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 of the dogs starting from rest if each dog exerts an average force of 185 N backward on the snow. (b) Calculate the force in the coupling between the dogs and the sled.

Answers

Answer 1

Answer:

The acceleration of the dogs as they pull the sled on the snow is $\boxed{4.09\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}$ .

The force experienced by the sled due to the dogs is $\boxed{859\,{\text{N}}}$ .

Further Explanation:

As the dogs collectively full the sled on the wet snow surface, the sled will experience a net force due to all the dogs acting on it.

Part (a):

The net force applied by the 8 dogs on the sled is:

$\begin{aligned}{F_{{\text{net}}}}&=8*F\n&=8*185\n&=1480\,{\text{N}}\n\end{aligned}$

The net mass of the dog and the sled combined is:

$\begin{aligned}M&=\left({8*{m_d}}\right)+{m_s}\n&=\left({8*19}\right)+210\n&=362\,{\text{kg}}\n\end{aligned}$

The net acceleration of the dogs and the sled combined is:

$a=\frac{{{F_(net)}}}{M}$

Substitute the values of net force and mass in above expression.

$\begin{aligned}a&=\frac{{1480}}{{362}}\n&=4.088\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\n&\approx4.09\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\n\end{aligned}$

Thus, the acceleration of the dogs as they pull the sled on the snow is $\boxed{4.09\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}$ .

Part (b):

Since the sled is moving with an acceleration of $4.09\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}$ , the net force experienced by the sled due to the pull of the dogs is:

${F_(sled)}={m_s}* a$

Substitute the values in above expression:

$\begin{aligned}{F_(sled)}&=210*4.09\n&=858.9\,{\text{N}}\n&\approx{\text{859}}\,{\text{N}}\n\end{aligned}$

Thus, the force experienced by the sled due to the dogs is $\boxed{859\,{\text{N}}}$ .

Learn More:

1. A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 N brainly.com/question/7031524

2. A horizontal rope is tied to a 50 kg box brainly.com/question/2416204

3. Why is it important to define a frame of reference brainly.com/question/526888

Answer Details:

Grade: High School

Subject: Physics

Chapter: Newton’s law of Motion

Keywords:

Team of eight dogs, waxed wood runners, loaded sled, with its rider, acceleration, average force 185 N, coupling between dog and sled, average masses of 19 kg each.

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A Carnot heat engine has an efficiency of 0.400. If it operates between a deep lake with a constant temperature of 298.0k and a hot reservoir, what is the temperature of the hot reservoir?

Answers

Answer:

496.7 K

Explanation:

The efficiency of a Carnot engine is given by the equation:

$\eta = 1 - (T_H)/(T_L)$

where:

$T_H$ is the temperature of the hot reservoir

$T_C$ is the temperature of the cold reservoir

For the engine in the problem, we know that

$\eta = 0.400$ is the efficiency

$T_C = 298.0 K$ is the temperature of the cold reservoir

Solving for $T_H$ , we find:

$(T_C)/(T_H)=1-\eta\nT_H = (T_C)/(1-\eta) =(298.0)/(1-0.400)=496.7 K$

a 63 kg object needs to be lifted 7 meters in a matter of 5 seconds. approximately how much horsepower is required to achieve this task

Answers

Power is defined as the rate of doing work or the work per unit of time. The first step to solve this problem is by calculating the work which can be determined by the equation:

W = Fd

where:

F = force exerted = ma
d = distance traveled
m = mass of object
a = acceleration

Acceleration is equivalent to the gravitational constant (9.81 m/s^2) if the force exerted has a vertical direction such as lifting.

W = Fd = mad = 63(9.81)(7) = 4326.21 Joules

Now that we have work, we can calculate power.

P = W/t = 4325.21 J / 5 seconds = 865.242 J/s or watts

Convert watts to horsepower (1 hp = 745.7 watts)

P = 865.242 watts (1hp/745.7 watts) = 1.16 hp

work = mgh

power = mgh/t = 63•9.8•7/5 = 864 watts

1 HP = 746 watts

so that is 834/746 = 1.2 HP

A body with a mass of 5 Kg has the same kinetic energy as a second body. The second body has a mass of 10 kg and is moving at a speed of 20 m/s. How fast is the second body moving?

Answers

First body:

$E_(k_1)=(1)/(2).m.v^2\n \n E_(k_1)=(1)/(2).5.v_1^2=(5)/(2).v_1^2$

Second body:

$E_(k_2)=(1)/(2).m.v_2^2\n \n E_(k_2)=(1)/(2).10.(20)^2=2.000 \ J$

From description of the task we have:

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What happens to a reflection of light that has struck a surface?A) It bounces off.
B) It becomes absorbed.
C) It gets transmitted.
D) It gains energy.

Answers

Answer:

A) It bounces off.

Explanation:

Bouncing back of light into the same medium is known as reflection of light

As per law of reflection we know that light incident on the surface at a given angle with respect to the normal then the angle of reflected light with the normal will be same

angle of incidence of light = angle of reflection of light

Also we know that incident light, reflected light and normal always lie in the same plane

So we can say that here correct answer will be

A) It bounces off.

Answer:

The answer is A) It bounces off

Explanation:

Refraction bends and Reflection bounces off. (I just took the test).

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Answers

Answer:

The velocity of the sound waves across the plains of Sneedville is 500 m/s.

Step-by-Step Explanation:

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Let the velocity of the sound waves be v, such that:

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⇒ v = 250/0.5

⇒ v = 500 m/s

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Answers

Answer:gravity

Explanation:

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