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3 years ago

If your runnin at full speed for an average human, and lept over a 10 foot brook would you make it?

2 answers:

7 0

Unless you are a mutant....I don't think you would make it. I'm 5'8" and most people are 6'7" or less. It's bsically impossible to do. :)

LiRa [457]3 years ago

6 0

This seems a bit too trivial for this site but I'll bite anyways. An average speed for a human would be around 8 mph, so running would probably be around 12 mph. The longest long jump ever is 26 feet, but the average is 7' 3" to 7' 6.5", which is quite far away from the 10 ft brook. Therefore nope!

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Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.

balu736 [363]

Answer:

1. False

2. True

3. False

4. True

Explanation:

Conservation of Momentum

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

$p_a=m_a.v_a$

The total momentum of both pucks at the initial condition is

$p_1=m_a.v_a+m_b.v_b$

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

$m_a=m$

$m_b=2m$

We are given

$v_a=6\ m/s\\v_b=2\ m/s$

The total initial momentum is

$p_1=6m+2(2m)=10m$

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

$p_2=3m.v'$

Equating the initial and final momentum

$10m=3m.v'$

Solving for v'

$v'=10/3\ m/s=3.33\ m/s$

1. Compute the initial kinetic energy:

$\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2$

$\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2$

$K_1=18m+4m=22m$

The final kinetic energy is

$\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2$

$\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2$

$K_2=16.63m$

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

$p_1=10m$

$p_2=3m.v'=3m(10/3)=10m$

Proven

5 0

3 years ago

One type of slingshot can be made from a length of rope and a leather pocket for holding the stone. The stone can be thrown by w

grigory [225]

Answer:

15.66 rad/s

Explanation:

The vertical motion and horizontal motion are independent of each other.

t = √ ( 2 s/ g) where t = time for the ball to reach the ground and s is the height of the cliff = 18.0 m

t = √ ( 36 / 9.81 ) = 1.916 secs

horizontal distance travel = ut where u is the horizontal velocity of the stone = 30 × r (radius)

tangential velocity V = angular velocity ( ω) × radius

distance traveled = ω × r × t = 30 × r

radius cancelled on both side

ω = 30 / 1.9156 = 15.66 rad/s

4 0

3 years ago

N capacitors are connected in parallel to form a "capacitor circuit". The capacitance of first capacitor is C, second one is C/2

liberstina [14]

Answer:

Explanation:

The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.

So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.

Using the formula for the sum of the infinite terms of a geometric series, we have:

Sum = First term / (1 - rate)

Sum = C / (1 - 0.5)

Sum = C / 0.5 = 2C

So the equivalent capacitance of this parallel connection is 2C.

5 0

3 years ago

Formula for percentage error

GarryVolchara [31]

Answer:

PE = (|accepted value – experimental value| \ accepted value) x 100%

Explanation:

7 0

2 years ago

A Carnot engine's operating temperatures are 240 ∘C and 20 ∘C. The engine's power output is 910 W . Part A Calculate the rate of

scoray [572]

Answer:1200

Explanation:

Given data

Upper Temprature $\left ( T_H\right )=240^{\circ}\approx 513$

Lower Temprature $\left ( T_L\right )=20^{\circ}\approx 293$

Engine power ouput $\left ( W\right )=910 W$

Efficiency of carnot cycle is given by

$\eta =1-\frac{T_L}{T_H}$

$\eta =\frac{W_s}{Q_s}$

$1-\frac{293}{513}=\frac{910}{Q_s}$

$Q_s=2121.954 W$

$Q_r=1211.954 W$

rounding off to two significant figures

$Q_r=1200 W$

5 0

3 years ago