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MAXImum [283]
3 years ago
11

G A dragster starts from rest and accelerates at 35 m/s2 m / s 2 . How fast is it going after t t

Physics
1 answer:
bija089 [108]3 years ago
4 0

Complete question:

A dragster starts from rest and accelerates at 35 m/s2 m / s 2 . How fast is it going after 7s.

Answer:

The final velocity of the dragster is 245 m/s.

Explanation:

Given;

acceleration of the dragster, a = 35 m/s²

initial velocity of the dragster, u = 0

time of motion, t = 7 s

the final velocity of the dragster after 7s is given by;

v = u + at

where;

v is the final velocity

u is the initial velocity

v = 0 + (35 x 7)

v = 245 m/s

Therefore, the final velocity of the dragster is 245 m/s.

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2 equal charges, 27 micro Coulomb each, are separated by 5 cm. Find force between those.
padilas [110]

Answer:

The force between charges is  F= 2.624*10^3N.

Explanation:

The Coulomb force F between the two charges q_1 and q_2 separated by distance d is given by the equation

F = k\dfrac{q_1q_2}{d^2}

where k is the coulombs constant, and has the value

k= 9*10^9N\cdot C\cdot m^2.

Now in our case

q_1=q_2=27\mu C =27*10^{-6}C

and

d= 5cm =0.05m,

therefore, the Coulomb force between the charges is

F =( 9*10^9N\cdot C\cdot m^2)*\dfrac{(27*10^{-6}C)(27*10^{-6}C)}{(0.05m)^2}

\boxed{ F= 2.624*10^3N}

3 0
4 years ago
this is how i made the map , i’m not sure where to put the symbols but can someone help me now answer the last few questions i p
N76 [4]

Answer:

hi my name is bill im here to help no problem 1+1=2

8 0
3 years ago
Two long parallel wires are placed side-by-side on a horizontal table. if the wires carry current in opposite directions,
STatiana [176]
\vec F = I (\vec L \times  \vec B)

if currents go in opposite directions, wires repel
3 0
3 years ago
Halley's comet orbits the sun roughly once every 76 years. It comes very close to the surface of the Sun on its closest approach
Licemer1 [7]

Answer:

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

3 0
3 years ago
If a star with an absolute magnitude of -5 has an apparent magnitude of +5 ,then its distance is
klio [65]
You asked a question.  I'm about to answer it. 
Sadly, I can almost guarantee that you won't understand the solution. 
This realization grieves me, but there is little I can do to change it. 
My explanation will be the best of which I'm capable.


Here are the Physics facts I'll use in the solution:

-- "Apparent magnitude" means how bright the star appears to us.

-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).

-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by  ⁵√100 .
That's about  2.512... .  

-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .

-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).

That's all the Physics.  The rest of the solution is just arithmetic.
____________________________________________________

-- The star in the question would appear M(-5) at a distance of
32.6 light years. 

-- It actually appears as a M(+5).  That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.

-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .

-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is

(32.6) · (100)^(1) light years

= (32.6) · (100) light years

=  approx.  3,260 light years .   (roughly 1,000 parsecs)


I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
6 0
3 years ago
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