What is the amplitude of a wave?

A large cube (mass = 25 kg) being accelerated across a horizontal frictionless surface by a horizontal force . A small cube mass

belka [17]

Answer:F=400.28 N

Explanation:

Given

mass of large cube $M=25 kg$

mass of small cube $m=4 kg$

coefficient of static friction $\mu =0.71$

acceleration a of the system is given by

$F=(M+m)a$

$a=\frac{F}{M+m}$

Now Normal reaction between two blocks is given by

$N=ma$

friction tries to balance weight

thus $f_r=mg$

and $f_r=\mu N$

$f_r=\mu ma$

$\mu ma=mg$

$\mu \frac{F}{M+m}=g$

$F=\frac{(M+m)g}{\mu }$

$F=\frac{29\times 9.8}{0.71}$

$F=400.28 N$

7 0

3 years ago

An unstable atomic nucleus of mass 1.80 10-26 kg initially at rest disintegrates into three particles. One of the particles, of

beks73 [17]

Answer:

a) v = (-7.73 i ^ - 6.95j ^) 10⁶ m/s and b) ΔK = 3.96 10⁻¹³ J

Explanation:

We can work this process of disintegration as a conservation problem of the moment.

We create a system formed by the initial nucleus and the three final particles, in this case all the forces involved are internal and the amount of movement is conserved. We will write the equations each axis

X axis

p₀ = 0

pf = m1 v1x + m vfx

m₁ = 8.50 10-27 kg

vf₁ = 4.00 106 m / s

p₀ = pf

0 = m₁ v₁ₓ + m₃ vfₓ

vfₓ = -m₁ / m₃ v1ₓ

Y Axis

P₀ = 0

Pf = m₂ v₂ + m₃ vfy

m₂ = 5.10 10⁻²⁷ kg

v₂ = 6.00 10⁶ m / s

p₀ = pf

0 = m₂ v₂ + m₃ vfy

vfy = -m₂ / m3₃v₂

We have the initial particle mass and it decomposes into three parts after disintegration

m = m₁ + m₂ + m₃

m₃ = m-m₁-m₂

m₃ = 1.80 10⁻²⁶ - 5.10 10⁻²⁷ - 8.50 10⁻²⁷ = (1.80 -0.510 -0.850) 10⁻²⁶

m3 = 0.44 10⁻²⁶ kg = 4.4 10⁻²⁷ kg

Let's replace and calculate

vfₓ = -m₁ / m₃ v₁ₓ

vfₓ = - 8.50 10⁻²⁷/4.4 10⁻²⁷ 4.00 10⁶

vfₓ = -7.73 10⁶ m / s

vfy = -m₂ / m₃ v₂

vfy = - 5.10 10⁻²⁷ /4.4 10⁻²⁷ 6.00 10⁶

vfy = -6.95 10⁶ m/s

We set the speed vector

v = (-7.73 i ^ - 6.95j ^) 10⁶ m/s

b) Let's calculate the kinetic energy

Initial

K₀ = 0

Final

Kf = K₁ + K₂ + K₃

Kf = ½ m₁ v₁² + ½ m₂ v₂² + ½ m₃ v₃²

v₃² = (7.73 10⁶)²+ (6.95 10⁶)² = 108.5 10⁻¹²

Kf = ½ 8.50 10⁻²⁷(4 10⁶) 2 + ½ 5.1 10⁻²⁷ (6 10⁶) 2 + ½ 4.4 10⁻²⁷ 108.05 10⁻¹²

Kf = 68 10⁻¹⁵ + 91.8 10⁻¹⁵ + 237.7 10⁻¹⁵

Kf = 396.8 10⁻¹⁵ J

Kf = 3.96 10⁻¹³ J

ΔK = Kf -K₀

ΔK = 3.96 10⁻¹³ J

7 0

4 years ago

John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2

Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function $cos(\omega t +\phi)$ is defined between -1 and 1, So it is not possible obtain a value $2\pi$ greater.

In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have $f=1+cos(\omega t +\phi)$ , the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the $f=1+cos(\omega t +\phi)$ , and its mean that the function $f=cos(\omega t +\phi)$ , in general, is not greater than $cos(\omega t +\phi)$ .