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2 years ago

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A block is suspended from a scale and then lowered into a bucket of water. The density of the water is 1 gm/cm3. The initial rea

ding on the scale is 19N and one complete revolution of the scale is a change of 10N. 1) When the block is lowered into the water, the mass of the block:

1 answer:

miv72 [106K]2 years ago

6 0

Answer:

1.94 kg

Explanation:

we have given density of water = 1 gm/ $cm^3$

the initial reading of the scale = 19 N

change in reading of scale =10 N

we have to find the mass of the block when the block is lowered in water

mass of the block when it is lowered in the water is given by

$w_b=\frac{f}{g}$

here g is the gravity of acceleration which value is given by 9.8 $m/sec^2$

so the mass of the block = $\frac{19}{9.8}=1.94\ kg$

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Lilli suggests that they explore the simulation starting with varying only a single parameter in order to understand the role of

mrs_skeptik [129]

Answer:

B.

Explanation:

One of the ways to address this issue is through the options given by the statement. The concepts related to the continuity equation and the Bernoulli equation.

Through these two equations it is possible to observe the behavior of the fluid, specifically the velocity at a constant height.

By definition the equation of continuity is,

$A_1V_1=A_2V_2$

In the problem $A_2$ is $2A_1$ , then

$A_1V_1=2A_1V_2$

$V_2 = \frac{V_1}{2}$

<em>Here we can conclude that by means of the continuity when increasing the Area, a decrease will be obtained - in the diminished times in the area - in the speed.</em>

For the particular case of Bernoulli we have to

$P_1 + \frac{1}{2}\rho V_1^2 = P_2 +\frac{1}{2}\rho V_2^2$

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-V_2^2)$

For the previous definition we can now replace,

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-(\frac{V_1}{2})^2)$

$\Delta P = \frac{3}{8} \rho V_1^2$

<em>Expressed from Bernoulli's equation we can identify that the greater the change that exists in pressure, fluid velocity will tend to decrease</em>

The correct answer is B: "If we increase A2 then by the continuity equation the speed of the fluid should decrease. Bernoulli's equation then shows that if the velocity of the fluid decreases (at constant height conditions) then the pressure of the fluid should increase"

4 0

2 years ago

At what condition does a body become weightless at equator

salantis [7]

Hope this is helpful <span>Weightlessness

</span>

3 0

2 years ago

29. 2072 Set C Q.No. 10c

Annette [7]

Answer:

$90.2^{\circ}C$

Explanation:

Considering the thermal conductivity of aluminium and brass as $k_{al}=205 W/mK$ and $k_{br}=109 W/mk$ respectively

The temperature at the end of aluminium and brass are given as $T_{al}=150^{\circ}C$ and $T_{br}=20^{\circ}C$ respectively with length of rod L=1.3 m , Length of aluminium $L_{al}=0.8 m$ , length of brass $L_{br}=0.5 m$ and letting temperature at steady state be T

At steady state, thermal conductivity of both aluminium and brass are same hence

$H_{br}=H_{al}$

$k_{al}A\frac {T_H-T}{L_{al}}= k_{br}A\frac {T-T_H}{L_{br}}$

Upon re-arranging

$T=\frac {k_{al}L_{al}T_{br}+k_{al}L_{br}T_{al}}{k_{br}L_{al}+k_{al}L_{br}}$

$(205)\frac {150-T}{0.8}=109\frac {T-20}{0.5}$

$T=\frac {(109*0.8*20)+(205*0.5*150)}{(109*0.8)+(205*0.5)}$

$T=90.2^{\circ}C$

Therefore, the temperatures at which the metals are joined is $90.2^{\circ}C$

6 0

2 years ago

How many calories are equal to one BTU? (One calorie = 4.186 J, one BTU = 1 054 J.)

I am Lyosha [343]

<h2>Option C is the correct answer.</h2>

Explanation:

We need to find how many calories is 1 BTU.

Given

1 BTU = 1054 J

1 calorie = 4.186 J

So we have

1 BTU = 4.186 x 251.79 J

1 BTU =251.79 calorie

1 BTU = 252 calorie.

Option C is the correct answer.

3 0

2 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

a)

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

2 years ago