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Mumz [18]
3 years ago
10

A twig from a tree drops from a 200m high cliff on to a beach below.during its fall 40% twig's energy is converted into thermal

energy.what is the speed with which the twig hits the beach?
Physics
1 answer:
kolezko [41]3 years ago
7 0
<span>Calculating for potential energy P.E=mgh
m=mass
g=acceleration due to gravity
h=height
</span>Let the twig's mass be M<span> 
</span><span>
therefore at the start, the twig has 200*9.8*M potential energy. 

</span>During its fall 40% twig's energy is converted into thermal energy and the remaining <span>60% of this will turn into kinetic energy during the fall, giving us 1176M
Now the equation for kinetic energy, K.E is 0.5*m*v^2
m is mass
v is velocity

Therefore 1176M=0.5*M*v^2 (mass will divide out)
velocity is </span><span>48.49 m/s</span>
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To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
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Explanation:

LD₁ = 10⁵ mm⁻²

LD₂ = 10⁴mm⁻²

V = 1000 mm³

Distance = (LD)(V)

Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m

Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m

Conversion to miles:

Distance₁ = 10×10⁴ m / 1609m = 62 miles

Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.

7 0
3 years ago
Read 2 more answers
A truck has shock absorbers with a spring constant of 24200 N/m. When it hits a bump, it oscillates at 0.429 Hz. What is the mas
siniylev [52]

Answer:

3331.5 kg

Explanation:

Given:

Spring constant of the spring (k) = 24200 N/m

Frequency of oscillation (f) = 0.429 Hz

Let the mass be 'm' kg.

Now, we know that, a spring-mass system undergoes Simple Harmonic Motion (SHM). The frequency of oscillation of SHM is given as:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

Rewrite the above equation in terms of 'm'. This gives,

2\pi f=\sqrt{\frac{k}{m}}\\\\Squaring\ both\ sides,\ we\ get:\\\\(2\pi f)^2=\frac{k}{m}\\\\m=\frac{k}{4\pi^2 f^2}

Now, plug in the given values and solve for 'm'. This gives,

m=\frac{24200\ N/m}{4\pi^2\times (0.429\ Hz)^2 }\\\\m=\frac{24200\ N/m}{4\pi^2\times 0.184\ Hz^2}\\\\m\approx3331.5\ kg

Therefore, the mass of the truck is 3331.5 kg.

3 0
3 years ago
How can you increase the potential energy of a bouncing ball
ad-work [718]

Answer:

When you lift the ball, you are doing work to increase its gravitational potential energy. When you then release the ball, gravitational energy is transformed into kinetic energy as the ball falls. When the ball hits the floor, the ball's shape changes as it flattens against the floor.

Explanation:thats should be the way^^ in explaining

7 0
3 years ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
3 years ago
A mass m = 550 g is hung from a spring with spring constant k = 2.8 N/m and set into oscillation at time t = 0. A second, identi
riadik2000 [5.3K]

Answer:

The second system must be set in motion t=0.70s seconds later

Explanation:

The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:

T=2\pi\sqrt(\frac{m}{k} )

One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.

If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s) seconds later

6 0
3 years ago
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