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tatyana61 [14]
3 years ago
6

A large group of people get together. each one rolls a die 180 times, and counts the number of 1's. about what percentage of the

se people should get counts in the range of 15-45?
Physics
1 answer:
Arte-miy333 [17]3 years ago
5 0
<span>Around 99% - 100%. This is because a die is six sided, which gives the odds of a one coming up being roughly 17% independent of every roll. 17% of 180 trials comes out to 30-31 times a one will show up every 180 trials. This puts you right in the middle of the 15-45 range which means that somebody will almost ALWAYS reach 15-45 one's in a trial of 180 rolls.</span>
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A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le
saw5 [17]

Answer:

Momentum of the bowling pin is 17kgm/s

Explanation:

P(initial of ball)=+30 kg m/s

P(initial of pin)=0  because its velocity is 0

P initial = P final

30= P(ball) + P(pin)

30=13+P(pin)

P(final of pin)=+13 kg m/s and P(final of ball)= 30-13=17kgm/s

8 0
3 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.8 km/h due east. Runner B is initia
vampirchik [111]

Answer:

0.39 km west

Explanation:

The position of Runner A is:

x = -5.7 + 8.8 t

The position of Runner B is:

x = 4.2 − 7.6 t

When the positions are equal:

-5.7 + 8.8 t = 4.2 − 7.6 t

16.4 t = 9.9

t = 0.604

Plug into either equation to find the position at this time:

x = -5.7 + 8.8 (0.604)

x = -0.39

The runners are 0.39 km west of the flagpole when they meet.

4 0
3 years ago
What does the measurement unit N stand for?
bonufazy [111]

Explanation:

The newton (symbol: N) is the International System of Units (SI) derived unit of force.

3 0
3 years ago
Read 2 more answers
Can rock undergo compression, tension, and shear stress all at once?<br> explain
Katena32 [7]
This causes reverse faults<span>, which are the reverse of </span>normal faults<span>, because in this case, the hanging wall slides upward relative to the footwall. Shear </span>stress<span> is when rock slabs slide past each other horizontally. There is no vertical movement of either the hanging wall or footwall, and we get a strike-slip </span>fault<span>.</span>
7 0
3 years ago
. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110
7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x 10^{5} J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, v_{0} = 110 km/h = 110 x 1000/3600 = 30.6 m/s\frac{30.6^{2} }{2x9.8}

initial height, h_{0} = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = \frac{1}{2} mv_{0} ^{2}

gh = \frac{1}{2} v_{0} ^{2}

h = \frac{v_{0} ^{2} }{2g}

  = 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

                         = mg (h - h_{0})

                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

 = 373 N

8 0
3 years ago
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