Physics, 30.10.2019 02:31 1tzM3

Aman with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. at the bottom of the hill the terrain levels out. as the man reaches the horizontal section, he grabs a 20.0-kg backpack and skis off a 2.00-m-high ledge. at what horizontal distance from the edge of the ledge does the man land (the man starts at rest)?

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Answer from: nataliahenderso

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

$0 + mgH = \frac{1}{2}mv^{2} + 0$

$2gH = v^{2}$

$v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s$

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

$65.0\times 9.89 = (65.0 + 20.0)v'$

$v' = 7.56\ m/s$

Now, time taken for the fall:

$h = \frac{1}{2}gt^{2}$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s$

Now, the horizontal distance is given by:

x = v't = $7.56\times 0.638 = 4.823\ m$

Answer from: Dhejxbdbbxbdbbx

Answer

given,

mass of the man = 65 kg

height = 5 m

mass of the back pack = 20 kg

skis off to 2.00 m high ledge

horizontal distance =

speed of the person before they grab back pack is equal to potential and kinetic energy

$mgh= \dfrac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8 \times 5}$

v = 9.89 m/s

now he perform elastic collision

$v = \dfrac{m_1v_1}{m_1+m_2}$

$v = \dfrac{65\times 9.89}{65+20}$

v = 7.57 m/s

time taken by the skies to fall is

$h = \dfrac{1}{2}gt^2$

$t = \sqrt{\dfrac{2h}{g}}$

$t = \sqrt{\dfrac{2\times 2}{9.8}}$

t = 0.6388 s

distance

d = v x t

d = 7.57 x 0.6388

d = 4.84 m

Answer from: Quest

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm. the wire diameter is d = 0.9 mm. the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2) in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

Answer from: Quest

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation:

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Aman with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. at the bottom of the...