All categories

3 years ago

How should you draw the field lines for Earth's magnetic field?

Physics

2 answers:

Juliette [100K]3 years ago

7 0

Answer:

Coming out near the South Pole and going in near the North Pole

Explanation:

yuradex [85]3 years ago

3 0

A. Coming out near the South Pole and going in near the North Pole

You might be interested in

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

It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft

Paul [167]

Answer: $1.23\ m/s^2$

Explanation:

Given

At an elevation of $y=34.7\ km$ , spacecraft is dropping vertically at a speed of $u=293\ m/s$

Final velocity of the spacecraft is $v=0$

using equation of motion i.e. $v^2-u^2=2as$

Insert the values

$\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2$

Therefore, magnitude of acceleration is $1.23\ m/s^2$ .

8 0

3 years ago

When a liquid is cooled, the kinetic energy of the particles ?

vodka [1.7K]

When a liquid is cooled, the kinetic energy of the particles, DECREASES..

4 0

3 years ago

A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force

VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0

3 years ago

g A 4-foot spring is elongated167feet long after a mass weighing 16 pounds is attached to it. The medium throughwhich the mass m

marta [7]

Answer: hello question b is incomplete attached below is the missing question

a) attached below

b) V = 0.336 ft/s

Explanation:

Elongation ( Xo) = 16/ 7 feet

mass attached to 4-foot spring = 16 pounds

medium has 9/2 times instanteous velocity

<u>a) Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s</u>

The motion is an underdamped motion because the value of β < Wo

Wo = 3.741 s^-1

attached below is a detailed solution of the question

3 0

3 years ago