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Blababa [14]
3 years ago
9

A runner runs around the track consisting of two parallel lines 96 m long connected at the ends by two semi circles with a radiu

s of 49 m she completes one lap in 100 seconds what is her average velocity?
Physics
1 answer:
Zielflug [23.3K]3 years ago
7 0
-0 m/s
- average velocity=displacement/time
- the runners displacement is zero so her average velocity must be zero
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A container of hydrogen gas has a volume of 2.3m³. What is the mass of hydrogen contained? The density of hydrogen = 90kg/m³.
Solnce55 [7]
Mass= volume x density
Mass= 90kg/m^3 x 2.3m^3

Therefore, Mass= 207 kg
3 0
2 years ago
Read 2 more answers
A short current element dl⃗ = (0.500 mm)j^ carries a current of 4.80 A in the same direction as dl⃗ . Point P is located at r⃗ =
sammy [17]

Answer:

Magnetic field in Tesla: 1.65*10^−10 T,   0,   3.09*10^−10 T

Explanation:

Given data:

i = 4.8 A ,

dBx = 1.65*10^-10 T

dBy = 0 T

dBz = 3.09 * 10^-10 T

Attached below is the detailed solution

7 0
3 years ago
A 2.6 kg mass attached to a light string rotates on a horizontal,
Ainat [17]

The maximum speed the mass can have before it breaks is 2.27 m/s.

The given parameters:

  • <em>maximum mass the string can support before breaking, m = 17.9 kg</em>
  • <em>radius of the circle, r = 0.525 m</em>

The maximum speed the mass can have before it breaks is calculated as follows;

T = ma_c\\\\Mg = \frac{Mv^2}{r} \\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v_{max} = \sqrt{0.525 \times 9.8} \\\\v_{max} = 2.27 \ m/s

Thus, the maximum speed the mass can have before it breaks is 2.27 m/s.

Learn more about maximum speed of horizontal circle here:brainly.com/question/21971127

8 0
2 years ago
Consider an optical cavity of length 40 cm. Assume the refractive index is 1, and use the formula for Icavity vs wavelength to p
Bad White [126]

Answer:

Diode Lasers  

Consider a InGaAsP-InP laser diode which has an optical cavity of length 250  

microns. The peak radiation is at 1550 nm and the refractive index of InGaAsP is  

4. The optical gain bandwidth (as measured between half intensity points) will  

normally depend on the pumping current (diode current) but for this problem  

assume that it is 2 nm.  

(a) What is the mode integer m of the peak radiation?  

(b) What is the separation between the modes of the cavity? Please express your  

answer as Δλ.  

(c) How many modes are within the gain band of the laser?  

(d) What is the reflection coefficient and reflectance at the ends of the optical  

cavity (faces of the InGaAsP crystal)?  

(e) The beam divergence full angles are 20° in y-direction and 5° in x-direction  

respectively. Estimate the x and y dimensions of the laser cavity. (Assume the  

beam is a Gaussian beam with the waist located at the output. And the beam  

waist size is approximately the x-y dimensions of the cavity.)  

Solution:  

(a) The wavelength λ of a cavity mode and length L are related by  

n

mL

2

λ = , where m is the mode number, and n is the refractive index.  

So the mode integer of the peak radiation is  

1290

1055.1

10250422

6

6

= ×

××× == −

−

λ

nL

m .  

(b) The mode spacing is given by nL

c f 2

=Δ . As

λ

c f = , λ

λ

Δ−=Δ 2

c f .  

Therefore, we have nm

nL f

c

20.1

)10250(42

)1055.1(

2 || 6

2 2 26

= ×××

× ==Δ=Δ −

− λλ λ .  

(c) Since the optical gain bandwidth is 2nm and the mode spacing is 1.2nm, the  

bandwidth could fit in two possible modes.  

For mode integer of 1290, nm

m

nL 39.1550

1290

10250422 6

= ××× ==

−

λ

Take m = 1291, nm

m

nL 18.1549

1291

10250422 6

= ××× ==

−

λ

Or take m = 1289, nm

m

nL 59.1551

1289

10250422 6

= ××× ==

−

λ .

Explanation:

8 0
3 years ago
To see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 400-loop-per
gtnhenbr [62]

Answer:

B = 4.059 x 10¹⁵ T

Explanation:

Given,

Number of loop, N = 400

radius of loop, r = 0.65 x 10⁻¹⁵ m

Current, I = 1.05 x 10⁴ A

Magnetic field at the center of the loop

B = \dfrac{\mu_0NI}{2R}

B = \dfrac{4\pi\times 10^{-7}\times 400 \times 1.05 \times 10^4}{2\times 0.65\times 10^{-15}}

B = 4.059 x 10¹⁵ T

7 0
3 years ago
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