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3 years ago

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A dart leaves a toy dart gun with initial velocity of 7.76 m/s, regardless of the angle it is fired. What is the maximum horizon

tal distance that the dart can travel? Assume that the dart gun is fired at ground level.

1 answer:

In-s [12.5K]3 years ago

4 0

Vf=vi plus 2 ad
0=7.76 + 2(9.8)d
d=0.395m

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Water flows at 0.850 m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an up

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Answer:

The velocity is $v_2= 0.45 \ m/s$

Explanation:

From the question we are told that

The initial speed of the hot water is $v_1 = 0.85 \ m/s$

The pressure from the heater $P_1 = 450 \ KPa = 450 *10^{3} \ Pa$

The height of the hot water before flowing is $h_1 = 0 \ m$

The height of bathtub above the heater is $h_2 = 3.70 \ m$

The pressure in the pipe is $P_2 = 414 KPa = 414 *10^{3} \ Pa$

The density of water is $\rho = 1000 \ kg/m^3$

Apply Bernoulli equation

$P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2$

Substituting values

$(450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )$

=> $v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}$

=> $v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}$

=> $v_2= 0.45 \ m/s$

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Answer:

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Explanation:

Using:

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Force on particle due to electric field:

$F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$

$(6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)$

$v = (F_{xnet} - F_{x1}) / (F_{x2} )$

$=((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})$

$= (- 104.25) m/s$

$v_{y} = -104 m/s$

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