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4 years ago

What mass of water will fill a tank that is 100.0 cm long, 50.0 cm wide, and 30.0 cm high? Express the answer in grams.

Physics

2 answers:

horsena [70]4 years ago

5 0

Since the density of water is 1g/cm^3, The mass of water needed to fill the tank is 150000 grams

allochka39001 [22]4 years ago

4 0

The volume of the tank is (100cm x 50cm x 30cm) = 150,000 cm³ .

If you're working with water, then water has just about 1 gram of mass in each 1 cm³ . So when this tank is full, it would hold about 150,000 grams of mass in the form of water.

150,000 grams is the same as 150 kilograms.

If you filled the same tank with a different substance, then there would be more or less mass in it. Different substances have different amounts of mass in each cm³ . The number of grams of mass in each cm³ of the substance is called the density of the substance. The density of water is very nearly ' 1 '. Substances with greater density feel heavier when you pick them up, because they have more mass in each unit of volume than other substances have.

For example, an 8-oz drinking glass holds about 237 grams of mass in the form of water when it's full, and that weighs about 8.3 ounces. The density of gold is about 19, so if you could fill up the same drinking glass with gold, it would hold about 4,520 grams of mass in the form of gold, and that would weigh about 10 pounds !

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You are standing on the ground. Describe two ways that you could increase the pressure between your feet and the ground.

ch4aika [34]

Answer:

1:by standing with one feet

2:by standing with your toes

Explanation:

1: by standing with one feet

2:by standing with ur toes

Pressure is inversely proportional to area, the little the area, the more the pressure.the bigger the area,the lesser the pressure

3 0

4 years ago

Your laundry basket Please 22 is in your room is 3.0 M above you on the second floor it takes you 6.0 seconds to carry the laund

Rainbow [258]

Power = (work done) / (time to do the work)

Work done = (force to lift the object) x (distance lifted)

In this question, the force is the (weight of the basket)+(your weight).

Work done = (weight of basket+you) x (3 meters)

Time to do the work = 6 seconds.

Power = (weights x 3 meters) / (6 seconds)

Power = (1/2)·(weight of the basket+you, in Newtons) watts

8 0

4 years ago

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Light from a laser strikes a diffraction grating that has 5500 lines per centimeter. The central and first-order maxima are sepa

WINSTONCH [101]

.Answer:

491.4 nm

Explanation:

The distance between central and first maxima is,

$y=0.455m$

And the distance between screen abnd grating is,

$L=1.62 m$

Now the angle can be find as,

$tan\theta=\frac{y}{L} \\\theta=tan^{-1}(\frac{0.455}{1.62}) \\\theta=15.68^{\circ}$

Now the grating distance is,

$d=\frac{1}{5500} cm\\d=1.82\times 10^{-6}m$

Now with m=1 condition will become,

$\lambda=dsin\theta$

So,

$\lambda=1.82\times 10^{-6}m\times sin(15.68^{\circ})\\\lambda=1.82\times 10^{-6}m\times 0.270\\\lambda=491.4\times 10^{-9}m\\\lambda=491.4 nm$

Therefore the wavelength of laser light is 491.4 nm.

3 0

3 years ago

You are at a furniture store and notice that a Grandfather clock has its time regulated by a physical pendulum that consists of

artcher [175]

Answer:

The distance is 1.026 m.

Explanation:

mass of rod, M = 1.23 kg

Length, L = 1.25 m

mass, m = 10 kg

Time period, T = 2 s

Let the distance is d.

The formula of the time period is given by

$T = 2\pi\sqrt\frac{\frac{1}{3}ML^2+md^2}{(M +m)g}\\\\2\times 2 = 4\pi^2\times \frac{\frac{1}{3}\times1.23\times1.25\times 1.25+ 10d^2}{(1.23 + 10)\times9.8}\\\\11.16 = 0.64 + 10d^2\\\\d= 1.026 m$

3 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

3 0

4 years ago

Read 2 more answers