Why can a solution be classified as a mixture?

A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

3 years ago

in the figure shown, if the mass of the block is 4kg and the coefficient of static friction is 0.5 and the coefficient of kineti

Elan Coil [88]

Answer:

Ff = 19.6 N

Explanation:

So since its saying whats the minimum F to move the block, we will use static friction (0.5).

We will use the equation for force of friction, which is Ff = uFn

Ff = (0.5)(4)(9.8)

Ff = 19.6 N

this is the minumum force needed to move the block, as that is the frictional force. You would need to apply a minimum force of 19.6 N to move the block

3 0

3 years ago

Name some of the mediums that sound can travel through

Brums [2.3K]

Answer:

gas, liquid, and solid

Explanation:

6 0

3 years ago

Read 2 more answers

A 295-kg object and a 595-kg object are separated by 4.10 m.

kodGreya [7K]

Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

r=d/2 = 2.05 m

The force between the mass m₁ and m₃

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

by putting the values

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

$F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}$

F₁₃=2.94 x 10⁻⁷ N

The force between the mass m₂ and m₃

by putting the values

$F_{23}=\dfrac{Gm_2m_3}{r^2}$

$F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}$

F₂₃=5.94 x 10⁻⁷ N

The net force F

F= F₂₃- F₁₃

F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

b)

Lest take at distance x from mass m₂ net force is zero.

$F_{23}=\dfrac{Gm_2m_3}{x^2}$

$F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}$

Form above two equation

$\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}$

$\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}$

$\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}$

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

4 0

3 years ago

A car with a mass of 900 kg has an initial velocity of 20 m/s. After 5 s, it has a velocity of 32 m/s.

joja [24]

See attachment file below.

Hope it helped!

3 0

3 years ago