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vesna_86 [32]
3 years ago
10

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.
Physics
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
grin007 [14]

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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A car travels west at 40 km/h
Nesterboy [21]

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Explanation:

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3 years ago
A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,
Romashka [77]

Answer:

                    v_{y}  = -104 m/s

Explanation:

Using:

Force = electric field * charge

F=e*q

Force = magnitude of charge * velocity * magnetic field * sin tither

F_{x2}= |q|*v*B*sin \alpha

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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3 years ago
Can someone help please I’ll mark brainless
morpeh [17]

Answer:

Synthesis Reaction

Explanation:

6 0
2 years ago
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