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Veseljchak [2.6K]

4 years ago

6

A force vector F1 points due east and has a magnitude of 200N. A second force F2 is added to F1. The resultant of the two vector

s has a magnitude of 400N and points along the east/west line. Find the magnitude and direction of F2Source https://www.physicsforums.com/threads/physics-trigonometry.61160/

1 answer:

PilotLPTM [1.2K]4 years ago

3 0

Answer:

The second vector $\vec{F_2}$ points due West with a magnitude of 600N

Explanation:

The original vector $\vec{F_1}$ points with a magnitude of 200N due east, the Resultant vector $\vec{R}$ points due west (that's how east/west direction can be interpreted, from east to west) with a magnitude of 400N. If we choose East as the positive direction and West as the negative one, we can write the following vectorial equation:

$\vec{F_1}+\vec{F_2}=\vec{R}\implies\vec{F_2}=\vec{R}-\vec{F_1}=-400N-200N=-600N$

With the negative sign signifying that the vector points west.

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The automobile has a weight of 2700 lb and is traveling forward at 4 ft>s when it crashes into the wall. If the impact occurs

alex41 [277]

Answer:

$F_b=153918\, lb.ft.s^{-2}$

Explanation:

Given:

mass, $m=2700 \,lb$

time, $t=0.06\,s$

velocity, $v=4\,ft.s^{-1}$

coefficient of kinetic friction between wheels & pavement, $\mu=0.3$

According to first condition,

$F\times \Delta t=m\times v$

$F\times 0.06= 2700\times 4$

$F=180000\,lb.ft.s^{-2}$

According to second condition,

<u>Magnitude of frictional force (which acts opposite to the direction of motion):</u>

$f=\mu.N$

where N is the normal reaction.

$f=0.3\times 2700\times 32.2$

$f=26082 \,lb.ft.s^{-2}$

Now, the impulsive force on the wall if the brakes were applied during the crash:

$F_b= F-f$

$F_b=180000-26082$

$F_b=153918\, lb.ft.s^{-2}$

3 0

3 years ago

1. A 12 kilogram block is sitting on a platform 24 m high. How much Potential energy does

Mazyrski [523]

The amount of potential energy the block contains is 2,822.4 Joules

<u>Given the following data:</u>

Mass of block = 12 kg

Height of platform = 24 meters.

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 $m/s^2$ .

To determine the amount of potential energy the block contains:

Mathematically, potential energy (P.E) is given by the formula;

$P.E = mgh$

Where:

m is the mass of object.

g is the acceleration due to gravity.

h is the height of an object.

Substituting the parameters into the formula, we have;

$P.E = 12 \times 9.8 \times 24$

Potential energy (P.E) = 2,822.4 Joules

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3 years ago

A mountain lion jumps to a height of 3.25 m when leaving the ground at an angle of 43.2°. What is its initial speed (in m/s) as

miss Akunina [59]

Recall that

${v_f}^2={v_i}^2+2a\Delta y$

where $v_i$ and $v_f$ are the lion's initial and final vertical velocities, $a$ is its acceleration, and $\Delta y$ is the vertical displacement.

At its maximum height, the lion has 0 vertical velocity, so we have

$0={v_i}^2-2gy_{\rm max}$

where <em>g</em> is the acceleration due to gravity, 9.80 m/s², and we take the starting position of the lion on the ground to be the origin so that $\Delta y=y_{\rm max}-0=y_{\rm max}$ .

Let <em>v</em> denote the initial speed of the jump. Then

$v_i=v\sin(43.2^\circ)=\sqrt{2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(3.25\,\mathrm m)}\implies\boxed{v\approx11.7\dfrac{\rm m}{\rm s}}$

6 0

3 years ago

Please can anybody tell me what are these lab equipments

olga nikolaevna [1]

Answer:

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4 0

3 years ago

Read 2 more answers

A worker at the top of a 588-m-tall television transmitting tower accidentally drops a heavy tool. If air resistance is negligib

liq [111]

The final velocity of the tool is 107.4 m/s

Explanation:

We can solve this problem by using the principle of conservation of energy.

In fact, if air resistance is negligible, the total mechanical energy of the tool is conserved during the fall, so we can write:

$K_i + U_i = K_f + U_f$

where

$K_i = 0$ is the kinetic energy of the tool at the top (zero since it is at rest)

$U_i = mgh$ is the gravitational potential energy of the tool at the top, where

m is the mass of the tool

g is the acceleration of gravity

h is the heigth of the tool

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the tool just before hitting the ground, where

v is the final speed of the tool

$U_f = 0$ is the gravitational potential energy of the tool at the bottom (zero since the height is zero)

Re-arranging the equation,

$mgh=\frac{1}{2}mv^2$

where we have

$g=9.8 m/s^2\\h=588 m$

And solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(588)}=107.4 m/s$

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3 0

3 years ago