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4 years ago

15

A dog sledding team is comprised of a 80.0 kg musher (the person driving the sled), a 21.0 kg sled and four dogs. assume each do

g\'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/driver system at a constant speed of 8.80 m/s where the coefficient of friction between the sled and the snow is 0.150? (the acceleration due to gravity is 9.81 m/s2.)

1 answer:

Xelga [282]4 years ago

8 0

You would divide or multiply by 9.8

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What describes the movement of a fluid during convection?

damaskus [11]

Convection is the movement<span> of groups of molecules within </span>fluids<span> such as gases and liquids, including molten rock (rheid).</span>

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3 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode

sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

$v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

$v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

$v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

$s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

$x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

$x_{0}=0$

we know that,

$a_{x}=0$

or,

$v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

$x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

$y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}$

we know, $y_{0}=0$ and $a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , so,

$y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)$

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0

3 years ago

HELP PLSSSSSSSSSS<br> lithium has three protons. which model shows a neutral atom of lithium?

wolverine [178]

Answer:

B

Explanation:

the 3 electrons makes it neutral

5 0

3 years ago

Read 2 more answers

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago

If you were to drive an average velocity of 65 mi/hr to Traverse City, which is

Amanda [17]

Find the number of hours by dividing the distance by mph. The number of hours will be to the left of the decimal point:

250 miles / 65 mph
= 3.846153846
= 3 hours

2) Find the number of minutes by multiplying what is remaining from step 1 by 60 minutes. The minutes will be to the left of the decimal point:

0.846153846 x 60
= 50.76923076
= 50 minutes

3) Find the number of seconds by multiplying what is remaining from step 2 by 60 seconds. The seconds will be to the left of the decimal point:

0.76923076 x 60
= 46.1538456
= 46 seconds

So 3 hours 50 mins and 46 seconds

3 0

3 years ago