A plane has a mass of 2,000 kg and is traveling 100 meters above the ground. What is the plane’s potential energy?

You walk to the north, then turn 90° to your left and walk another How far are you from where you originally started?

rodikova [14]

Whatever distance north and then west you walked, you are then

(1.41 x that distance)

northwest of where you started.

3 0

2 years ago

A green plant absorbs light a frog eats flies these are both examples of how organisms

Llana [10]

I think I can answer your question since I've worked on this before. Your answer should be obtain energy. If your answer choices were; obtain energy escape predators produce offspring excrete waste

7 0

2 years ago

What is the Kinetic Energy of a 1200 kg object that is moving with a speed of 24 m/s

SVEN [57.7K]

Kinetic Energy = 1/2mv^2

m= 1200kg

v= 24 m/s

KE = 1/2 (1200kg)(24m/s)^2 = 345,600 N

8 0

3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

3 years ago

Franklin’s hometown is flooding due to a severe storm. He is evacuating the area in his vehicle. What should he do if he comes a

Mrac [35]

Franklin must not drive through a flood for there may no road at all under the water, unless he is familiar with the road and the flowing water is below one foot.

Moreover, if negotiating a flooded section of road, he must drive in the middle where the water will be at its shallowest and he must not drive through water against approaching vehicle to consider other drivers.

4 0

3 years ago

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