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2 years ago

Please help me with 1 and 2

Physics

1 answer:

nikitadnepr [17]2 years ago

8 0

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A box of unknown mass is sliding with an initial speed vi = 5.60 m/s across a horizontal frictionless warehouse floor when it en

Maksim231197 [3]

We know that the Delta E + W(Work done by non-conservative forces) = 0 (change of energy)

In here, the non-conservative force is the friction force where f = uN (u =kinetic friction coefficient)

W= f x d = uNd ; N=mg
Delta E = 1/2 mV^2 -1/2mVi^2

umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term)

This will then give us:

1/2Vi^2-ugd = 1/2V^2

V^2 = Vi^2 - 2ugd

So plugging in our values, will give us:

V= Sqrt (5.6^2 -2.3^2)

=sqrt (26.07)

= 5.11 m/s

6 0

3 years ago

Which of the following is a health problem associated with obesity in children?

mihalych1998 [28]

Risk of not being able to reduce their weight

5 0

2 years ago

Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13

lyudmila [28]

Answer:

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

$y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right )$

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

$y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right )$

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$

3 0

2 years ago

It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon

trapecia [35]

Answer: $8.76\times 10^{-3} min^{-1}$

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

$y=1-e^{-kt^n}$

where y=fraction Transformed

k=constant

t=time

$0.3=1-e^{-k(100)^5}$

$e^{-k(100)^5} =0.7$

Taking log both sides

$-k\cdot (10^{10}=\ln 0.7$

$k=3.566\times 10^{-11}$

At this Point we want to compute $t_{0.5}\ i.e.\ y=0.5$

$0.5=1-e^{-kt^n}$

$0.5=e^{-kt^n}$

$0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}$

taking log both sides

$\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5$

$t^5=1.943\times 10^{10}$

$t=114.2 min$

Rate of Re crystallization at this temperature

$t^{-1}=8.76\times 10^{-3} min^{-1}$

3 0

2 years ago

Which statement about activation energy is true? A. It is not affected by catalysts. B. It has no effect on the rate of the reac

HACTEHA [7]

C it is the energy required to break existing chemical bonds, it is the amount of energy that a reaction requires in order for the reactants to successfully collide and react

6 0

3 years ago

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