanation:

We have given mass of the block m = 0.5 kg

The spring is compressed by length x = 0.2 m

Spring constant of the sprig k = 100 N/m

Blocks moves a horizontal distance of s = 1 m

Work done in stretching the spring is equal to $W=\frac{1}{2}kx^2=\frac{1}{2}\times 100\times 0.2^2=2J$

This energy will be equal to kinetic energy of the block

And this kinetic energy must be equal to work done by the frictional force

So $\mu mg\times s=2$

$\mu\times 0.5\times 9.8\times 1=2$

$\mu =0.4081$

So coefficient of kinetic friction will be equal to 0.4081

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3 years ago

A standard quantity of a physical property used as a factor to compare other occurring quantities of that property is called a(n

Brrunno [24]

The correct answer that would complete the given statement above would be option B. UNIT (of measurement). <span>A standard quantity of a physical property used as a factor to compare other occurring quantities of that property is called a unit of measurement. Hope this answers your question. </span>

3 0

3 years ago

The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI

OverLord2011 [107]

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

<u>Explanation: </u>

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

$\text { Work done }=\text { Force } \times \text { displacement }$

$\text { Work done }=0.010 \mathrm{N} \times 0.032 \mathrm{m}=0.00032 \mathrm{Nm}$

It is known that $1 N=1 \mathrm{kg} \mathrm{ms}^{-2}$

So, the work done can be expressed in $k g m s^{-2}$ as,

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2}$

It is known that $1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^{2} / \mathrm{s}^{2}$ , so the conversion of units from Nm to erg will be done as follows:

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2} \times \frac{1000 \mathrm{g}}{1 \mathrm{kg}} \times \frac{100 * 100 \mathrm{cm}^{2}}{m^{2}}=3200 \mathrm{g} \mathrm{cm}^{2} \mathrm{s}^{-2}$

Thus, work done in ergs is 3200 ergs.

6 0

3 years ago

dalvyx [7]

Answer:

$4.5\ \text{N}$

Explanation:

$F_1$ = Gravitational force between the objects = $18\ \text{N}$

$r_1$ = Initial distance between the two objects

$r_2$ = Final distance between the two objects = $2r_1$

Gravitational force between two objects is given by

$F=\dfrac{Gm_1m_2}{r^2}$

So

$F\propto \dfrac{1}{r^2}$

$\dfrac{F_2}{F_1}=\dfrac{r_1^2}{r_2^2}\\\Rightarrow \dfrac{F_2}{F_1}=\dfrac{r_1^2}{(2r_1)^2}\\\Rightarrow \dfrac{F_2}{F_1}=\dfrac{1}{4}\\\Rightarrow F_2=\dfrac{F_1}{4}\\\Rightarrow F_2=\dfrac{18}{4}\\\Rightarrow F_2=4.5\ \text{N}$

The new force of attraction between the objects is $4.5\ \text{N}$ .

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2 years ago

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Phantasy [73]

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