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Dmitry_Shevchenko [17]
4 years ago
9

A piece of paper has a density of 13.29g/cm 3 and a volume of 25cm, what is the mass of the liquid?

Physics
1 answer:
Julli [10]4 years ago
5 0
25cm is not a volume 
if you are using 25cm^{3} then 
you can say 
\frac{13.29g}{cm^3} *25cm^3=332.25g
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Explain the process that causes dew to form on blades of grass. ​
AysviL [449]

Answer:

Condensation causes dew to form on blades of grass. This is because the water molecules react to the grass when it is heated, and the surroundings became cool. The molecules would bring to calm and turn to liquid.

Explanation:

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8 0
3 years ago
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a 2kg object is moving horizontally with a speed of 4 m/s.how much net force is required to keep the object moving at this speed
Elza [17]
If the object is moving in a straight line with constant speed,
that's a description of " acceleration = zero ".

Zero acceleration means zero net force on the object.

NO net force is 'required' to keep an object moving in a straight line
at constant speed.  In fact, if there IS any net force on the object,
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None of this depends on the object's mass, or on the speed or direction
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8 0
4 years ago
A van accelerates from 4m/s to 20m/s in 8s. How far does it travel in this time?​
zubka84 [21]

Answer:

96 m

Explanation:

Given,

Initial velocity ( u ) = 4 m/s

Final velocity ( v ) = 20 m/s

Time ( t ) = 8 s

Let Acceleration be " a ".

Formula : -

a = ( v - u ) / t

a = ( 20 - 4 ) / 8

= 16 / 8

a = 2 m/s²

Let displacement be " s ".

Formula : -

s = ut + at² / 2

s = ( 4 ) ( 8 ) + ( 2 ) ( 8² ) / 2

= 32 + ( 2 ) ( 64 ) / 2

= 32 + ( 2 ) ( 32 )

= 32 + 64

s = 96 m

Therefore, it travels 96 m in time 8 s.

3 0
3 years ago
The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an
Mrrafil [7]

Answer:

v_1  = 3.5 \ m/s

Explanation:

Given that :

mass of the SUV is  = 2140 kg

moment of inertia about G , i.e I_G = 875 kg.m²

We know from the conservation of angular momentum that:

H_1= H_2

mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2

2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2

1637.1 v_1 = 3841.575 \omega_2

\omega_2 = \frac{1637.1 v_1}{3841.575}

\omega _2 = 0.4626 \ v_1

From the conservation of energy as well;we have :

T_2 +V_{2  \to 3} = T_3 \\ \\ \\  \frac{1}{2} I_A \omega_2^2 - mgh =0

[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0

706.93 \ v_1^2 - 8657.49 =0

706.93 \ v_1^2  = 8657.49

v_1^2  =  \frac{8657.49}{706.93 }

v_1 ^2 =  12.25

v_1  = \sqrt{ 12.25

v_1  = 3.5 \ m/s

6 0
3 years ago
When you ask you friend for hoemwork answers and he answers like a year later hahahah
Hitman42 [59]
G that’s a nice friend
5 0
3 years ago
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