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2 years ago

11

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.8 m/s rel

ative to the ground.

1 answer:

Svetach [21]2 years ago

4 0

I'm assuming the question is what is the robin's speed relative to to the ground...

Create an equation that describes its relative motion.

rVg = rVa + aVg

Substitute values.

rVg = 12 m/s [N] + 6.8 m/s [E]

Use vector addition.

| rVg | = √ | rVa |² + | aVg |²

| rVg | = √ 144 m²/s² + 46.24 m²/s²

| rVg | = √ 19<u>0</u>.24 m²/s²

| rVg | = 1<u>3</u>.78 m/s

Find direction.

tanФ = aVg / rVa

tanФ = 6.8 m/s / 12 m/s

Ф = 29°

Therefore, the velocity of the robin relative to the ground is 14 m/s [N29°E]

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A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.

Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

7 0

2 years ago

A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does

astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus = 14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

= 24.25 × 10⁶ N/m²

Volume = $\dfrac{4}{3} \pi r^3$

$\dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}$

Bulk modulus is equal to

$B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }$

now

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }$

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }$

$(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}$

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

7 0

2 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

1 year ago

A block of mass 2.00 kg is initially at rest at x=0 on a slippery horizontal surface for which there is no friction. Starting at

Allisa [31]

Answer:

x = 1,185 m , t = 4/3 s , F = - 4 N

Explanation:

For this exercise we use Newton's second law

F = m a = m dv /dt

β - α t = m dv / dt

dv = (β – α t) dt

We integrate

v = β t - ½ α t²

We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t

v-v₀ = β t - ½ α t²

the farthest point of the body is when v = v₀ = 0

0 = β t - ½ α t²

t = 2 β / α

t = 2 4/6

t = 4/3 s

Let's find the distance at this time

v = dx / dt

dx / dt = v₀ + β t - ½ α t2

dx = (v₀ + β t - ½ α t2) dt

We integrate

x = v₀ t + ½ β t - ½ 1/3 α t³

x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³

The body comes out of rest

x = 3.5556 - 2.37

x = 1,185 m

The value of force is

F = β - α t

F = 4 - 6 4/3

F = - 4 N

8 0

2 years ago

The graph below shows the sunspot number observed between 1750 and 2000. The graph shows sunspot number on the y axis and years

OlgaM077 [116]

1850 to 1900 because the slope would be 105. It says what is the greatest fall, so the upward slope of 120 wouldn't count.

3 0

2 years ago

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