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svetoff [14.1K]
2 years ago
9

It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft

is dropping vertically at a speed of 293 m/s. The spacecraft is to make a soft landing -- that is, at the instant it reaches the surface of Mars, its velocity is zero. Assume the spacecraft undergoes constant acceleration from the elevation of 34.7 km until it reaches the surface of Mars. What is the magnitude of the acceleration
Physics
1 answer:
Paul [167]2 years ago
8 0

Answer: 1.23\ m/s^2

Explanation:

Given

At an elevation of y=34.7\ km, spacecraft is dropping vertically at a speed of u=293\ m/s

Final velocity of the spacecraft is v=0

using equation of motion i.e. v^2-u^2=2as

Insert the values

\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2

Therefore, magnitude of acceleration is 1.23\ m/s^2.

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Two forces one of 12N and another of 5N act on a body in such away that they makes an angle of 90 digre with each other,what is
ELEN [110]

Answer:

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6 0
3 years ago
Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine
TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because,  the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by tanθ = y/D

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ =  λD/d and y₋₁ =  -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ =  3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

7 0
3 years ago
An object of mass m moving at a speed of v1 possesses kinetic energy that is equal to KE1. When the object's speed is doubled, t
Vera_Pavlovna [14]

Answer:

KE₂ = 4KE₁

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7 0
3 years ago
A flow is isentropically expanded to supersonic speeds in a convergent-divergent nozzle. The reservoir and exit pressures are 1.
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Answer:

Ae/A* = 1.115

Explanation:

Let the reservoir pressure be  p_0

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Ratio of reservoir pressure and exit pressure

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The value of pressure ratio is Ae/A* = 1.115

6 0
3 years ago
An object falls from a high building and hits the ground in 8.0 seconds. Ignoring air resistance, what is the distance that it f
d1i1m1o1n [39]

Answer:

310 meters

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Δy = -313.6

Rounded to two significant figures, the object fell 310 meters.

4 0
3 years ago
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