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yarga [219]
3 years ago
13

Alexandra drops an egg from 30 m above the ground to hit Vanessa on the head. If Vanessa stands 3.5 meters tall. How fast is the

egg moving when it strike her head? (Hint: Draw a picture and you will see why you have to consider her height and find the time first!)
Physics
1 answer:
I am Lyosha [343]3 years ago
7 0

This is problem of free falling objects, which can be solved using the formula:

V = sqrt(2gy)

Where v is the velocity upon impact

G is the acceleration due to gravity ( 9.81 m/s2)

Y is the height

Since Venessa is 3.5 m

Y = 30 -3.5 = 26.5 m

V = sqrt(2 (9.81 m/s2) ( 26.5 m))

<span>V = 22.8 m/s</span>

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Car A has a mass of 1,200 kg and is traveling at a rate of 22km/hr. It collides with car B has a mass of 1,900 kg and is traveli
Zina [86]
<span>The momentum before the collision is equal to the momentum after the collision</span>
7 0
3 years ago
camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Usin
son4ous [18]

The question is incomplete. Here is the complete question.

The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined  for this image, so the cars you see are in fact the same car, but photographed at differene times.

Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?

Answer: v = 6.5 m/s

Explanation: The question asks for velocity of the car. Velocity is given by:

v=\frac{\Delta x}{\Delta t}

The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:

Δx = 7(5.3)

Δx = 37.1 m

The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

Then, the car was driving:

v=\frac{37.1}{5.4}

v = 6.87 m/s

The car drove at, approximately, a velocity of 6.87 m/s

7 0
3 years ago
Three parallel wires of length l each carry current Iin the same direction. They’re positioned at the vertices of an equilateral
cluponka [151]

Answer:

F = μi²l/πa

Explanation:

The magnetic force F on a length of wire, l carrying a current i in a magnetic field B is given by

F = Bilsinθ      

The magnetic field due to one wire of length, l carrying a current, i at a distance a from it is given by B = μi/2πa

So, the force on the first wire due to the second wire is F₁ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

the force on the first wire due to the third wire is F₂ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

Since the magnetic field due to the one wire is perpendicular to the length of the other wire its field acts upon, θ = 90

So, F₁ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa and

F₂ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa

Since the angle between F₁ and F₂ is 60° (since it is an equilateral triangle)

The resultant force F is thus

F = √(F₁² + F₂² + 2F₁F₂cos60°)

F = √(F₁² + F₂² + 2F₁F₂ × 0.5)

F = √(F₁² + F₂² + 2F₁F₂)   (since F₁ = F₂)

F = √(2F₁² + 2F₁²) = √(4F₁²)

F = 2F₁

F = 2μi²l/2πa

F = μi²l/πa

6 0
3 years ago
Read 2 more answers
A student wishes to work out how much power she uses to lift her body when climbing a
Bess [88]

Answer

i'm not 100% sure but 1764

Explanation:

Work done = gravitational potential energy

Gravitational potential energy = mass(kg) × height(m) × gravitational field strength(N/kg)

We can assume that the student is on earth so the gravitational field strength is 9.8N/kg

So work done = 60 × 3×9.8

=1764

(if you need help calculating power but if you do just divide your answer by 12 and you will get 147)

4 0
2 years ago
WHAT ARE THREE TECHNICAL ADVANCEMENTS
Snezhnost [94]

the iphone

the airplane

5G

6 0
3 years ago
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