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Ulleksa [173]
3 years ago
9

1.4 The inside of a sports hall measures 80 m

Physics
1 answer:
NISA [10]3 years ago
6 0

a) 48000 m^3

The volume of the sports hall can be calculated using the equation

V=a\cdot b \cdot c

where

a, b, c are the measures of the sizes of the hall

For the sport hall in this problem, we have

a = 80 m

b = 40 m

c = 15 m

Substituting into the equation, we find

V=(80)(40)(15)=48,000 m^3

b) 62400 kg

Being a gas, the air fills the whole sport halls, therefore its volume is equal to the volume of the sports hall.

The relationship between the mass, the volume and the density of the air is

\rho = \frac{m}{V}

where

\rho is the density

m is the mass

V is the volume

Here we have:

\rho = 1.3 kg/m^3

V=48000 m^3

Solving for m, we find the mass of air:

m=\rho V = (1.3)(48000)=62,400 kg

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Answer:

mass conservation is valid for all closed system where the mass will remain in the system always

Explanation:

Conservation of mass is applicable everywhere in classical physics

Here we can also apply mass conservation as we know that the initially the beaker and its water content is having total mass of 109.44 g

m_1 = 109.44 g

now when we heated it to higher temperature then its total mass will be lesser than the initial mass this is because some of the water may evaporate from the system.

Here if we repeat the same experiment with closed boundary then we can see that total mass will be conserved

So here mass conservation is valid for all closed system where the mass will remain in the system always

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3 years ago
Ce unitate de masura are indicele de REFRACTIE?
ivolga24 [154]

Answer:se obesrva ca indicele de refractie NU are unitate de masura, este adimensional. Se exprima print-un raport, de exemplu 4/3, 1,5 etc.

Explanation:

3 0
2 years ago
Refer to Concept Simulation 4.4 for background relating to this problem. The drawing shows a large cube (mass = 28.9 kg) being a
Usimov [2.4K]

Answer:

smallest magnitud is P=33.3 N

Explanation:

We are analyze the situation as an external force is applied and there is a friction force. We have a problem with Newton's second law.

          F = ma

As the two blocks go together they must have the same acceleration, so we can calculate this for the entire system

        P = (m1 + m2) a

        a = P / (m1 + m2)

In this case there is no friction force because the small block does not touch the ground.

In order to calculate the friction force, we must analyze each system component separately.

The large block on the X axis has an applied P force and as it moves feels a force from the small block.  In the Y axis has the weight (W1) and the reaction to normal (N1)

For the small block on the X axis, the force it feels is the thrust of the large block, note that this is an action and reaction force between the two blocks, it is the same definition we have of the normal one, so we can call this force (N)

Y axis it has the weight (W2) down, the force of friction (fr) that opposes the movement, so it is directed upwards. we write these equations

       N = m2 a

       fr -W2 = 0    

       fr = W2

       

The definition of friction force is

       fr = μ N

       

Let's replace and calculate

       μ (m2 a) = m2 g

       μ (P / (m1 + m2)) = g

       P = g /μ  (m1 + m2)

Let's calculate the value of this force

       P = 9.8 / 0.710 (28.9 +4.4)

       P = 13.80 (33.3)

       P = 33.3 N

This is the minimum friction force that prevents the block from sliding down

6 0
3 years ago
when in object moves stop moving changes speed or changes direction how do scientist describe that condition?
Kruka [31]

im taking test rn nd this question was on there, im saying unbalanced force, if it not correct i will put right answer, but im pretty sure the answer is unblanced

6 0
3 years ago
Read 2 more answers
You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 12-m-hi
Vlada [557]

Answer:

Vmax=11.53 m/s

Explanation:

from conservation of energy

      E_A} =E_{B}

     Spring potential energy =potential energy due to elevation

   0.5*k*x²= mg(h_{B}-h_{A} )=mgh

   0.5*k*2.3²= 430*9.81*6

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For safety reason

                                 k"=1.13 *k= 1.13*9568.92

                                    k"=10812.88 N/m

agsin from conservation of energy

      E_A} =E_{C}

    spring potential energy=change in kinetic energy

   0.5*k"*x²=0.5*m*V_{max}^{2}

      10812.88 *2.3²=430*V_{max}^{2}

           V_{max}=11.53 m/s

5 0
3 years ago
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