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lisabon 2012 [21]
3 years ago
15

A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 19.0 kg, and the

loaded sled with its rider has a mass of 210 kg.
(a) Calculate the acceleration starting from rest if each dog exerts an average force of 185 N backward on the snow.
(b) What is the acceleration once the sled starts to move?
The coefficient of kinetic friction is 0.1 and static friction is 0.14
Physics
1 answer:
aliina [53]3 years ago
3 0

Answer:

a) a = 3.29 m/s²

b) a = 3.51 m/s²

Explanation:

mass of dogs = 19 kg

loaded sled mass = 210 Kg

a)writing all the forces in x direction

\sum F_x=8F_d-f_{s(max)}= (m_s+8m_d)a\\a=\dfrac{8F_d-\mu_sm_sg}{m_s+8m_d}\\a=\dfrac{8\times 185-(0.14)(210)(9.81)}{210+8(19)}

a = 3.29 m/s²

b) acceleration  when sled start to move the friction will now be acting will be kinetic friction.  

a=\dfrac{8F_d-\mu_sm_sg}{m_s+8m_d}\\a=\dfrac{8\times 185-(0.1)(210)(9.81)}{210+8(19)}

a = 3.51 m/s²

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An ideal gas initially at 4.00atm and 350 K is permitted
Nuetrik [128]

Explanation:

It is given that initially pressure of ideal gas is 4.00 atm and its temperature is 350 K. Let us assume that the final pressure is P_{2} and final temperature is T_{2}.

(a)   We know that for a monoatomic gas, value of \gamma is \frac{5}{3}[/tex].

And, in case of adiabatic process,

                PV^{\gamma} = constant              

also,         PV = nRT

So, here    T_{1} = 350 K,    V_{1} = V,  and   V_{2} = 1.5 V

Hence,      \frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}

         \frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{5}{3} -1}

          T_{2} = 267 K

Also,   P_{1} = 4.0 atm,   V_{1} = V,  and   V_{2} = 1.5 V

        \frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}

        \frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{5}{3}}

            P_{2} = 2.04 atm

Hence, for monoatomic gas final pressure is 2.04 atm and final temperature is 267 K.

(b) For diatomic gas, value of \gamma is \frac{7}{5}[/tex].

As,        PV^{\gamma} = constant              

also,         PV = nRT

T_{1} = 350 K,    V_{1} = V,  and   V_{2} = 1.5 V

              \frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}

         \frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{7}{5} -1}

          T_{2} = 289 K

And,   P_{1} = 4.0 atm,   V_{1} = V,  and   V_{2} = 1.5 V

                \frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}

        \frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{7}{5}}

            P_{2} = 2.27 atm

Hence, for diatomic gas final pressure is 2.27 atm and final temperature is 289 K.

6 0
3 years ago
What is question 22?
Fynjy0 [20]

the answer is a!! its pretty simple I just read the graph.
4 0
3 years ago
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A snail is traveling from garden a to garden b which are 2 meters apart. It takes 4 hours for the snail to make it to garden b.
lys-0071 [83]

The snail was going at 0.5

7 0
3 years ago
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A 0.30 kg yo-yo consists of two solid disks of radius 5.10 cm joined together by a massless rod of radius 1.0 cm and a string wr
docker41 [41]

Answer:

0.70046 m/s²

2.732862 N

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of yo-yo = 0.3 kg

R = Radius of rolling disk = 5.1 cm

r = Radius of rod = 1 cm

For a rolling disks the acceleration is given by

a=\frac{g}{1+\frac{R^2}{2r}}\\\Rightarrow a=\frac{9.81}{1+\frac{0.051^2}{2\times 0.01^2}}\\\Rightarrow a=0.70046\ m/s^2

The acceleration of the yo-yo is 0.70046 m/s²

The tension in the string will be

T=m(g-a)\\\Rightarrow T=0.3\times (9.81-0.70046)\\\Rightarrow T=2.732862\ N

The tension in the string is 2.732862 N

8 0
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BRainliest if correct
kolbaska11 [484]

Answer:

C.

hope this helps

have a good day :)

Explanation:

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