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3 years ago

9

Covalent bonds form when electrons are

2 answers:

NeX [460]3 years ago

8 0

It would be A) shared

WINSTONCH [101]3 years ago

4 0

the answer is A) shared

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On a distance vs. time graph, how do you know when the object is moving away from its starting position?

Ainat [17]

U know by if they are in first place

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2 years ago

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Two trains travel toward each other on the same track, beginning 100 miles apart. One train travels at 40 miles per hour; the ot

Paladinen [302]

D = distance between th two trains at the start of the motion = 100 miles

V = speed of the faster train towards slower train = 60 mph

v = speed of the slower train towards faster train = 40 mph

t = time taken by the two trains to collide = ?

time taken by the two trains to collide is given as

t = D/(V + v)

t = 100/(60 + 40) = 1 h

v' = speed of the bird = 90 mph

d = distance traveled by the bird

distance traveled by the bird is given as

d = v' t

d = 90 x 1

d = 90 miles

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3 years ago

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In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

3 years ago

ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the

antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = $M_A=5 kg$

Velocity of the ball A before collision: $U_A=20 m/s$

Velocity of ball A after collision= $V_A=10 m/s$

Mass of ball B= $M_B$

Velocity of the ball B before collision: $U_B=10 m/s$

Velocity of ball B after collision= $V_B=15 m/s$

$M_AV_A+M_BV_B=M_AU_A+M_BU_B$

$5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s$

$M_B=10kg$

The mass of ball B is 10 kg.

8 0

2 years ago

Two automobiles traveling at right angles to each other collide and stick together. Car A has a mass of 1200 kg and had a speed

sergij07 [2.7K]

Answer:

$v_{B0}=15.73 m/s$

Explanation:

We can use the conservation of momentum. The initial momentum is equal to the final momentum:

x-coordinate

$p_{0x}=p_{fx}$

$m_{A}v_{A0}=(m_{A}+m_{B})v_{cx}$

$m_{A}v_{A0}=(m_{A}+m_{B})v_{c}cos(40)$ (1)

y-coordinate

$p_{0y}=p_{fy}$

$m_{B}v_{B0}=(m_{A}+m_{B})v_{cy}$

$m_{B}v_{B0}=(m_{A}+m_{B})v_{c}sin(40)$ (2)

We can divide equations (2) and (1):

$\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=\frac{sin(40)}{cos(40)}$

$\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=tan(40)$

$v_{B0}=\frac{m_{A}v_{A0}}{m_{B}}*tan(40)$

$v_{B0}=\frac{1200*25}{1600}*tan(40)$

$v_{B0}=15.73 m/s$

I hope it helps you!

4 0

3 years ago

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