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Ksenya-84 [330]

3 years ago

7

I need help please ASAP

1 answer:

RSB [31]3 years ago

3 0

Answer:

it's D

Explanation:

you divide the miles by the hours

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The British attempted to break the land-speed record and the sound barrier using a jet-powered car. They made their attempt in t

jolli1 [7]

Your answer is B.

The relationship between medium temperature and speed of sound is a direct relationship: when one factor increases, the other increases as is shown in graph B. The British would choose the the time of day which would give the lowest speed of sound, because this would be easiest to break. Graph B shows that the lowest speed of sound would occur with the lowest air temperature - in the morning.

4 0

3 years ago

Read 2 more answers

A particle of mass 6.3 x 10-8kg and charge 7.1 muC is traveling due east. It enters perpendicularly a magnetic field whose magni

ASHA 777 [7]

Answer:

t=0.016s

Explanation:

we can use

$\frac{v}{r}=\omega=\frac{qB}{m}=\frac{7.1*10^{-6}C*1.7T}{6.3*10^{-8}kg}=191.58s^{-1}$

the time that the particle is in the magnetic field is one half oa period. Hence

$t=\frac{T}{2}=\frac{\pi}{\omega}=0.016s$

I hope this is useful for you

regards

8 0

3 years ago

On an asteroid, the density of dust particles at a height of 3 cm is ~30% of its value just above the surface of the asteroid. A

Anvisha [2.4K]

From the law of atmosphere

$N_v(y) = n_0*e^{-\frac{mgy}{Kb*T}}$

Where

$n_0$ = constant and is number density where the height y = 0cm

$n_V$ = Number density at height y=3cm

Kb = Boltzmann constant $= 1.38*10^{-23}J/K$

$T=20K$

$m = 10^{-19}kg$

Re-arranging the equation to have the value of the gravity,

$\frac{N_v(y)}{n_0} = e^{-\frac{mghy}{KbT}}$

$ln(\frac{N_v(y)}{n_0}) = -\frac{mgy}{KbT}$

Since it is 30% of value above surface, therefore $N_v = 0.3n_0$

$ln(\frac{0.3n_0}{n_0}) = -\frac{mgy}{KbT}$

$g = -\frac{KbT ln(0.3)}{my}$

$g = -\frac{(1.38*10^{-23}J/K)(20K)(Ln(0.3))}{10^{-19}(3*10^{-2})}$

$g = \frac{1.38*2*ln(0.3)*10^{-22}}{3*10^{-4}}$

$g = 1.104*10^{-1}m/s^2$

$g = 0.1m/s^2$

Therefore the correct answer is C.

4 0

4 years ago

The maximum lift-to-drag ratio of the World War I Sopwith Camel was 7.7. If the aircraft is in flight at 5000 ft when the engine

andre [41]

The related concepts to solve this problem is the Glide Ratio. This can be defined as the product between the height of fall and the lift-to-drag ratio. Mathematically, this expression can be written as,

$R = h (\frac{L}{D})_{max}$

Replacing,

$R = 5000ft (7.7)$

$R = 38500ft$

Converting this units to miles.

$R = 38500ft (\frac{1mile}{5280ft})$

$R = 7.2916miles$

Therefore the glide in terms of distance measured along the ground is 7.2916miles

3 0

4 years ago

What are some of the objects found in space and their characteristics?

RideAnS [48]

Answer:

meteor

Explanation:

big rock or metal maybe

6 0

3 years ago