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Aleks04 [339]
3 years ago
11

How can we calculate the density of a sustance that has irregular volume

Physics
1 answer:
Helen [10]3 years ago
3 0
Use the water displacement method. Fill the a tube with a certain amount of water. put the object in a tube. now measure it. subtract the two measurements to find the volume. then put the object on a balance beam. find the mass. Divide mass by the volume.
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The physical and chemical properties of a substance are used to identify the substance and to see how it will behave in the pres
pentagon [3]
Im going to guess on this one, and say that it is chemical.
8 0
3 years ago
Read 2 more answers
9. A statue is to be scaled down isomorphically (It will have its size changed without changing its shape). It starts with an in
Fynjy0 [20]

Answer:

   m = 1.45 kg

Explanation:

For this exercise we look for size reduction in height

              Reduction = y / y₀

              Reduction = 2.15 / 6.75

              Reduction = 0.3185

As the statue should not be deformed, all reduction has the same factor.

Let's use the concept of density

       ρ = m / V

Initial statue

         ρ = m₀ / V₀

         

It is reduced

         V = x y z

         V = 0.3185 x₀ 0.3185 y₀ 0.3185 z₀

         V = 0.3185³ V₀

       

Density is

         ρ = m / V

         ρ = m / 0.3185³ V₀

As the density remains constant we can match them

         m₀ / Vo = m / 0.3185³ V₀

         m = 0.3185³ m₀

Let's calculate

        m = 0.3185³ 45

        m = 0.03231   45

        m = 1.45 kg

5 0
3 years ago
Two students are watching a person riding a skateboard up and down a ramp. Each student shares what they think about the energy
avanturin [10]

Answer:

Explanation:

We know that , If the frictional force on a system is zero , then the total energy of a system will be conserved.

By using energy conservation

KE₁ +  U₁ = KE₂ + U₂

KE₁=Kinetic energy at location 1

U₁ =Potential energy at location 1

KE₂=Kinetic energy at location 2

U₂=Potential energy at location 2

Therefore, Raymond is thinking in a right way.

7 0
3 years ago
A concave mirror produces a real image that is three times as large as the object. If the object is 20 cm in front of the mirror
TiliK225 [7]

Answer:

The image is produced 60 cm behind the mirror

The focal length of the mirror is 30 cm

Explanation:

u = Object distance =  20 cm

v = Image distance

f = Focal length

m = Magnification = 3

m=-\frac{v}{u}\\\Rightarrow 3=-\frac{v}{20}\\\Rightarrow v=-3\times 20\\\Rightarrow v=-60\ cm

The image is produced 60 cm behind the mirror

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{20}+\frac{1}{-60}\\\Rightarrow \frac{1}{f}=\frac{1}{30}\\\Rightarrow f=\frac{30}{1}=30\ cm

The focal length of the mirror is 30 cm

8 0
3 years ago
One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the f
Andreas93 [3]

Answer:

Part a)

\theta_2 = 15 degree

Part b)

\Delta t = 2.88 s

Explanation:

Part a)

In order to have same range for same initial speed we can say

R_1 = R_2

\frac{v^2 sin2\theta_1}{g} = \frac{v^2 sin2\theta_2}{g}

so after comparing above we will have

\theta_1 = 90 - \theta

so we have

75 = 90 - \theta_2

\theta_2 = 15 degree

Part b)

Time of flight for the first ball is given as

T_1 = \frac{2vsin\theta}{g}

T_1 = \frac{2(20)sin75}{9.81}

T_1 = 3.94 s

Now for other angle of projection time is given as

T_2 = \frac{2(20)sin15}{9.81}

T_2 = 1.05 s

So here the time lag between two is given as

\Delta t = T_1 - T_2

\Delta t = 3.94 - 1.05

\Delta t = 2.88 s

5 0
3 years ago
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