What are the x and y components of a 100m/s velocity acting at an angle of 37° from the positive x axis

The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of

Ahat [919]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is $=4.51m/s$

Explanation:

The kinetic energy of the 9 kg can be determined by these expression

Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction

Now to obtain the initial energy stored

Let U denote the initial energy stored and

$U = \frac{1}{2} kx^2$

Where x is the length the spring is displaced

k is the force constant of the string

$U = \frac{1}{2} * 627 * (0.6)^2$

$= 112.86 J$

Now referring to the formula above

i.e Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction

$\frac{1}{2} mv^2 = 112.86 - \mu_kmgx$

$v^2 = \frac{2(112.86 -\mu_kmgx)}{m}$

$v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}$

and we are told that coefficient of friction = 0.4 and the mass is 9 kg ,the acceleration due to gravity $= 9.8m/s^2$ this displacement length of spring = 0.6

Therefore $v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }$

$=4.51m/s$

8 0

3 years ago

Why are common measurement systems important

kenny6666 [7]

So that international people don't mistake units for others. ex 1 inch could be mistaken for 1 cm in non American countries

6 0

3 years ago

A student suspends a chain consisting of three links, each of mass m 0.250 kg, from a light rope. The rope is attached to the to

klemol [59]

Answer:

B) a= 2.2 m/s², T₁ '= 5.45 N, T₂ ’= 3 N

8) F> T₁ ’> T₂’

Explanation:

A) In the adjoint we can see the free-body diagrams of each element and of the set

B)

i) The acceleration of the chain is

F - W = M a

M = m₁ + m₂ + m₃

a = $\frac{F - Mg}{M}$

a = $\frac{F}{M}$ - g (1)

a = 9 / (3 0.250) - 9.8

a = 2.2 m / s²

the positive sign indicates that the system is rising

ii) the outside of the top link over the middle

T₁ '- W₂ -T₃ = m a

the acceleration of all the links is the same because they are united

Tension T3 is the lower link force that must be equal to its weight

T₃ = W₃

we substitute

T₁ ’= m a + W₂ + W₃

T₁ ’= m a + m g + m g

T₁ ’= m (a + 2g) (2)

T₁ ’= 0.250 (2.2 + 2 9.8)

T₁ '= 5.45 N

iii) the force on the lower link

T₂ ' -W₃ = m a

T₂ '= m a + m g

T₂ '= m (a + g) (3)

T₂ '= 0.25 (2.2 + 9.8)

T₂ ’= 3 N

1) The external forces are

set

* the force of the cueda (F) upwards

* the force of the drawer (W) down

Top link

* the strength of the rope (F)

* the pso of the link (W1)

* the tension of the second link (T2)

Middle link

* The tension of the first link (T1 ')

* Weight (W2)

* the tension of the last link (T3)

Lower link

* The tension of the intermediate link (T2 ')

* Weight (W3)

2) the action and reaction forces are

* T₂ and T₁ '

* T₃ and T₂ '

this are forces of equal magnitude and direction, applied in two bodies

3) in the attachment you have the free body diagram of the body, the vertical upward direction is considered positive

4) The inconnitas are the acceleration of the body and the internal tensions between each link bone 2 tensions

5, 6 and 7)

F - W₁ -T₂ = m a

T₁'- W₂ -T₃ = m a

T₂ '- W₃ = m a

T₂ = T₁ '

T₃ = T₂ '

One way to check that the sign is correct is that the action and reaction force between two bodies can cancel out when adding the equations

F -W₁ - W₂ - W₃ = (m + m + m) a

for which we can evaluate and find the acceleration of the systems

8) order the strengths from greatest to least

F = 9 N,

T₁ '= 5.45 N

T₂ ’= 3 N

F> T₁ ’> T₂’

9) repeat the problem for an exeran force of F = 7.35 N

the acceleration we substitute in equation 1

a = F / m -g

a = 7.35 / 0.75 - 9.8

a = 0

the system is in equilibrium

the voltages using 2 and 3 are

T₁ ’= m (a + 2g)

T₁ ’= 0.25 (0+ 2 9.8)

T₁ ’= 4.9 n

T₂ '= m (a + g)

T₂ '= 0.25 (0 +9.8)

T₂ ’= 2.45 N

the order of force is

F> T₁ ’> T₂’

8 0

3 years ago

Light of wavelength 656 nm and 410 nm emitted from a hot gas of hydrogen atoms strikes a grating with 5300 lines per centimeter.

padilas [110]

Answer:

$20.32^{\circ}$ and $44.08^{\circ}$

$12.56^{\circ}$ and $25.77^{\circ}$

Explanation:

$\lambda$ = Wavelength

$\theta$ = Angle

m = Order

Distance between grating is given by

$d=\dfrac{1}{5300}\\\Rightarrow d=0.0001886\ \text{cm}$

$\lambda=656\ \text{nm}$

We have the relation

$d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}$

m = 1

$\theta=\sin^{-1}\dfrac{1\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=20.35^{\circ}$

m = 2

$\theta=\sin^{-1}\dfrac{2\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=44.08^{\circ}$

The first and second order angular deflection is $20.32^{\circ}$ and $44.08^{\circ}$

$\lambda=410\ \text{nm}$

m = 1

$\theta=\sin^{-1}\dfrac{1\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=12.56^{\circ}$

m = 2

$\theta=\sin^{-1}\dfrac{2\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=25.77^{\circ}$

The first and second order angular deflection is $12.56^{\circ}$ and $25.77^{\circ}$ .

4 0

3 years ago

Answer circled questions.<br> It's about plate tectonics

Alik [6]

A fault line?

A volcano?

5 0

3 years ago

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What are the x and y components of a 100m/s velocity acting at an angle of 37° from the positive x axis​

What are the x and y components of a 100m/s velocity acting at an angle of 37° from the positive x axis