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3 months ago

A bus slows down uniformly from 75.0 km/h to 0 km/h in 21 s. How far does it travel before stopping?

2 answers:

7 0

1 hour = 3,600 seconds
1 km = 1,000 meters

75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s

The mean speed during the deceleration is

(1/2)(20-5/6 + 0) = 10-5/12 m/s.

Traveling at this average speed for 21 seconds,
the bus covers

(10-5/12) × (21) = 218.75 meters.

kicyunya [3.2K]3 months ago

4 0

According to the problem, the bus decelerates uniformly from 75 km/h to 0 km/h.

Thus the initial speed u = 75 km/h and the final speed v = 0 km/h.

The time to stop is t = 21 s.

Convert units: 1 km = 1000 m and 1 hour = 3600 s.

Therefore u = $75 km/h =75*\frac{1000 m}{3600s}$ = 20.83 m/s.

We need the stopping distance.

From kinematics, v = u + a t, where a is the acceleration.

With v = 0, u = 20.83 m/s, and t = 21 s,

0 = 20.83 + a × 21

so a × 21 = -20.83

hence a = -[20.83] ÷ 21 = $= -0.9919 m/s^2$ (the negative sign shows deceleration)

Using the displacement formula $s = ut +\frac{1}{2} at^2$ and substituting the known values gives

$s = 20.83 *21 -\frac{1}{2} 0.9919 *[21^2]$

which results in s = 218.7 metre. [ans]

Therefore the bus stops after 218.7 m.

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Yuliya22 [3333]

Answer:

Explanation:

a )

Each blade resembles a rod with its axis positioned near one end.

The moment of inertia for one blade is:

= 1/3 x m l²

where m stands for the mass of the blade

l represents the length of each blade.

Total moment of inertia for 3 blades is:

= 3 x $\frac{1}{3}$ x m l²

ml²

2 )

Details provided include:

m = 5500 kg

l = 45 m

Substituting these values produces:

moment of inertia of one blade:

= 1/3 x 5500 x 45 x 45

= 37.125 x 10⁵ kg.m²

Moment of inertia for 3 blades:

= 3 x 37.125 x 10⁵ kg.m²

= 111.375 x 10⁵ kg.m²

c )

Angular momentum

= I x ω

I denotes the moment of inertia of the turbine

ω symbolizes angular velocity

ω = 2π f

f indicates the rotational frequency of the blades

d )

We have I = 111.375 x 10⁵ kg.m² (Calculated)

f = 11 rpm (revolutions per minute)

= 11 / 60 revolutions per second

ω = 2π f

= 2π x 11 / 60 rad / s

Calculating angular momentum yields

= I x ω

111.375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s

= 128.23 x 10⁵ kgm² s⁻¹.

4 0

3 months ago

The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the

kicyunya [3294]

Answer:

Explanation:

An image of the bond resulting from the search is attached.

Consider the force directed towards thymine as negative.

For the O-H-N combination:

The resulting force from this combination is:

$F=-F_{OH}+F_{ON}\\\\=\frac{Ke^2}{r^2}+\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[(0.280-0.110)\times 10^{-9}m]^2}+\frac{1}{0.280\times 10^{-9}m)^2}]\\\\=-5.03354\times 10^{-9}N$

In the case of the N-H-N combination:

The total force acting from this combination is:

$F'=F_{NN}-F_{HN}\\\\=\frac{Ke^2}{r^2}-\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[0.300\times 10^{-9}m]^2}-\frac{1}{((0300-0..110)\times 10^{-9})m)^2}]\\\\=-3.822\times 10^{-9}N$

The force that thymine applies on adenine is:

$F_{net}=F+F'\\\\=-5.03354\times 10^{-9}N-3.822\times 10^{-9}N\\\\=-8.8558\times 10^{-9}N$

When rounded to three significant figures, the net force is $8.86\times 10^{-9}N$

The negative value indicates that the force is attractive, as it is aimed towards thymi.

The force acting on the electron due to the proton is:

$F=\frac{Ke^2}{r^2}\\\\=\frac{(9.0\times 10^9Nm^2/C^2)(1.6\times 10^{-9}C)^2}{(5.29\times 10^{-11}m)^2}\\\\=8.233\times 10^{-8}N$

Since the electron and proton carry opposite charges, the force on the electron points towards the proton.

The ratio of the above forces is:

$\frac{F}{F_{net}}=\frac{8.233\times 10^{-8}N}{8.233\times 10^{-8}N}\\\\=9.3$

Therefore, the bonding strength of the electron in the hydrogen atom is 9.3 times greater than the bonding force between adenine and thymine molecules.

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3 months ago