What is the period of an objects motion?

Light from an argon laser strikes a diffraction grating that has 4,917 lines per cm. The first-order principal maxima are separa

miss Akunina [59]

Answer:

Wavelength is 4.8x10^-7m

Explanation:

See attached file

3 0

3 years ago

Consult Multiple-Concept Example 15 to review the concepts on which this problem depends. Water flowing out of a horizontal pipe

kiruha [24]

Answer:

The pressure of the water in the pipe is 129554 Pa.

Explanation:

<em>There are wrongly written values on the proposal, the atmospheric pressure must be 101105 Pa, and the density of water 1001.03 kg/m3, those values are the ones that make sense with the known ones.</em>

We start usign the continuity equation, and always considering point 1 a point inside the pipe and point 2 a point in the nozzle:

$A_1v_1=A_2v_2$

We want $v_2$ , and take into account that the areas are circular:

$v_2=\frac{A_1v_1}{A_2}=\frac{\pi r_1^2 v_1}{\pi r_2^2}=\frac{r_1^2 v_1}{r_2^2}$

Substituting values we have (we don't need to convert the cm because they cancel out between them anyway):

$v_2=\frac{r_1^2 v_1}{r_2^2}=\frac{(1.8cm)^2 (0.56m/s)}{(0.49cm)^2}=7.56m/s$

For determining the absolute pressure of the water in the pipe we use the Bernoulli equation:

$P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2$

Since the tube is horizontal $h_1=h_2$ and those terms cancel out, so the pressure of the water in the pipe will be:

$P_1=P_2+\frac{\rho v_2^2}{2}-\frac{\rho v_1^2}{2}=P_2+\frac{\rho (v_2^2-v_1^2)}{2}$

And substituting for the values we have, considering the pressure in the nozzle is the atmosphere pressure since it is exposed to it we obtain:

$P_1=101105 Pa+\frac{1001.03Kg/m^3 ((7.56m/s)^2-(0.56m/2)^2)}{2}=129554 Pa$

3 0

3 years ago

A square loop of wire is held in a uniform 0.24 T magnetic field directed perpendicular to the plane of the loop. The length of

NNADVOKAT [17]

Answer:

Explanation:

Given that,

Magnetic field of 0.24T

B = 0.24T

Field perpendicular to plane i.e 90°

Rate of decrease of length of side of square is 5.4cm/s

dL/dt = 5.4cm/s = 0.054m/s

Since it is decreasing

Then, dL/dt = -0.054m/s

When L is 14cm, what is the EMF induced?

L = 14cm = 0.14m

EMF is give as

ε = - dΦ/dt

Where flux is given as

Φ = BA

Where A is the area of the square

A = L²

Then, Φ = BL²

Substituting this into the EMF

ε = - dΦ/dt

ε = - d(BL²)/dt

B is constant

ε = - Bd(L²)/dt

ε = -2BL dL/dr

ε = -2 × 0.24 × 0.14 × -0.054

ε = 3.63 × 10^-3 V

ε = 3.63mV

8 0

4 years ago

a car with a mass of 1.2×10^3 kilograms is moving at a speed of 18 meters/second. it is hit by a truck from behind and it's spee

garik1379 [7]

I = Delta p

I = Impulse
Delta p = Momentum variation

Delta p = pf - po

Pf = Final momentum
Po = Initial momentum

p = m x v

M = Mass
V = Velocity

Delta p = (m x Vf) - (m x Vo)

I = 1.2 x 10^3 x 20 - 12 x 10^2 x 18
I = 2.4 x 10^4 - 216 x 10^2
I = 2.4 x 10^4 - 2.16 x 10^4
I = 0.24 x 10^4
I = 24 x 10^2 Newtons x seconds (N x s)

Answer: The value of impulse is 24 x 10^2 Newtons x seconds (N x s).

8 0

3 years ago

A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.

Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mb*g*H = 0.5*mb*vb^2

vb = √2*g*H

vb = ( 2*9.81*15 ) ^0.5

vb = 17.15517 m/s

- The angular momentum of system before collision is:

M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

M1 = M2

vo = 247.03448 / (5.10*3)

vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mo*g*h = 0.5*mo*vo^2

h = vo^2 / 2*g

h = 16.14604^2 / 2*(9.81)

h = 13.3 m

3 0

3 years ago