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3 years ago

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An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive xx direction

and has a magnitude of 6.9 N; a second force has a magnitude of 4.5 N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object.

1 answer:

Rudik [331]3 years ago

8 0

Answer with Explanation:

We are given that

$F_1=6.9 N$

$F_2=4.5 N$

We have to find the direction and magnitude of the third force acting on the object.

Resultant force,F= $\sqrt{F^2_1+F^2_2}$

$F=\sqrt{(6.9)^2+(4.5)^2}=8.24 N$

The object moves with constant velocity .Therefore, net force on object is equal to zero

So,Third force, $F_3=F=8.24 N$

Direction, $\theta=tan^{-1}(\frac{F_2}{F_1})$

$\theta=tan^{-1}(\frac{4.5}{6.9})=33.02^{\circ}$

Angle lies in second quadrant because the direction of third force is opposite to the direction of the resultant force of F1 and F2.

Therefore, $\theta=\pi-\theta=180-33.02=146.98^{\circ}$

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Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .

Plugging this value in

$t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}$

$t=1.06\ seconds$

So, the time of flight of the ball is 1.06 seconds.

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3 years ago

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xz_007 [3.2K]

Answer:

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Además de los jugadores, el equipo esta conformado por masajistas, medicos, Entrenador, director y asistente técnico, los cuales tienen funciones específicas para prestar asistencia y dirección a los jugadores.

Explanation:

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2 years ago

hichkok12 [17]

The correct answer is 195.6 N

Explanation:

Different from the mass (total of matter) the weight is affected by gravity. Due to this, the weight changes according to the location of a body in the universe as gravity is not the same in all planets or celestial bodies. Moreover, this factor is measured in Newtons and it can be calculated using this simple formula W (Weight) = m (mass) x g (force of gravity). Now, leps calculate the weigh of someone whose mass is 120 kg and it is located on the moon:

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F= 195.6 N

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3 years ago

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Igoryamba

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