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Crank
2 years 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:
Elina [12.6K]2 years ago
3 0
<span>You are given an applied force of 110 n with an angle of 30</span>°<span> with the ground. Since the  force is not perpendicular or parallel to the sled then you will have two components. These components are in sine and cosine form.

for parallel component
x = rcos</span>β
<span>x = 110cos30</span>°
<span>x = 95.26

for the perpendicular component
y = rsin</span>β
<span>y = 110sin30</span>°
<span>y = 55</span>
Soloha48 [4]2 years ago
3 0

Explanation:

It is given that,

Force exerted on the sled, F = 110 N

The angle force makes with the ground is 30 degrees.

There are two components of force i.e. the parallel and perpendicular components. They are termed 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

So, the parallel and perpendicular components of this force with respect to the ground are 86.6 N and 50 N respectively. Hence, this is the required solution.

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klasskru [66]

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The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.

In the first situation, the horizontal distance covered is

d=v_0 t = x_0

in the second case, the horizontal velocity is increased to

v_x' = 3v_0

And so the new distance travelled will be

d' = v_x' t = 3 v_0 t = 3 x_0

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5 0
3 years ago
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Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
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Answer:

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Let I be the moment of inertia of the both disk after the welding,\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

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A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)

Net torque on the smaller disk,

\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)

From eqn (1) and (2), we get,

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let 'a_2' be the acceleration of the block in the second case,

From the above expression,

\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2

5 0
3 years ago
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PtichkaEL [24]

Answer:

The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.                    

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u=v-at

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So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

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Oduvanchick [21]

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-- Its direction is the direction from the start-point to the end-point.

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So what's the magnitude of the displacement in exactly one year ?
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The Earth covered a huge distance in that year, but the displacement
is zero.

5 0
2 years ago
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