Help pls i give brainielest

WITCHER [35]

The constant in Newton's law of gravitation relating gravity to the masses and separation of particles, equal to 6.67 × 10-11N m2 kg-2.

6 0

3 years ago

A helicopter is ascending vertically. a passenger accidentally drops her wallet out the sides of the helicopter when it is 160 m

Advocard [28]

Answer:

(E)56.0 m/s

Explanation:

Height =h=-160 m

Because the wallet moving in downward direction

Time=t=7 s

Final speed of wallet=v=0

We have to find the speed of helicopter ascending at the moment when the passenger let go of the wallet.

$v^2-u^2=2gh$

Where $g=9.8 m/s^2$

Substitute the values

$0-u^2=2(-160)\times 9.8$

$u^2=3136$

$u=\sqrt{3136}=56m/s$

Option (E) is true

8 0

4 years ago

Beginning with earth, summarize the structure of the universe. Also Andromeda galaxy is located appromaxily 2.5 million light-ye

Leokris [45]

Because of the power of the light it is very strong so that is why light reaches all the way to earth

4 0

3 years ago

a golf ball is hit with a velocity of 30.0 m/s at an angle of 25 degrees about the horizontal. how long is the ball in the air a

Pie

Ok so use trigonometry to work out the vertical component of velocity.

sin(25) =opp/hyp
rearrange to:
30*sin(25) which equals 12.67ms^-1

now use SUVAT to get the time of flight from the vertical component,

V=U+at

Where V is velocity, U is the initial velocity, a is acceleration due to gravity or g. and t is the time.

rearranges to t= (V+u)/a

plug in some numbers and do some maths and we get 2.583s

this is the total air time of the golf ball.

now we can use Pythagoras to get the horizontal component of velocity.

30^2-12.67^2= 739.29
sqrt739.29 = 27.19ms^-1

and finally speed = distance/time

so--- 27.19ms^-1*2.583s= 70.24m

The ball makes it to the green, and the air time is 2.58s

4 0

3 years ago

A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th

saveliy_v [14]

Answer :

Velocity will be $3.28\times 10^{-11}m/sec$

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere = $=1.6+\frac{1.5}{2}=2.35mm = 0.00235m$

Force on electron is given by $F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}$ , here q is charge on sphere and e is charge on electron

$F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N$

This force work as centripetal force

So $F=\frac{mv^2}{r}$

$4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}$

v = $=0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec$

6 0

3 years ago