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3 years ago

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A dog sledding team is composed of a 82.8 kg musher (the person driving the sled), a 21.4 kg sled, and four dogs. Assume that ea

ch dog's pull is fully transmitted to the sled, applied parallel to the ground, and the sled travels along level ground. How much power does each dog supply to pull the sled and driver system at a constant speed of 8.90 m/s if the coefficient of friction between the sled and the snow is 0.150

2 answers:

vlabodo [156]3 years ago

6 0

Answer: 340.8W

Explanation: Please see the attachments below

alisha [4.7K]3 years ago

4 0

Answer:

340.8 watt for each dog

Explanation:

Now, Work = Force x Displacement

We have to calculate power,

Now for power we will divide equation with time "t"

Work/time = Force x Displacement/time

So, equation becomes,

Power = Force x velocity

Now we have to calculate net force acting upon system.

Net force = Force - Frictional force

Since, Coefficient of friction = F/N

Where F = force acting to slide and N = Normal force acting towards center of earth

Net Force = ma - (coefficient of friction x normal force)

now 0 = ma - 0.15 x {9.8 x (82.8 + 21.4)}

0 = (Force - 153.17) N

Force = 153.17 N

Notice that "net force" will become "0" because velocity is constant.

So,

Power = Force x Velocity

Power = 153.17 x 8.9

Power = 1363.2 watt

Now for each dog

Power = 1363.2/4

Power = 340.8 watt

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A 45 kg wagon is being pulled with a rope that makes an angle of 38 degrees with the horizontal. The applied force is 410 N and

mel-nik [20]

Answer:

7.35m/s²

Explanation:

From the question we are not told what to find but we can as well find the acceleration of the wagon.

According to newton second law of motion

$\sum F_x = ma_x\\Fm - Ff = ma_x\\Fm - \mu R = ma_x\\Fm- \mu mg = ma_x\\$

Fm is the moving force = 410N

$\mu$ is the coefficient of friction = 0.18

m is the mass = 45kg

g is the acceleration dur to gravity = 9.8m/s²

a is the acceleration of the wagon

Substitute the given data into the equation ang get ax

$Fm- \mu mg = ma_x\\410 - (0.18)(9.8)(45) = 45a_x\\410 - 79.38 = 45a_x\\330.62 = 45a_x\\a_x = 330.62/45\\a_x = 7.35m/s^2\\$

Hence the acceleration of the wagon is 7.35m/s²

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2 years ago

What is the name of the small early planets, which formed through gravitational attraction reaching sizes of a few miles to even

Svetradugi [14.3K]

Answer:

Planetesimals (Ex, Mercury, Venus , Mars & Earth)

Explanation:

They are known as Planetesimals , They are small in size and have rocky surface. The examples of Planetesimals are Mercury, Venus Mars and Earth.

They are small early planets , which also has gravitational attraction.

Thanks

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3 years ago

The table and graph below show the distances traveled by two different objects. (3 points)

Aleks04 [339]

Answer:c

Explanation: the speed of object a changes but b travels at constant speed

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2 years ago

A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of n

Oduvanchick [21]

Answer:

15.0 m/s^2

Explanation:

3 0

3 years ago

Read 2 more answers

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

cupoosta [38]

Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × $10^{-31}$ kg ×a = 1.18 × $10^{4}$ × 1.602 × $10^{-19}$

a = 20.75 × $10^{14}$ m/s²

so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × $10^{7}$ = 0 + 20.75 × $10^{14}$ (t)

t = 11.80 × $10^{-9}$ s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 × $10^{-9}$

time is 23.6 × $10^{-9}$ s

time will elapse before it return to its staring point is 23.6 ns

7 0

2 years ago