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Ivan
3 years ago
10

Jenny and Betty are having a great time at Busch Gardens riding the Ubanga Banga bumper cars. Jenny, who is traveling southward

in her bumper car, aims her car toward Betty, who is traveling northward in her bumper car. The cars collide and briefly come to a stop. What can you say about the magnitude of the force that Jenny's car exerts on Betty's car versus the magnitude of the force that Betty's car exerts on Jenny's car?
Physics
1 answer:
lutik1710 [3]3 years ago
7 0
They both exert an equal amount of force onto each other

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A girl is standing 150m in front of a tall building, fires a shot with starting pistol. A boy standing 350m behind her, hears tw
Natasha2012 [34]

Answer:

300 m/s

Explanation:

The difference in time between the two bangs is 1 s.

Thus;

t2 - t1 = 1

We know that distance/time = speed.

Thus;

d2/v - d1/v = 1

Multiply through by v to get;

d2 - d1 = v

Where v is speed of sound in air.

d1 = 350 m

d2 = (150 × 2) + 350 = 650 m

Thus;

v = d2 - d1 = 650 - 350 = 300 m/s

8 0
3 years ago
3. A simple way to state Newton’s first law is:
MrRissso [65]

Explanation:

A simple way to state Newton's first law is:

For every action force, there is a reaction force which is equal in magnitude and opposite in direction.

4 0
3 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
Which sediments typically make up the C horizon?
Nastasia [14]
B) sand, silt, and clay!
5 0
3 years ago
Read 2 more answers
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Kruka [31]

Answer:

The student can conclude that as the bulb is not lighting up so the object can be insulator

Explanation:

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7 0
2 years ago
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