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Yuki888
6 days ago
7

Jack pulls a sled across a level field by exerting a force of 110 n at an angle of 30 with the ground. what are the parallel and

perpendicular components, respectively, of this force with respect to the ground?
Physics
2 answers:
Softa [913]6 days ago
3 0

Explanation:

It is stated that:

The force acting on the sled is F = 110 N.

The angle with the ground is 30 degrees.

This force has two components: one parallel and the other perpendicular. They are referred to as F_x\ and\ F_y

Parallel component:

F_x=F\ cos\theta

F_x=100\ cos(30)

F_x=86.6\ N

Perpendicular component:

F_y=F\ sin\theta

F_y=100\ sin(30)

F_y=50\ N

Thus, the components of this force relative to the ground are 86.6 N for the parallel and 50 N for the perpendicular. This is the final answer.

Softa [913]6 days ago
3 0
<span>A force of 110 N is applied at an angle of 30</span>°<span> to the horizontal. Because the force does not align directly either vertically or horizontally with the sled, it can be broken down into two components based on sine and cosine.

For the component parallel to the ground:
x = rcos</span>β
<span>x = 110cos30</span>°
<span>x = 95.26

For the component perpendicular to the ground:
y = rsin</span>β
<span>y = 110sin30</span>°
<span>y = 55</span>
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Answer:

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The mass flow rate can be expressed as follows:

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Based on the problem conditions, as the fluid's density remains constant, we can write:

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

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Given the conditions in the problem, A_{2} > A_{1}, leading from the formula to v_{2} < v_{1}.

Furthermore, fluid pressure arises from the fluid's movement through any specific area. When the fluid accelerates, part of its energy increases its speed in the direction of flow, resulting in lower pressure.

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6 0
4 days ago
An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.
Softa [913]

Answer:

The plane's speed in relation to the ground is 300.79 km/h.

Explanation:

Provided details include:

Wind speed = 75.0 km/hr

Plane's airspeed = 310 km/hr

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Using the angle formula

\sin\theta=\dfrac{v'}{v}

Where v' represents the wind speed

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\theta=\sin^{-1}(\dfrac{75}{310})

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Now, we must find the resultant speed

Using the resultant speed formula

\cos\theta=\dfrac{v''}{v}

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\cos14=\dfrac{v''}{310}

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Consequently, the plane's speed in relation to the ground equals 300.79 km/h.

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7 days ago
The magnitude of the electrical force acting between a +2.4 × 10–8 C charge and a +1.8 × 10–6 C charge that are separated by 0.0
kicyunya [1011]
To solve this problem, Coulomb's law will be applied as follows:
F = k*q1*q2 / r^2 where:
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</span>
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Answer:

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