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2 years ago

4

The diffraction phenomenon can be observed whenever the wavelength is comparable in magnitude to the size of the slit opening. T

o be "diffracted," how fast must a person weighing 84 kg move through a door 1 m wide

1 answer:

Bond [772]2 years ago

5 0

Answer:

Velocity, $v=7.89\times10^{-36}m/s$

Explanation:

Diffraction will only occur when the wavelength, $\lambda$ , is <em>comparable in size</em> to the slit opening. Given the, $\lambda=1m,$ ,and mass of the person as $84kg$ , we can apply

De-Broglie equation, to obtain the velocity.

$\lambda=\frac{h}{mv}\\v=\frac{6.626\times10^{-34}Js}{84kg\times1m}\\v=7.89\times10^{-36}m/s$

The velocity of the person is $7.89\times10^{-36}m/s$

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A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h

Brilliant_brown [7]

Answer:

Explanation:

Impulse = change in momentum

mv - mu , v and u are final and initial velocity during impact at surface

For downward motion of baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes final velocity after impact

change in momentum

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

= .7056 N.s.

Impulse by floor in upward direction

= .7056 N.s

6 0

2 years ago

If a body is moving in the horizontal axis with a velocity Vx= 6m/s and in the vertical axis Vy=8m/s What is the angle Theta abo

cluponka [151]

Answer: C

Explanation: It's a lot of math.

7 0

2 years ago

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

2 years ago

If a scale on Earth reads 650 N, what is your mass?

OlgaM077 [116]

If the scale reads 650N, then the mass of whoever it is standing on the scale is

(weight) / (gravity) = (650N) / (9.8 m/s²) = 66.3 kilograms .

It's not MY mass, even if I'm the one standing on the scale.
If I stand on a scale and it reads 650 N, the scale is broken.

4 0

2 years ago

A 0.900 kg ornament is hanging by a 1.50 m wire when the ornament is suddenly hit by a 0.400 kg missile traveling horizontally a

just olya [345]

Explanation:

The given data is as follows.

Mass of the ornament ( $m_{1}$ ) = 0.9 kg

Length of the wire (l) = 1.5 m

Mass of missile ( $m_{2}$ ) = 0.4 kg

Initial speed of missile ( $u_{2}$ ) = 12 m/s

r = 1.5 m

According to the law of conservation of momentum,

$p_{i} = p_{f}$

$m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$

Putting the given values into the above formula as follows.

$m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$

$0.9 \times 0 + 0.4 \times 12 = (0.9 + 0.4)v$

0 + 4.8 = 1.3v

v = 3.69 m/s

Now, the centrifugal force produced is calculated as follows.

$F_{c} = (m_{1} + m_{2}) \times \frac{v^{2}}{r}$

= $(0.9 + 0.4) \times \frac{(3.69)^{2}}{1.5}$

= 11.80 N

Hence, tension in the wire is calculated as follows.

T = $F_{c} + (m_{1} + m_{2})g$

= $11.80 N + (0.9 + 0.4) \times 9.8$

= 24.54 N

Thus, we can conclude that tension in the wire immediately after the collision is 24.54 N.

4 0

2 years ago