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alekssr [168]
3 years ago
10

1. An asphalt block has a mass of 90 kg and a volume of 0.075 m. Determines the density of the asphalt.

Physics
1 answer:
Kisachek [45]3 years ago
7 0

Answer:

1. Density = 1200[kg/m^3]; 2. Volume= 0.005775[m^3], mass= 15.59[kg]

Explanation:

1. We know that the density is defined by the following expression.

Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]

2. First we need to convert the units to meters.

wide = 35[cm] = 35/100 = 0.35[m]

long = 11 [dm] =  11 decimeters = 11/10 = 1.1[m]

Thick = 15[mm] = 15/1000 = 0.015[m]

Now we can find the density using the expression for the density.

density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]

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adelina 88 [10]

Answer:

0.003333 s to 0.000125s or from 3.33ms to 0.125ms wher m is for milli

1.1m to 0.04125 m

Explanation:

T= 1/f=

if f= 300Hz then T = 1/300 =0.003333 s

if f= 8000 then T= 1/8000 = 0.000125s

now v=f×wave length

or wavelength = speed/ frequency

when f = 300 Hz

wavelength = 330/300=1.1 m

wavelength = 330/8000 = 0.04125m

note : i have taken speed of sound as 330 m/s you can take any value given in between 330m/s to 340m/s

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3 years ago
If vector A =i+2j-k and vec A cross vec B =3i-j+5k. find vec B​​​
postnew [5]

Let <em>B</em> = <em>a</em> <em>i</em> + <em>b</em> <em>j</em> + <em>c</em> <em>k</em>. Then the cross product of <em>A</em> = <em>i</em> + 2<em>j</em> - <em>k</em> with <em>B</em> is

<em>A</em> × <em>B</em> = ( <em>i</em> + 2<em>j</em> - <em>k </em>) × ( <em>a</em> <em>i</em> + <em>b</em> <em>j</em> + <em>c</em> <em>k</em> )

<em>A</em> × <em>B</em>  = <em>a</em> ( <em>i</em> × <em>i</em> ) + 2<em>a</em> ( <em>j</em> × <em>i</em> ) - <em>a</em> ( <em>k</em> × <em>i </em>)

… … … + <em>b</em> ( <em>i</em> × <em>j</em> ) + 2<em>b</em> ( <em>j </em>× <em>j</em> ) - <em>b</em> ( <em>k</em> × <em>j</em> )

… … … + <em>c</em> ( <em>i</em> × <em>k</em> ) + 2<em>c</em> ( <em>j</em> × <em>k</em> ) - <em>c</em> ( <em>k</em> × <em>k</em> )

<em>A</em> × <em>B</em> = 0 - 2<em>a</em> <em>k </em>- <em>a</em> <em>j</em>

… … … + <em>b</em> <em>k</em> + 0 + <em>b</em> <em>i</em>

… … … - <em>c</em> <em>j</em> + 2<em>c</em> <em>i</em> - 0

<em>A</em> × <em>B</em> = (<em>b</em> + 2<em>c</em>) <em>i</em> + (-<em>a</em> - <em>c</em>) <em>j</em> + (<em>b</em> - 2<em>a</em>) <em>k</em>

So we have

3 <em>i</em> - <em>j</em> + 5 <em>k </em>= (<em>b</em> + 2<em>c</em>) <em>i</em> + (<em>c</em> - <em>a</em>) <em>j</em> + (<em>b</em> - 2<em>a</em>) <em>k</em>

which gives us the system of equations,

{ <em>b</em> + 2<em>c</em> = 3

{ -<em>a</em> - <em>c</em> = -1

{ -2<em>a</em> + <em>b</em> = 5

Solve for <em>a</em>, <em>b</em>, and <em>c</em>.

• Eliminate <em>c</em> from the first two equations:

(<em>b</em> + 2<em>c</em>) + 2 (-<em>a</em> - <em>c</em>) = 3 + 2 (-1)

-2<em>a</em> + <em>b</em> = 1

But -2<em>a</em> + <em>b</em> = 5, and 5 ≠ 1, so there is no such vector <em>B</em> that satisfies the cross product!

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3 years ago
There are two blocks: one large one initially at rest, and a smaller one, initially moving to the right withsome speed. The smal
KATRIN_1 [288]

Answer:

Explanation:

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by applying conservation of momentum we can find velocity of common mass

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V = v / 3

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1/2 m v² = mgh

1/2  x 75 x V² = 75 x g x 10

V² = 2g x 10

v² / 9 = 2 x 9.8 x 10

v² = 9 x 2 x 9.8 x 10

v = 42 m /s

small block must have velocity of 42 m /s .

Impulse by small block on large block

= change in momentum of large block

= 75 x V

= 75  x 42 / 3

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6 0
3 years ago
In a football game, a receiver is standing still, having just caught a pass. Before he can move, a tackler, running at a velocit
ivolga24 [154]

Answer:

m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg    

Explanation:

The football players collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V=1.6m/s.

We need to use the conservation of momentum Law, the total momentum is the same before and after the collision, at the initial point the receiver does not have any speed

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m_{receiver}={m_{tackler}*v_{tackler}/V}-m_{tackler}    

m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg    

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3 years ago
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