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e-lub [12.9K]
3 years ago
11

An ion in a mass spectrometer follows a semicircular path of radius 14.8. What is the distance it travels?

Physics
2 answers:
aleksklad [387]3 years ago
7 0

The circumference of a circle is (2π · the circle's radius).

The length of a semi-circle is  (1π · the circle's radius) =

                                                 (π · 14.8) = 46.5 (rounded)

(The unit is the same as whatever the unit of the  14.8  is.)

Y_Kistochka [10]3 years ago
4 0
Is there a time frame in which it travels?

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What is the primary force help suspension bridges use cables to hold there spans up
snow_tiger [21]

Explanation:

Suspension bridges, like the Golden Gate Bridge or the Brooklyn Bridge, use tension force as the primary source of force that cables use to hold their spans up. The supporting cables receive the tension forces of the bridge, and this same force passes to the anchorages and into the ground

3 0
2 years ago
Read 2 more answers
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
A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an
Nadya [2.5K]
Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is 
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
   = (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
   = 4.5 cm

Answer: 4.5 cm
7 0
3 years ago
Read 2 more answers
Salmon often jump waterfalls to reach their
natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

v_x = u cos \theta

where u is the initial speed and \theta=37.7^{\circ}. The horizontal distance travelled by the salmon is

d=v_x t = (ucos \theta)t

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

t=\frac{d}{v_x}=\frac{d}{u cos \theta} (1)

Along the vertical direction, the equation of motion is

y=h+u_y t -\frac{1}{2}gt^2

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

u_y = u sin \theta is the vertical component of the initial velocity of the salmon

g=9.81 m/s^2 is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}

and we can solve this formula for u, the initial speed of the salmon:

y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s

5 0
3 years ago
Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates <0, 4.00 cm> and <0, -4.00 cm
Dmitry [639]

Answer:

Explanation:

Potential energy of the system of charges

=  9 x 10⁹ x [  q₁q₂ / r₁₂ +  q₂q₃ / r₂₃ +  q₁q₃ / r₁₃ ]

here  r₁₂ ,  r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.

r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C

Potential energy  =  9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08  +  -200x10⁻¹⁸ / .04 +  200 x 10¹⁸ / .04 ]

= 9 x 10⁹  x  - 400 x 10⁻¹⁸ / .08

= 45 x 10⁻⁶ J .

b)

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

= 9 x 10⁹ x 10 x 10⁻⁹ / .03

= 3000 V .

potential energy of fourth particle = charge x potential

= 3000 x 40 x 10⁻⁹ = 12 x 10⁻⁵ J .

kinetic energy at infinity = 12 x 10⁻⁵ J

1/2 m v² = 12 x 10⁻⁵ J

.5 x 2 x 10⁻¹³ x v² = 12 x 10⁻⁵

v² = 12 x 10⁸

v = 3.46 x 10⁴ m/s

= 9 x 10⁹

5 0
3 years ago
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