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schepotkina [342]

4 years ago

13

A particle (m = 3.8 × 10-28 kg) starting from rest, experiences an acceleration of 2.4 × 107 m/s2 for 5.0 s. What is its de Brog

lie wavelength λ at the end of this period?

1 answer:

Mazyrski [523]4 years ago

4 0

Answer: The De broglie's wavelength is 1.45×10^-14m

Explanation:

Using De broglie's equation

λ= h/mv

Where h= Planck's constant = 6.6×10^-34Js

m= mass in kilogram = 3.8×10^-28kg

V= velocity = acceleration ×time =

2.4×10^7 ×5

V= 1.2×10^8m/s

λ= is the De broglie's wavelength

λ= 6.6×10^-34Js /(3.8×10^-28kg ×1.2×10^8)

λ= 1.45×10^-14m

Therefore ,De broglie's wavelength=

1.45×10^-14m

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5 0

3 years ago

A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40

Vinvika [58]

Answer:

Ball hit the tall building 50 m away below 10.20 m its original level

Explanation:

Horizontal speed = 20 cos40 = 15.32 m/s

Horizontal displacement = 50 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

50 = 15.32 t + 0.5 x 0 x t²

t = 3.26 s

Now we need to find how much vertical distance ball travels in 3.26 s.

Initial vertical speed = 20 sin40 = 12.86 m/s

Time = 3.26 s

Vertical acceleration = -9.81 m/s²

Substituting in s = ut + 0.5at²

s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²

s = -10.20 m

So ball hit the tall building 50 m away below 10.20 m its original level

5 0

3 years ago

A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg

saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

$x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2$

$F =ma$

$V(t) = V_{o} +at$

$x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2$

(a) Calculating velocity right before going up the ramp.

Wooden block is going on a straightaway and has net for on it.

$F_{n} =F-F_{s} = F-uF_{n} = 100N-0.4*9.8m/s^2*5kg$ = $81N$

and this force produces acceleration of

$a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .$

With this acceleration, wooden block would reach at the foot of ramp in.

$x(t) = 12m = 16.2m/s^2*t^2$

$t = 1.217s$

and final velocity will be

$v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.$

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

$V= V_{h}cos(20) = 18.5288m/s$

and deceleration along the ramp is

$a = \frac{F_{s} }{m}$

Where $F_{s}$ force of friction along the inclined plane.

$F_s = uF_n = u*m*a$

$a = 9.8m/s^2*cos(20) = 9.2089m/s^2$

is a component of g along normal of the inclined plane.

$F_{s} = 0.25*5kg*9.2089m/s^2$

$= 11.5112N$

$a = \frac{11.5112N}{5kg} = 2.3022m/s^2$

And with this deceleration time needed to get wooded block to stop is.

$v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0$

$t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s$

and in that time wooden block would travel

$x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m$

This is how up wooden box will go before coming to stop.

3 0

3 years ago

Assume you push a 20 kg child in a little red wagon with a force of 100 N. However, the dirt and rocks below the wheels

GenaCL600 [577]

Answer:

a

Explanation:

4 0

3 years ago

Please help me to solve this and give me the summary of answer

Neko [114]

Using the principle of floatation.

u = w............(a)

Upthrust of fluid is equal to the weight of the object.

Let the volume of the wood be V.

The upthrust u, is related to the volume submerged in water, and that is 1/5 of it volume, that is (1/5)V = 0.2V

Formula for upthrust, u = vdg

where v = volume of fluid displaced
d = density of fluid
g = acceleration due to gravity

weight, w = mg
where m = mass
g = acceleration due to gravity

From (a)

   u = w

   vdg = mg Cancel out g

   vd = m

The v is equal to 0.2V, which is the submerged volume. Notice that the small letter v is volume of fluid displaced, and capital V is the volume of the solid.

d is density of fluid which is water in this case, 1000 kg/m³

   0.2V * 1000 = m

   200V = m

Hence the mass of the object is 200V kg.

But Density of solid = Mass of solid / Volume of solid

   = 200V / V

   = 200 kg/m³

Density of solid = 200 kg/m³

7 0

3 years ago