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3 years ago

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A hiker walks 200 m west and then walks 100 m north. In what direction is her resulting displacement?a. northb. westc. northwest

d. None of the answers is correct.

1 answer:

emmainna [20.7K]3 years ago

7 0

Answer:

Option (C) is correct.

Explanation:

Assuming that the hiker starts walking from the origin O as shown in the figure.

First, he walks 200 m west to point A (say), then he walks 100 m north to the final point B (say) as shown in the figure.

The final point B is in the north-west direction, therefore, the resulting point is in the north-west direction.

Hence, option (C) is correct.

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An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis

Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

$W=\Delta K$ (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

$W=\Delta K + E_{lost}$

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane ( $\Delta K$ ) and part is lost because of the air resistance ( $E_{lost}$ ).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

$F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N$

Now we can calculate the acceleration of the plane, by using Newton's second law:

$a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2$

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

$v^2- u^2 = 2aS$

where

$v=?$ is the final velocity

$u=66.0 m/s$ is the initial velocity

$a=1.69 m/s^2$ is the acceleration

$S=5.00 \cdot 10^2 m$ is the distance travelled

Solving for v, we find

$v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s$

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The sun’s___and the planet’s___keeps planets moving is___orbits.

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The sun’s gravitational attraction and the planet’s inertia keeps planets moving is circular orbits.

Explanation:

The planets in the Solar System move around the Sun in a circular orbit. This motion can be explained as a combination of two effects:

1) The gravitational attraction of the Sun. The Sun exerts a force of gravitational attraction on every planet. This force is directed towards the Sun, and its magnitude is

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

M is the mass of the Sun

m is the mass of the planet

r is the distance between the Sun and the planet

This force acts as centripetal force, continuously "pulling" the planet towards the centre of its circular orbit.

2) The inertia of the planet. In fact, according to Newton's first law, an object in motion at constant velocity will continue moving at its velocity, unless acted upon an external unbalanced force. Therefore, the planet tends to continue its motion in a straight line (tangential to the circular orbit), however it turns in a circle due to the presence of the gravitational attraction of the Sun.

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3 years ago

What is the the steadiest firing position?

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The closer you are to the ground the more accurate you'll be. That's why most snipers are in the "prone" position.

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A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

gavmur [86]

Answer:

a. ρ $_\beta=1.996J/m^3$

b. $U_E=9.445x10^{-15} J/m^3$

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

ρ $_\beta=\frac{\beta^2}{2*u_o}$

ρ $_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

ρ $_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}$

ρ $_\beta=1.996J/m^3$

b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$

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A woman pushes a 27 kg lawnmower at a steady speed. She exerts a 120 N force in a direction 35◦ below the horizontal. The accele

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The vertical force exerted on the lawn is 68.8 N downward

Explanation:

The vertical force exerted by the lawnmower on the lawn is equal to the vertical component of the force applied, therefore:

$F_y = F sin \theta$

where

F is the magnitude of the force applied

$\theta$ is the angle between the direction of the force and the horizontal

In this problem:

F = 120 N

$\theta=-35^{\circ}$

Substituting,

$F_y = (120)(sin 30)=-68.8 N$

where the negative sign means the direction of the force is downward.

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