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3 years ago

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A puppy finds a rawhide bone and begins to pull it with a force, Ft. The free-body diagram is shown. Which describes what happen

s to the bone?
It begins to move toward the right.
It stays in static equilibrium.
It stays in dynamic equilibrium.
It begins to move toward the left.

2 answers:

Anastasy [175]3 years ago

8 0

It begins to move towards the right

GalinKa [24]3 years ago

4 0

It begins to move towards the right

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The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

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Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s

Explanation:

Given data:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = time to collision = 0.5 s

Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It is neccesary calculate the force:

$F=(m+m)*g*sin\theta$

Here, g = gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

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You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b

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Answer:

$\theta=34 \textdegree$

Explanation:

From the question we are told that:

Mass $m=55kg$

Angle $\theta =28.0$

Coefficient of static friction $\alpha =0.680$

Generally, the equation for Newtons second Law is mathematically given by

For

$\sum_y=0$

$N=mgcos \theta$

for

$\sum_x=0$

$F_{s}=mgsin\theta$

Where

$F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta$

$F_{s}=0.68*55*9.8*cos 28$

$F_{s}=323.62N$

Therefore

$\alpha mgcos \theta=mg sin \theta$

$\theta=tan^{-1}(0.68)$

$\theta=34 \textdegree$

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2 years ago

four children pull on the same stuffed toy at the same time , yet there is no net force on the toy.how is this possible?

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An 8.50 m long ladder leans against the side of a building. The ladder is initially inclined at an angle of 47.0° to the horizon

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The angle of the ladder inclined with respect to the horizontal after being moved a distance of 0.82 m closer to the building is 53.84°

cos θ = Adjacent side / Hypotenuse

θ $_{1}$ = 47°

Hypotenuse = Length of ladder = 8.5 m

cos 47° = Adjacent side / 8.5

Adjacent side = Initial distance of base of ladder from the building = 5.8 m

Adjacent side 2 = Final distance of base of ladder from the building

Adjacent side 2 = 5.8 - 0.82 = 4.98 m

cos θ $_{2}$ = Adjacent side 2 / Hypotenuse

cos θ $_{2}$ = 4.98 / 8.5 = 0.59

θ $_{2}$ = $cos^{-1}$ ( 0.59 )

θ $_{2}$ = 53.84°

The formula used above is one of trigonometric ratios. Trigonometric ratios can used only in a right angled triangle where one of the angles in at 90 degrees and the other two angles are less than 90 degrees.

Therefore, the angle of the ladder inclined with respect to the horizontal after being moved is 53.84°

To know more about trigonometric ratios

brainly.com/question/1201366

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