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ale4655 [162]
3 years ago
7

A 50 kg rock climber accidentally falls from the side of a rock and freefalls until she is stopped by her 9 m saftey rope. Assum

ing that the rope stops her completely with no rebound, what is the impulse impared on her body by the rope? If the rock climber's rope took 0.224s to stop her completely after she fell, what was the average force felt by the rock climber as she stopped?
Physics
1 answer:
olchik [2.2K]3 years ago
8 0

Answer:

a) 665 kg.m/s

b) 2.97 kN

Explanation:

The impulse is given by:

J=\Delta p\\J=m(v_a-v_b)

where

va=velocity after

vb=velocity before

The velocity just before is given by:

v_b=\sqrt{2*9.81m/s^2*9m}=13.3m/s(-\hat{j})

The velocity just after is zero because there wasn't a rebound.

So the impulse is:

J=50kg*(0-13.3m/s(-\hat{j}))\\J=50kg*(0+13.3m/s(\hat{j}))=665kg.m/s

The impulse is also given by:

J=\Delta p=F* \Delta t\\so:\\F=\frac{J}{\Delta t}\\\\F=\frac{665kg.m/s}{0.224s}=2.97kN

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Force F = (-6.0 N)ihat + (2.0 N) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.
finlep [7]

Answer:

\tau=(-32k)\ N-m

\theta=116.55^{\circ}

Explanation:

Given that.

Force acting on the particle, F=(-6i + 2.0j)\ N

Position of the particle, r=(4i+4j)\ m

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

\tau=F\times r

\tau=(-6i + 2.0j)\times (4i+4j)

\tau=\begin{pmatrix}0&0&-32\end{pmatrix}

\tau=(-32k)\ N-m

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r, |r|=\sqrt{4^2+4^2}=5.65\ m

Magnitude of F, |F|=\sqrt{(-6)^2+2^2}=6.324\ m

Using dot product formula,

F{\circ}\ r=|F|.|r|\ cos\theta

cos\theta=\dfrac{F{\circ} r}{|F|.|r|}

cos\theta=\dfrac{-24+8}{6.324\times 5.65}

\theta=116.55^{\circ}

Therefore, this is the required solution.

6 0
3 years ago
Read 2 more answers
PLEASE HELP! How much heat is absorbed by 57g iron skillet when its temperature rises from 11°c to 30°C?
Amiraneli [1.4K]

mass times specific heat times tem change.

57 x sh x (30-11)

convert to mks and look up sh iron

7 0
3 years ago
Someone please hellpp!!! Will mark brainliest
____ [38]

<em>Answer:</em>

<em>r=x+y</em>

<em>sorry if its not correct you can delete if you want.</em>

6 0
3 years ago
Human-Powered Flight Human-powered aircraft require a pilot to pedal, as in a bicycle, and produce a sustained power output of a
alukav5142 [94]

Answer:

# of Snickers bars 2

Explanation:

Power output= 0.30 HP

=0.3*746

= 0.30 HP (746 W=1.00 HP)

= 224 W

time required 2 h 49 m = 10140 seconds

Since power is work divided by time, then work is:

Work done by the jet = P*t

= 224 *(10140)

= 2.3 MJ (2.3 x 10^{6} J)

Converting MJ to Cal

2.3 MJ=549 Cal

# of Snickers bars = 549 Cal / 280 Cal

= 2.0 bars (rounded from 1.96)

8 0
3 years ago
A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu
Aloiza [94]
In other words a infinitesimal segment dV caries the charge 
<span>dQ = ρ dV </span>

<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>

<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>

<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
6 0
3 years ago
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