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AlekseyPX
3 years ago
12

Which spacecraft did john glenn pilot when he became the first man to orbit the earth in 1962?

Physics
1 answer:
Makovka662 [10]3 years ago
7 0
Spacecraft used is "Friendship 7". Hope it helps.
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Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart. Th
Dima020 [189]

Answer:

Explanation:

Let the thickness of the film is t and the refractive index of the material of film is n.

When light travels through a sheet of thickness t, the optical path traveled is nt.

When the path of one of slit is covered by a sheet of thickness t, the optical path becomes

x = ( n - 1) t

As the one fringe is shift, so the optical path changed by one wavelength.

i.e., x = λ

So, λ = ( n - 1) t

t=\frac{\lambda }{n-1}

7 0
3 years ago
20 points and brainliest‼️‼️‼️‼️
Anastasy [175]

Answer:

0 N

Explanation:

Applying,

F = qvBsin∅................. Equation 1

Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.

From the question,

Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°

Substitute these values into equation 1

F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)

Since sin0° = 0,

Therefore,

F = 0 N

3 0
2 years ago
Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
Read 2 more answers
During the internal examination, the pathologist drains the intestines, removes any undigested food and feces,and examines the c
DanielleElmas [232]
During the internal examination, the pathologist drains the intestines, removes any undigested food and feces, and examines the contents of the stomach. This examination could give the pathologist clues of the time of death, and the location of death.  The process of digesting and defecation vary from person to person, the entire process is generally considered to take approximately 40 hours in adults
8 0
3 years ago
A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the p
frez [133]

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

a(t)=(5-10t)\ m/s^2

(a) Since, a=\dfrac{dv}{dt}

v = velocity

dv=a.dt

v=\int(a.dt)

v=\int(5-10t)(dt)

v=5t-\dfrac{10t^2}{2}=5t-5t^2

(b) v=\dfrac{dx}{dt}

x = position

x=\int v.dt

x=\int (5t-5t^2)dt

x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3

(c) Velocity function is given by :

v=5t-5t^2

5t-5t^2=0

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

7 0
2 years ago
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