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Anit [1.1K]
2 years ago
14

Is Mercury a heavier element than tin

Physics
1 answer:
Kryger [21]2 years ago
4 0
Yes because mercury has more protons and electrons that tin. (30 more)
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Teams red and blue are having a tug-of-war. According to Newton's third law, the force with which the red team pulls on the blue
Zielflug [23.3K]
There are a few reasons that one teams could win. It depends if the on of the  teams has a better grip on the rope or not or maybe one off the teams has more friction between the ground and there shoes. So really it no matter what one of the teams will win.
7 0
3 years ago
Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

3 0
3 years ago
A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward th
inn [45]

Answer:

|\vec r|=339.82\ m

\theta=-6.67^o

Explanation:

<u>Displacement </u>

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

\vec r=\vec r_1+\vec r_2

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

x=rcos\theta

y=rsin\theta

And the vector is expressed as

\vec z==

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

\vec r_1==\ km=\ m

The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is

\vec r_2==\ m

The total displacement is

\vec r=\ m+\ m

\vec r=\ m

In (magnitude,angle) form:

|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m

\boxed{|\vec r|=339.82\ m}

\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169

\boxed{\theta=-6.67^o}

5 0
3 years ago
An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte
telo118 [61]

Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2

a. The angular acceleration is Torque divided by moments of inertia:

\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2

b. 5 revolution would be equals to 10\pi rad, or 31.4 rad. Since the engine just got started

\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5

\omega = \sqrt{1893.5} = 43.5 rad/s

c. Work done during the first 5 revolution would be torque times angular displacement:

W = T*\theta = 1930 * 31.4 = 60633 J

d. The time it takes to spin the first 5 revolutions is

t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s

The average power output is work per unit time

P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

P_i = T*\omega = 1930*43.5=83983 W or 84 kW

7 0
2 years ago
Your science teacher gives you three liquids to pour into a jar. After pouring, the liquids layer as seen here. What property of
Blizzard [7]
A. mass
b. color
c. density
d. texture

i believe A
6 0
3 years ago
Read 2 more answers
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