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ArbitrLikvidat [17]

2 years ago

13

A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40° and fly ano

ther 4.0 km. How far west are they of their initial position? What is the magnitude of their displacement? Knight, Randall D., (Professor Emeritus). College Physics (p. 100). Pearson Education. Kindle Edition.

1 answer:

Fudgin [204]2 years ago

7 0

Answer:

(a). The distance is 7.06 km due to west.

(b). The magnitude of their displacement is 7.51 km.

Explanation:

Given that,

The geese fly 4.0 km due west, then turn toward the north by 40° and fly another 4.0 km.

According to figure,

(a). We need to calculate the distance

Using formula of distance

$AD=AB+BD$

Put the value into the formula

$D= 4+4.0\cos40^{\circ}$

$D=7.06\ km$

The distance is 7.06 km due to west.

(b). We need to calculate the magnitude of their displacement

Using formula of displacement

$AC=\sqrt{(CD)^2+(AD)^2}$

Put the value into the formula

$AC=\sqrt{(4.0\sin40)^2+(7.06)^2}$

$AC=7.51$

The magnitude of their displacement is 7.51 km.

Hence, (a). The distance is 7.06 km due to west.

(b). The magnitude of their displacement is 7.51 km.

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A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision

TEA [102]

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

$0 = m(3.5 sin30) - mv_y$

$v_y = 1.75 m/s$

Now similarly in X direction we will have

$m(4.5) = m(3.5 cos30 ) + mv_x$

$v_x = 1.47 m/s$

now final kinetic energy of both puck after collision is given as

$KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)$

$KE_f = 8.73 m$

initial kinetic energy of both pucks is given as

$KE_i = \frac{1}{2}m(4.5^2) + 0$

$KE_i = 10.125 m$

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

5 0

2 years ago

A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

4 0

2 years ago

A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy

GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{5^2+8^2}=9.43\ m/s$

Let E is the kinetic energy of the plane. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2$

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0

2 years ago

Read 2 more answers

A balloon drifts 140m toward the west in 45s ; then the wind suddenly changes and the balloon flies 90m toward the east in the n

Bogdan [553]

Answer: 140 m

Explanation:

Let's begin by stating clear that motiont is the change of position of a body at a certain time. So, during this motion, the balloon will have a trajectory and a displacement, being both different:

The<u> trajectory</u> is <u>the path followed by the body, the distance it travelled</u> (is a scalar quantity).

The displacement is <u>the distance in a straight line between the initial and final position</u> (is a vector quantity).

So, according to this, the distance the balloon traveled during the first 45 s (its trajectory) is 140 m.

But, if we talk about displacement, we have to draw a straight line between the initial position of the balloon (point 0) to its final position (point 90 m). Being its displacement 95 m.

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2 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

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$T_i$ = Initial temperature of water and Aluminum = 30°C

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$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

2 years ago