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nadezda [96]
3 years ago
11

A system consists of electrons and protons only. It contains 150 electrons and has a total charge of +22e. What is the mass of t

he system? me=9.11 x 10^â31kg, mp=1.673 x 10^â27kg.
Physics
1 answer:
liubo4ka [24]3 years ago
5 0

According to the statements the number of electrons is 150, then

e = 150

But there is a positive charge of +22e, then the number of protons would be

p = 150+172

If the mass of the electrons is

m_e = 9.11*10^{-31} kg

And the mass of the protong is

m_p = 1.673*10^{-27}kg

We have that the total mass of the system would be

m = e*m_e +pm_p

m = 150 * (9.11*10^{-31})+170(1.673*10^{-27})

m = 2.84547*10^{-25} kg

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A highway patrol car traveling a constant speed of 105 km/h is passed by a speeding car traveling 140 km/h. Exactly 1.00 s after
vodka [1.7K]

Answer:

The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.

Explanation:

Hi there!

The position of the patrol car at a time "t" can be calculated using this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the patrol car at a time "t"

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:

x = x0 + v · t

Where "v" is the constant velocity.

First, let´s convert the velocity units into m/s:

140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s

105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s

We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.

When the patrol car catches the speeder, the position of both cars is the same:

position of the patrol car = position of the speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:

x = v · t

x = 38.9 m/s · 1 s = 38.9 m

When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t

Let´s agrupate terms and equalize to zero:

-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0

-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0

Solving the quadratic equation for t using the quadratic formula:

t = 8.24 s  (the other solution is discarded because it is negative)

The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.

3 0
3 years ago
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

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distance traveled by the bird is given as

d = v' t

d = 90 x 1

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7 0
3 years ago
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A car dropped from a height of 44 meters fall to a height of zero meters. How fast will the car be traveling as it hits the grou
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Vi=0m/s
Vf=?
A=9.81
D=44
T=not needed

Vf^2=Vi^2+2ad
Vf=2ad square rooted
Vf=2(9.81)(44) square root it
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gogolik [260]

Answer:

W = 100000 J = 100 KJ

Explanation:

Here we will use the most basic and general formula of work, which is as follows:

W = Fd

where,

W = Work Done = ?

F = Force Required = 200 N

d = Length of Track = 500 m

Therefore,

W = (200\ N)(500\ m)\\

<u>W = 100000 J = 100 KJ</u>

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If 2 objects had the same momentum, what must be true about the mass of the object that traveled the fastest?
julsineya [31]

Yes, the above-given statement is true

<u>Explanation:</u>

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  • <u>Momentum (p) = Mass (M) * Velocity (v)</u>
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