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RoseWind [281]
3 years ago
14

Iron ball sinks in water but not in mercury. Why?

Physics
2 answers:
Zigmanuir [339]3 years ago
8 0

Answer:

mercury is more dense

Explanation:

Anvisha [2.4K]3 years ago
7 0

Explanation:

Iron ball sinks in water but not in mercury because of the difference in the density iron used in iron ball. The iron ball has more density than water and less than the mercury.

Hence, Iron ball sinks in water but not in mercury.

<u>-TheUnknownScientist</u>

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Cells that remain as sources of other cells are called
inna [77]
They are called stem cells. This cells are undifferentiated which means it can specialize in other types when it receives the right stimuli. They can divide through mitoses and become more stem cell or become a bone, muscle, blood cell, etc.

They can have 2 origins: embryos or some human tissue; their function is to regenerate or substitute damaged cells
8 0
3 years ago
5. A person fishing from a pier observes that four wave crests pass by in 7.0 s and estimates that the distance between two succ
TiliK225 [7]

Answer:

v= 1.71 m/s

Explanation:

Given that

Distance between two successive crests = 4.0 m

 λ = 4 m

T= 7 sec

T is the time between 3 waves.

3 waves = 7 sec

1 wave = 7 /3 sec

So t= 7/3 s

We know that frequency f

f= 1/t= 3/7 Hz

Lets take speed of the wave is v

v= f λ

f=frequency

λ=wavelength

v= 3/7 x 4 = 12 /7

v= 1.71 m/s

3 0
3 years ago
3, A 4kg block is pushed 2m at an acceleration of 0.2m/s square up a vertical wall by a constant force F applied at an angle of
Andrews [41]

The work done by the applied force on the block against the frictional force is 15.75 J.

<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

  • F is applied force
  • Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

F = 9.86 N

W = Fdcosθ

W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

#SPJ1

5 0
2 years ago
Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

6 0
3 years ago
A 1300 kg car traveling at 35 mph rear-ends a 1000 kg car traveling 25 mph. Just after the collision (but before the driver’s sl
guajiro [1.7K]

Answer

given,

before collision

mass of car A = m_a = 1300 kg

velocity of car A = v_a  = 35 mph

mass of car B = m_b= 1000 kg

velocity of car B = v_b  = 25 mph

after collision

V_a = 30 mph

V_b = 31.5 mph

Initial momentum

P_1 = m_av_a + m_b v_b

P_1 = 1300 \times 35+ 1000 \times 25

P_1 =70500 Kg.m/s

final momentum

P_2 = m_aV_a + m_b V_b

P_2 = 1300 \times 30+ 1000 \times 31.5

P_2 =70500 Kg.m/s

here  initial momentum is equal to the final momentum of the car.

hence, momentum is conserved in the collision.

6 0
3 years ago
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