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3 years ago

Captain Kirk (80.0 kg) beams down to a planet that is the same size as Uranus and finds that he weighs

Physics

1 answer:

Archy [21]3 years ago

8 0

The mass of the planet is $1.51\cdot 10^{26}kg$

Explanation:

The weight of Captain Kirk on the surface of the planet is equal to the gravitational force between him and the planet, which is:

$F=G\frac{Mm}{R^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

M is the mass of the planet

m is the mass of Captain Kirk

R is the radius of the planet

In this problem, we have:

m = 80.0 kg is the mass of Kirk

$R=25,362 km = 2.54\cdot 10^7 m$ is the radius of the planet (same as Uranus)

F = 1250 N is the magnitude of the gravitational force between Kirk and the planet

Solving for M, we find the mass of the planet:

$M=\frac{FR^2}{Gm}=\frac{(1250)(2.54\cdot 10^7)^2}{(6.67\cdot 10^{-11})(80.0)}=1.51\cdot 10^{26}kg$

Learn more about gravitational force:

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brainly.com/question/12785992

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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

Strike441 [17]

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

6 0

3 years ago

A current of 5 A passes through a variable resistor set to 15 Ω. Calculate the voltage

OLEGan [10]

Answer:

75 volt

Explanation:

Current (I) = 5 A

Resistance (R) = 15 Ω

Voltage (V) = ?

We know

R = V/I

15 = v / 5

v = 75 Volt

3 0

3 years ago

Two parallel wires are separated by 6.10 cm, each carrying 2.85 A of current in the same direction. (a) What is the magnitude of

Westkost [7]

Answer:

The force per unit length is $2.66 \times 10^{-5} \ N/m$

Explanation:

The current carrying by each wires = 2.85 A

The current in both wires flows in same direction.

The gap between the wires = 6.10 cm

Now we will use the below expression for the force per unit length. Moreover, before using the below formula we have to change the unit centimetre into meter. So, we just divide the centimetre with 100.

$F/l = \frac{\mu _0i_1 i_2}{2\pi d} \\i_1 = 2.85 \\i_ 2 = 2.85 \\\mu _0 = 4\pi \times 10^{-7} \\d = 0.061 \\F/l = \frac{4\pi \times 10^{-7} \times 2.85 \times 2.85}{2 \pi \times 0.061} \\= 2.66 \times 10^{-5} \ N/m$