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umka21 [38]
3 years ago
5

Below are birds-eye views of six identical toy cars moving to the right at 2 m/s. Various forces act on the cars with magnitudes

and directions indicated below. All forces act in the horizontal plane and are either parallel or at 45 or 90 degrees to the car's motion. Rank these cars on the basis of their speed a short time (ie. before any car's speed can reach zero) after the forces are applied. Rank from largest to smallest. To rank items as equivalent, overlap them.

Physics
1 answer:
Firdavs [7]3 years ago
3 0
All cars have the same speed before the forces are applied ( v = 2 m/s ).
After that rank is:
1. Largest: 6th car ;  Fr = 10 N,
2. 1st car ;  Fr = 2.95 N.
3. After that: 2nd, 3rd and 5th car ;  Fr = 0 N
4. Smallest : 4th car ;  Fr = - 2.04 N. 

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zaharov [31]

B is the answer to the question

8 0
3 years ago
How many signifficant figures do the following number have 258​
Romashka [77]

Answer:

<u>Facts about 258</u>

<u>Sig Figs</u>

3

<u>258</u>

<u>Decimals</u>

0

<u>Scientific Notation</u>

2.58 × 102

<u>E-Notation</u>

2.58e+2

<u>Words</u>

two hundred fifty-eight

Explanation:

258 Rounded to Fewer Sig Figs

2 260                2.6 × 102

1 300           3 × 102

5 0
2 years ago
A uranium and iron atom reside a distance R = 44.10 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize
Ad libitum [116K]

Answer:

distance r from the uranium atom is 18.27 nm

Explanation:

given data

uranium and iron atom distance R = 44.10 nm

uranium atom = singly ionized

iron atom = doubly ionized

to find out

distance r from the uranium atom

solution

we consider here that uranium electron at distance = r

and electron between uranium and iron so here

so we can say electron and iron  distance = ( 44.10 - r ) nm

and we know single ionized uranium charge q2= 1.602 × 10^{-19} C

and charge on iron will be q3 = 2 × 1.602 × 10^{-19} C

so charge on electron is q1 =  - 1.602 × 10^{-19} C

and we know F = k\frac{q*q}{r^{2} }  

so now by equilibrium

Fu = Fi

k\frac{q*q}{r^{2} }  =  k\frac{q*q}{r^{2} }

put here k = 9*10^{9} and find r

9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{r^{2} }  =  9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{(44.10-r)^{2} }

\frac{1}{r^{2} } = \frac{2}{(44.10 -r)^2}

r = 18.27 nm

distance r from the uranium atom is 18.27 nm

6 0
3 years ago
A 1200 kg aircraft going 30 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and
Leno4ka [110]

Answer:

83.055  m

Explanation:

According to the given scenario, the calculation of skid distance is shown below:-

S = \frac{1}{2} \times (u + v) \times t

Where  

u = 11.3

v = 0

t = 14.7

Now placing these values to the above formula,

So,

S = \frac{1}{2} \times (11.3 + 0) \times 14.7

= 83.055  m

Therefore for computing the skid distance we simply applied the above formula i.e by considering the all items given in the question

4 0
3 years ago
A diamonds mass is 5.28g and its volume is 2cm cubed whatsthe density
podryga [215]
Density = mass/volume, so density=5.28g/2cm
5 0
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