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olchik [2.2K]
3 years ago
13

It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hook

e's law and is depressed 0.65 cm by its maximum load of 110 kg.
Physics
1 answer:
Molodets [167]3 years ago
7 0
<span>F = kx, where F is the force we apply, k is the spring constant, and x is the extension of the material 
so the </span><span>spring's effective spring constant k is:
k = F / x
k = ( 110 kg ) ( 9.81 m/s^2) / 0.65 cm
k = 1660 N/cm  is the </span>spring's effective spring constant
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When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these
Zigmanuir [339]

Answer:

a)  v = 0.7071 v₀, b) v= v₀, c)  v = 0.577 v₀, d)   v = 1.41 v₀, e)  v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

          v = \sqrt{ \frac{T}{ \mu } }

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

          m = 2m₀

let's find the new linear density

          μ = m / l

iinitial density

          μ₀ = m₀ / l

final density

          μ = 2m₀ / lo

          μ = 2 μ₀

we substitute in the equation for the velocity

initial            v₀ = \sqrt{ \frac{T_o}{ \mu_o} }

with the new dough

                    v = \sqrt{ \frac{T_o}{ 2 \mu_o} }

                    v = 1 /√2  \sqrt{ \frac{T_o}{ \mu_o} }

                    v = 1 /√2 v₀

                    v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

              μ = μ₀

   in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

           μ = 3m₀ / l₀

           μ = 3 m₀/l₀

           μ = 3 μ₀

            v = \sqrt{ \frac{T_o}{ 3 \mu_o} }

            v = 1 /√3  v₀

            v = 0.577 v₀

d) l = 2l₀

            μ = m₀ / 2l₀

            μ = μ₀/ 2

           v = \sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }

           v = √2 v₀

            v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

             μ = 10 m₀/2l₀

             μ = 5 μ₀

we look for speed

             v = \sqrt{ \frac{T_o}{5 \mu_o} }

             v = 1 /√5  v₀

             v = 0.447 v₀

5 0
3 years ago
A simple motor converts ________energy into________energy.
k0ka [10]
I think it is C. I hope I helped.
8 0
3 years ago
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The base of a pyramid covers an area of 13.0 acres (1 acre = 43,560 ft2) and has a height of 481 ft. If the volume of a pyramid
Allisa [31]
V = 1/3 Bh v = 1/3 (13 ac)(43560ft^2/ac)(481ft) v = 90793560 ft^3 * 0.3048m/ft * 0.3048m/ft * 0.3048m/ft = 2570987m^3
3 0
2 years ago
Read 2 more answers
A sealed container holding 0.0255 L of an ideal gas at 0.981 atm and 65 ∘ C is placed into a refrigerator and cooled to 41 ∘ C w
user100 [1]

Answer:

0.911 atm

Explanation:

In this problem, there is no change in volume of the gas, since the container is sealed.

Therefore, we can apply Gay-Lussac's law, which states that:

"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"

Mathematically:

p\propto T

where

p is the gas pressure

T is the absolute temperature

For a gas undergoing a transformation, the law can be rewritten as:

\frac{p_1}{T_1}=\frac{p_2}{T_2}

where in this problem:

p_1=0.981 atm is the initial pressure of the gas

T_1=65^{\circ}+273=338 K is the initial absolute temperature of the gas

T_2=41^{\circ}+273=314 K is the final temperature of the gas

Solving for p2, we find the final pressure of the gas:

p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm

3 0
3 years ago
Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o
Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

U=-\frac{GMm}{r}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

and so, substituting:

R=6370 km\\h_A = 5970 km\\h_B = 21200 km

We find

\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

K=\frac{1}{2}\frac{GMm}{r}

So, the ratio between the two kinetic energies is

\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}

For satellite A, we have

E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J

For satellite B, we have

E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J

So, satellite B has the greater total energy (since the energy is negative).

(d) -2.57\cdot 10^8 J

The difference between the energy of the two satellites is:

E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J

4 0
3 years ago
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