All categories

3 years ago

Please help

Physics

1 answer:

marishachu [46]3 years ago

3 0

Answer:

The two moments must be the same:

p1=p2

m1v1=m2v2

v2=(m1v1)/m2

v2=(90 kg x 0.9 m/s)/110kg=0.7 m/s

You might be interested in

A certain material has a mass of 565 g while occupying 50 cm3 of space. What is this material?

EastWind [94]

<u>Answer:</u>

Lead

<u>Explanation:</u>

To get the density of the material, the formula would be:

mass divided by volume which is given by $d = \frac { m } { v }$ .

Here in this problem, we are given a mass of $565 g$ which occupies a volume of $50 cm^3$ .

So plugging the data in the above formula to find the density:

Density = $\frac { 565 } { 50 } = 11.3$

From the table, we can see that the material is Lead which has a density of 11.3c/cm^3.

8 0

3 years ago

1. You discover that your iPod batteries keep dying after a few days. You think that the new 48 hour iBattery will work better.

stiks02 [169]

Answer:

Explanation:

It probably would

7 0

3 years ago

which of these is not a physical property of matter? select one: a. flammability b. color c. hardness d. density

netineya [11]

I believe the answer your looking for is A. Flammability. Because flammability is a chemical property, not a physical one. Hope this helps :)

6 0

3 years ago

What is the difference between collinear and coplanar

DIA [1.3K]

Answer:

Collinear is when points are in the same straight line but coplanar is when points are all on the same plane.

4 0

3 years ago

A 6.00 g lead bullet traveling at420 m/s is stopped by a large

Westkost [7]

Answer:

The increase in temperature of the bullet is 351.1 kelvin

Explanation:

First, we should find the kinetic energy of the bullet is:

$K=\frac{mv^{2}}{2}=\frac{(6\times10^{-3}\,kg)(420\,\frac{m}{s})^{2}}{2}$

with m the mass and v the velocity.

$K=529.2 J$

Now we know that half of the kinetic energy of the bullet is transformed into internal energy, by second's law of thermodynamics that means heat (Q) to raise bullet temperature (T), so:

$Q=\frac{K}{2}= 264.6\,J$

To know what the increase in temperature is, we should use specific heat of lead:

$c=125.604 \frac{J}{kg\.K}$

The equation that relates specific heat, change in temperature and mass is:

$Q=cm\varDelta T$

solving for $\varDelta T$ :

$\varDelta T=\frac{Q}{cm}=\frac{264.6}{(125.604)(6\times10^{-3})}$

$\varDelta T= 351.10\, K$

4 0

3 years ago