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Ilia_Sergeevich [38]
3 years ago
10

Lillian (mass 40.0 kg) standing at rest on slippery ice catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Wha

t is the speed of Lillian and her dog after the catch?
Physics
1 answer:
mestny [16]3 years ago
3 0

Answer:

0.82 m/s

Explanation:

From the law of conservation of momentum,

Total momentum = Final momentum.

mu+m'u' = V(m+m')......................... equation 1

Where m = mass of Lillian, u = initial velocity of Lillian before she catches the dog, m' = mass of the dog, u' = initial velocity of the dog, V = velocity of Lillian and the dog.

make V the subject of the equation,

V = (mu+m'u')/(m+m')................ Equation 2

Given: m = 40 kg, m' = 15 kg, u = 0 m/s (Lillian was standing at rest), u' = 3.0 m/s.

Substitute into equation 2

V = (40×0+15×3)/(40+15)

V = (0+45)/55

V = 45/55

V = 0.82 m/s

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A ski lift has a one-way length of 1 km and a vertical rise of200m. The chairs are spaced 20 m apart, and each chair can seatthr
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Answer:

a) 68.125 KW

b) 43.04 KW

Explanation:

Distance =d= 1 km

Height = h = 200 m

Spacing between chairs =D= 20 m

No. of people per chair = 3

Speed = V= 10 km/h= 10000m\3600 s=2.8 m/s

mass of chair = 250 kg

Work to operate sky lift

W= mgh

Number of chairs any moment operational = N= 1 km/20=1000m/20=50

So, total mass of chairs = 50 × 250 =12500kg

so, w= mgh=12500×9.8×200=2452500 j

Power is rate of work - we need time for operation time of lift

Time= t = distance/speedf= 1km/(10km/h)=1km/(10 km/3600s)=360s

So, Power P= Work/time=w/t=12500/360=68125 j/s=68.125 KW

Now calculating power for operating speed in 5 sec

We calculate accelleration=a for 5 sec

a= speed / time= V/52.8/5=0.28 m/sec2

for vertical acceleration we calculate θ angle first;

tanθ= height /distance= 200/1000= 0.2

==> θ=11.34°

Vertical acceleration = a₁=a sinθ= 0.28× sin 11.34=0.10835 m/sec2

to calculate height gained during startup use;

S=vit+1/2at2

here s=H

vi=0m/s

t=5 sec

==> H = 0+1/2a₁×t=0.5×0.10835 ×5²=0.5×0.10835×5×5=1.362 m

Total Work =mgH+0.5×m×V²=12500(9.8×1.362+0.5×2.8×2.8)=215240.56 j

Again power = work / time=215240.56/5=43048.112 W=43.04 KW

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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

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3 years ago
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