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kow [346]
3 years ago
9

Mass of a body is 210g and its density is 7.981g/cm^3 what will be its volume with regard to significant figures

Physics
1 answer:
aleksandrvk [35]3 years ago
3 0

Density is mass per unit volume,

\rho=\dfrac mV

so to get the volume, divide the mass by the density,

V=\dfrac m\rho

So the volume is

\dfrac{210\,\mathrm g}{7.981\frac{\rm g}{\mathrm{cm}^3}}\approx26.312\,\mathrm{cm}^3\approx\boxed{26\,\mathrm{cm}^3}

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Which force causes a wagon to speed up as it rolls down a hill?
Temka [501]
Gravity is the only one helping it.
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Alenkasestr [34]

Answer:

1/5 or .2

Explanation:

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3 years ago
300 ml of a gas at 27°C is Cooled at -3°c at Constant pressure the final volume is plzz answer fast i will mark brainliest ​
Scorpion4ik [409]

Answer:V₁=300ml

T₁=27°C

V₂=?

T₂= -3°C

as we know

V₁T₁=V₂T₂

By putting values in formula

300ml×27°C=V₂×(-3°C)

300ml×27°C/-3°C=V₂

8100ml/-3=V₂

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4 0
3 years ago
Physics Question: Determine the work that is being done by tension in pulling the box 193.0 cm along the table. Also determine t
lozanna [386]

1) The work done by tension is 12.0 J

2) The final speed of the box is 1.57 m/s

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

F = 6.2 N is the force applied (the tension in the string)

d = 193.0 cm = 1.93 m is the displacement

\theta=0 assuming that the string is horizontal

Substituting, we find the work done by the tension:

W=(6.2)(1.93)(cos 0)=12.0 J

2)

In order to determine the final speed of the box, we need to determine its acceleration first.

Beside the tension, acting forward, the other force acting horizontally on the box is the force of friction, whose magnitude is

F_f = \mu mg

where

\mu=0.16 is the coefficient of friction

m = 2.8 kg is the mass of the box

g=9.8 m/s^2 is the acceleration of gravity

Substituting,

F_f = (0.16)(2.8)(9.8)=4.4 N

Therefore, the net force on the box is

F=6.2  N - 4.4 N = 1.8 N

And the acceleration can be found by using Newton's second law:

F=ma

where

F = 1.8 N is the net force

m = 2.8 kg is the mass of the box

a is the acceleration

Solving for a,

a=\frac{F}{m}=\frac{1.8}{2.8}=0.64 m/s^2

Now we can finally find the final speed using the suvat equation:

v^2-u^2=2as

where

v is the final speed

u = 0 is the initial speed

a=0.64 m/s^2 is the acceleration

s = 1.93 m is the displacement

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(0.64)(1.93)}=1.57 m/s

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0
3 years ago
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