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3 years ago

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If a bowling ball ball was released from the 6th floor of a hotel room 22 meters from the ground, at what velocity would the bal

l be moving just before it hits the head of the man who is 1.5 meters tall? Solve using kinematic equations, unknown values, and SOHCAHTOA.

1 answer:

dexar [7]3 years ago

3 0

Explanation:

Given:

y₀ = 22 m

y = 1.5 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2 (-9.8 m/s²) (1.5 m − 22 m)

v = -20.0 m/s

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A submarine is 2.99 ✕ 102 m horizontally from shore and 1.00 ✕ 102 m beneath the surface of the water. A laser beam is sent from

Vsevolod [243]

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of incidence beam striking the water is $\theta = 49.63^o$

c the angle of refraction beam striking the water is $r = 59.7^o$

d The angle the refracted beam make with respect to the horizontal is $= 30.3^o$

e The height of the target above sea level is

$h= 125.05m$

Explanation:

From the diagram we see that the angle of the beam striking the water is

$tan \theta = \frac{100}{85}$

$\theta = tan^{-1}(\frac{100}{85} )$

$= 49.63^o$

According to Snell's law

$\mu_{water} *sin(i) = \mu_{air} *sin(r)$

Where $\mu_{water }$ is the refractive index of water = 1.333

$i$ is the angle of incidence

$\mu_{air}$ is the refractive index of air = 1

r is the angle of refraction

Substituting values accordingly

$1.33 * sin (40.37) = 1 * sin(r)$

Making r the subject of the formula

$r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})$

$= 59.7^o$

looking at the diagram we can see that to obtain the angle the refraction beam makes with the horizontal by subtracting the angle refraction from 90°

i.e $90 -59.7 = 30.3$ °

From the diagram we see that the height target above sea level can be obtained by this relation

$tan \theta = \frac{h}{214}\\$

Where h is the is the height

$tan(30.3) = \frac{h}{214}$

$h = 214 * tan (30.3)$

$=125.05m$

4 0

3 years ago

A water pipe tapers down from an initial radius of R1 = 0.21 m to a final radius of R2 = 0.11 m. The water flows at a velocity v

Aleks [24]

Answer:

$0.116 m^3/s$

Explanation:

The volume flow rate of a fluid in a pipe is given by:

$Q=Av$

where

A is the cross-sectional area of the pipe

v is the speed of the fluid

In this problem, at the initial point we have

v = 0.84 m/s is the speed of the water

r = 0.21 m is the radius of the pipe, so the cross-sectional area is

$A=\pi r^2 = \pi (0.21 m)^2 =0.138 m^2$

So, the volume flow rate is

$Q=(0.138 m^2)(0.84 m/s)=0.116 m^3/s$

8 0

3 years ago

Model a hydrogen atom as a three-dimensional potential well with Uo = 0 in the region 0 < x a. 283 eV <br> b. 339 eV <br> c.

denis23 [38]

This question is incomplete, the complete question is;

Model a hydrogen atom as a three-dimensional potential well with U₀ = 0 in the region 0 < x < L, 0 < y < L and 0 < z < L, and infinite otherwise, with L = 1.0 × 10⁻¹⁰ m.

Which of the following is NOT one of the lowest three energy levels of an electron in this model?

a. 283 eV

b. 339 eV

c. 113 eV

d. 226 eV

Answer:

the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

Explanation:

Given the data in the question;

Three dimension cube or particle in a cubic box

the energy value is given by;

$E_{nx,ny,nz$ = $( n_x^2 + n_y^2 + n_z^2 )$ × π²h"² / 2ml²

where h" = h/2π and h is Planck's constant ( 6.626 × 10⁻³⁴ m² kg / s )

m is mass of electron ( 9.1 × 10⁻³¹ kg )

l is length of side of box ( 1.0 × 10⁻¹⁰ m )

for ground level ( $n_x = n_y = n_z = 1$ )

so

$( n_x^2 + n_y^2 + n_z^2 )$ × π²h"² / 2ml²

since h" = h/2π

$( n_x^2 + n_y^2 + n_z^2 )$ × π²h² / (2π)²2ml²

so we substitute

$E_{111$ = ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]

$E_{111$ = 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]

$E_{111$ = 3 × [ 6.03082165 × 10⁻¹⁸ ]

Now, we know that electric charge = 1.602 x 10⁻¹⁹

so

$E_{111$ = 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]

$E_{111$ = 3 × [ 37.645578 ]

$E_{111$ = 112.9 ≈ 113 eV

$E_{211$ = $( n_x^2 + n_y^2 + n_z^2 )$ × π²h² / (2π)²2ml²

we substitute

$E_{211$ = ( 1² + 1² + 2² ) × [ 37.645578 ]

$E_{211$ = 6 × [ 37.645578 ]

$E_{211$ = 225.87 ≈ 226 eV

$E_{221$ = $( n_x^2 + n_y^2 + n_z^2 )$ × π²h² / (2π)²2ml²

we substitute

$E_{221$ = ( 2² + 2² + 1² ) × [ 37.645578 ]

$E_{211$ = 9 × [ 37.645578 ]

$E_{211$ = 338.8 ≈ 339 eV

Therefore, the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

8 0

2 years ago

If a hockey puck is made to slide along a straight line on a completely frictionless surface, which statement about the motion o

77julia77 [94]

C. the hockey puck will accelerate and continue moving in a straight line

8 0

3 years ago

Read 2 more answers

schepotkina [342]

Answer:

Explanation:

Length if the bar is 1m=100cm

The tip of the bar serves as fulcrum

A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum

The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.

Using moment

Sum of the moments of all forces about any point in the plane must be zero.

Let take moment about the fulcrum

100×20-F×2=0

2000-2F=0

2F=2000

Then, F=1000N

The force acting in the crate lid is 1000N

Option D is correct

7 0

3 years ago