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3 years ago

The black lines that wrap themselves around a basketball represent an example of what type of geometry?

1 answer:

5 0

In Euclidean geometry parallel lines never intersect. But in non-Euclidean geometry parallel lines can either curve away from each other, or curve towards each other. Example : the black lines that wrap themselves around the basketball.
Answer: B ) non-Euclidean

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a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u

n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

$a_{c}$ = $\frac{v^{2} }{r}$

$a_{c}$ = 23.04/10.3

$a_{c}$ = 2.23 m/s²

It continues to fly but now with some tangential acceleration

$a_{t}$ = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

$a_{net}$ = $\sqrt{a_{c} ^{2} +a_{t} ^{2} }$

$a_{net}$ = $\sqrt{4.97 + 0.396}$

$a_{net}$ = 2.316 m/s²

So the magnitude of net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

#SPJ4

8 0

1 year ago

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

cupoosta [38]

Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

put here all value

9.11 × $10^{-31}$ kg ×a = 1.18 × $10^{4}$ × 1.602 × $10^{-19}$

a = 20.75 × $10^{14}$ m/s²

so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × $10^{7}$ = 0 + 20.75 × $10^{14}$ (t)

t = 11.80 × $10^{-9}$ s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 × $10^{-9}$

time is 23.6 × $10^{-9}$ s

time will elapse before it return to its staring point is 23.6 ns

7 0

3 years ago

Please help I’ll give brainliest

lara [203]

I think the answer is B

6 0

3 years ago

14. If the spring constant of a simple harmonic oscillator is doubled, by what factor will the mass of the system need to change

klio [65]

Lets se

And

$\\ \rm\Rrightarrow T=2\pi\sqrt{\dfrac{m}{k}}$

$\\ \rm\Rrightarrow \sqrt{k}T=2\pi\sqrt{m}$

$\\ \rm\Rrightarrow k\propto m$

If spring constant is doubled mass must be doubled

8 0

2 years ago

Over a time interval of 1.99 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.7 km/

madam [21]

Answer:

(a) - 42700 m/s

(b) - 6.8 x 10^-4 m/s^2

Explanation:

initial velocity of star, u = 20.7 km/s

Final velocity of star, v = - 22 km/s

time, t = 1.99 years

Convert velocities into m/s and time into second

So, u = 20700 m / s

v = - 22000 m/s

t = 1.99 x 365.25 x 24 x 3600 = 62799624 second

(a) Change in planet's velocity = final velocity - initial velocity

= - 22000 - 20700 = - 42700 m/s

(b) Accelerate is defined as the rate of change of velocity.

Acceleration = change in velocity / time

= ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2

8 0

3 years ago