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nadezda [96]
3 years ago
9

You are asked to construct a mobile with four equal m = 141 kg masses, and three light rods of negligible mass and equal lengths

. The rods are of length 55 cm. (a) At what location on the level 1 rod should the free end of rod 2 be attached?
Physics
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

The free end must be attached at a distance of 27.5 cm

Solution:

Mass, m = 141 kg

Length of the rods, L= 55 cm

Now,

As clear from fig. 1:

The free end of the rod 2 must be attached at:

F = 2 W

WL = W(55 - L)

2L = 55

L = 27.5 cm

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A pebble is dropped from rest from the top of a tall cliff and falls 53.4 m after 3.3 s has elapsed. How much farther does it dr
jeka94

Answer:

426.84 m

Explanation:

initial velocity u = 0

time t = 3.3 s

distance travelled s = 53.4 m

acceleration due to gravity = g

s = ut + 1/2 g t²

53.4 = 0 + 1/2 g x 3.3²

g = 9.8 m /s²

For the whole length of fall

distance travelled = h

total time = 6.6 + 3.3 = 9.9 s

h = ut + 1/2 g t²

u again = 0

h = .5 x 9.8 x 9.9²

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distance travelled in last 6.6 s

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6 0
3 years ago
Match these items.
Nitella [24]

Explanation :

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(4) Voluntary muscle is the muscles that are not controlled by our will.  

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4 0
3 years ago
A spacecraft is fueled using hydrazine (N2H4; molecular weight of 32 grams per mole [g/mol]) and carries 1640 kilograms [kg] of
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Answer:

The value is t = 689.029 \  hours

Explanation:

From the question we are told that

The molar mass of hydrazine is Z =  32 g/mol = \frac{32}{1000} = 0.032 \  kg/mol

The initial temperature is T_i  =  -186 ^o F = (-186-32) *\frac{5}{9} +273.15 = 152\ K

The final temperature is T_f  =  78 ^o F = (78-32) *\frac{5}{9} +273.15 = 298.7 \ K

The specific heat capacity is c_h  =  0.099 [kJ/(mol K)] = 0.099 *10^3 J/(mol/K)

The power available is P = 300 \ W

The mass of the fuel is m =   1640 \  kg

Generally the number of moles of hydrazine present is

n  =  \frac{m}{Z}

=> n  =  \frac{1640}{= 0.032}

=> n  =  51250 \ mol

Generally the quantity of heat energy needed is mathematically represented as

Q =  n * c_h * (T_f -T_i)

=> Q =  51250  * 0.099 *10^3  * (298.7 - 152)

=> Q =  7.441516913 * 10^{8} \  J

Generally the time taken is mathematically represented as

t =  \frac{Q}{P}

=> t =  \frac{7.441516913 * 10^{8} }{300}

=> t = 2480505.6377 s

Converting to hours

t = \frac{2480505.6377}{3600}

=> t = 689.029 \  hours

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