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damaskus [11]
3 years ago
14

A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the

south relative to the ground. What is the bird's velocity relative to the ground? State your answer to one decimal place.
Physics
2 answers:
alexandr402 [8]3 years ago
7 0

Answer:

r=6.05km/hr

z=59.1 degree to the horizontal

Explanation:

A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the south relative to the ground. What is the bird's velocity relative to the ground? State your answer to one decimal place

can be solved using pythagoras theorem

r^2=o^2+a^2

r^2=5.2^2+3.1^2

r^2=36.65

r=6.1km/hr is te birds velocity relative to the ground

tanz=5.2/3.1

z=tan^-1(5,2/3.1)

z=59.1 degree to the horizontal

podryga [215]3 years ago
6 0

Answer:

6.0 km/hr

Explanation:

The wind's velocity relative to the ground is perpendicular to the bird's velocity relative to the air.

When you use Pythagorean theorem to add the two velocity vectors, we find that the bird's velocity relative to the ground is

6.0 kilometers/hour southeast.

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A man attempts to pick up his suitcase of weight ws by pulling straight up on the handle. However, he is unable to lift the suit
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Answer:

b)

Explanation:

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In this case, as the external force on the suitcase pulls upward, in order  to counteract the influence of gravity, normal force is less than the weight of the suitcase, as follows:

F + Fn = m*g

⇒ Fn = m*g - F

So, the normal force is equal to the magnitude of the weight of the suitcase (m*g) minus the magnitude of the force of the pull (F) which is the same expressed by the statement b.

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3 years ago
Which of the following is most closely related to an activated complex?
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3 years ago
A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the
Andrej [43]

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a  

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ  + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

6 0
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Oxana [17]
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