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DENIUS [597]
3 years ago
9

Andy took a bus and then walked from his home to downtown. For the first 1.6 hour, the bus drove at an average speed of 15 km/h

Physics
2 answers:
qwelly [4]3 years ago
6 0

Answer:

1.6x15=24

=0.4x4.5=1.8

=24+1.8=25.8

=1.6+0.4=2

=25.8/2=12.9

Yuliya22 [10]3 years ago
4 0

<em>1.6x15=24 </em>

<em> </em>

<em>0.4x4.5=1.8 </em>

<em> </em>

<em>24+1.8=25.8 </em>

<em> </em>

<em>1.6+0.4=2 </em>

<em> </em>

<em>25.8/2=12.9</em>

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A proton is confined within an atomic nucleus of diameter 3.60 fm. part a estimate the smallest range of speeds you might find f
Cerrena [4.2K]
The answer for this problem would be:
Assuming non-relativistic momentum, then you have: 
ΔxΔp = mΔxΔv = h / (4) 
Δv = h / (4πmΔx) 
m ~ 1.67e-27 h ~ 6.62e-34,Δx = 4e-15 --> 
Δv ~ 6.62e-34 / (4π * 1.67e-27 * 4e-15) ~ 7,886,270 m/s ~ 7.89e6 m/s 
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3 0
3 years ago
A car drives at a constant speed of 21 m/s around a circle of radius 100 m.
Arte-miy333 [17]

Answer:

Option D. 4.4 m/s²

Explanation:

The following data were obtained from the question:

Velocity (v) = 21 m/s

Radius (r) = 100 m

Centripetal acceleration (a) =.?

The centripetal acceleration of the car can be obtained as follow:

Centripetal acceleration (a) = Velocity square (v²) / radius (r)

a = v²/r

a = 21²/100

a = 441/100

a = 4.41 ≈ 4.4 m/s²

Therefore, the centripetal acceleration of the car is 4.4 m/s².

8 0
3 years ago
Pete is driving down 7th street. He drives 150 meters in 18 seconds. Assuming he does not speed up or slow down, what is his spe
Lubov Fominskaja [6]
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3 0
4 years ago
We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm . The cylinder rotates with
amm1812

Answer: 14.16

Explanation:

Given

d = 38cm

r = d/2 = 38/2 = 19cm = 0.19m

K.E = 510J

m = 10kg

I = 1/2mr²

I = 1/2*10*0.19²

I = 0.18kgm²

When it has 510J of Kinetic Energy then,

510J = 1/2Iω²

ω² = 1020/I

ω² = 1020/0.18

ω² = 5666.67

ω = √5666.67 = 75.28 rad/s

Velocity is the block, v = ωr

V = 75.28 * 0.19

V = 14.30m/s

The "effective mass" M of the system is

M = (14.0 + ½*10.0) kg = 19.0 kg

The motive force would be

F = ma

F = 14 * 9.8

F = 137.2N

so that the acceleration would be

a = F/m

a = 137.2/19

a = 7.22m/s²

Finally, using equation of motion.

V² = u² + 2as

14.3² = 0 + 2*7.22*s

204.49 = 14.44s

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6 0
3 years ago
How many swings would a 23 cm long pendulum make in 30 seconds?
iragen [17]

Is there any possible chance that at some point in your science
studies, sometime before you were given this question for your
homework, that maybe you might have encountered this formula
for the period of a simple pendulum ?

                   Period = (2 pi) √(length/gravity) .

If the length is 0.23 meter, and the
acceleration of gravity is 9.8 m/s²,
then the period is

                                 = (2 pi) √(0.23/9.8)

                                 =   0.963... second  (rounded)

That's how long it takes for a simple pendulum, 23cm long,
hanging on a massless string and not swinging too far to
the side, to complete one full swing left and right.

Now, if you can figure out how many periods of  0.963 second
there are in  30 seconds, you'll have your answer.  I'll leave
that part of it to you.

8 0
3 years ago
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