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3 years ago

14

Where is earth's magnetic north pole located?

2 answers:

scoundrel [369]3 years ago

7 0

Answer:ellesmere island, Canada

andre [41]3 years ago

4 0

Answer: it is located over Ellesmere Island, Canada

Explanation

it is now drifting away from North America and toward Siberia.

You might be interested in

What is the diffference between distance and displacement

Jobisdone [24]

Answer:

distance is how far away something has travelled from another object, while displacement is how far something is from the other object. Displacement is a vector quantity, unlike distance.

5 0

3 years ago

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he

lana [24]

Answer:8.28 km

Explanation:

Given

First it drifts $45^{\circ}$ 2.5 km

$r_1=2.5cos45 i+2.5sin45 j$

Secondly it drifts $60^{\circ}$ 4.70 km

$r_{12}=4.7cos60 i-4.7sin60 j$

After that it drifted along east direction 5.1 km

$r_{23}=5.1 i$

After that it drifts $55^{\circ}$ 7.2 km

$r_{34}=-7.2cos55 i-7.2sin55 j$

After that it drifts $5^{\circ}$ 2.8 km

$r_{54}=-2.8cos5 i+2.8sin5 j$

$r_{5O}$ = $\left [ 2.5cos45+4.7cos60+5.1-7.2cos55-2.8cos5\right ]\hat{i}$ + $\left [ 2.5sin45-4.7sin60-7.2sin55+2.8sin5\right ]\hat{j}$

$r_{5O}=2.299\hat{i}-7.95\hat{j}$

$|r_{5O}|=8.28 km$

for direction

$tan\theta =\frac{7.95}{2.299}=3.4580$

$\theta =73.87^{\circ}$ south of east

7 0

3 years ago

Read 2 more answers

If 1.8 1016 electrons enter a light bulb in 3 milliseconds, what is the magnitude of the electron current at that point in the c

Mice21 [21]

Answer:

I = 0.96 A

Explanation:

No of electrons, $n=1.8\times 10^{16}$

Time, t = 3 ms = $3\times 10^{-3}\ s$

We need to find the electric current. We know that electric charge per unit time is equal to the electric current.

$I=\dfrac{q}{t}$

q = ne (Quantization of electric charge)

$I=\dfrac{ne}{t}\\\\I=\dfrac{1.8\times 10^{16}\times 1.6\times 10^{-19}}{3\times 10^{-3}}\\\\I=0.96\ A$

So, the electric current is 0.96 A.

7 0

3 years ago

A 0.0427 kg racquet-ball is moving

Gwar [14]

Answer:

Mass of the box = 0.9433 kg

Explanation:

Mass of racket-ball $(m_1)$ = 0.00427 kg

Velocity of racket-ball before collision $(v_{1i})$ = 22.3 m/s

Velocity of racket-ball after collision with box $(v_{1f})$ = -11.5 m/s

[Since ball is bouncing back, so velocity is taken negative.]

Velocity of the box before collision $v_{2i}$ = 0 m/s

<em>[Since the box is stationary, so velocity is taken zero]</em>

Velocity of box moving forward after collision $v_{2f}$ = 1.53 m/s

To find the mas of the box $m_2$ .

By law of conservation of momentum we have:

Momentum before collision = Momentum after collision

This can be written as:

$p_i=p_f$

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

We can plugin the given value to find $m_2$

$(0.0427\times 22.3)+(m_2\times 0)=(0.0427\times (-11.5))(m_2\times 1.53)$

$0.9522+0=-0.4911+1.53m_2$

Adding both sides by 0.4911

$0.9522+0.4911=-0.4911+0.4911+1.53m_2$

$1.4433=1.53m_2$

Dividing both sides by 1.53.

$\frac{1.4433}{1.53}=\frac{1.53m_2}{1.53}$

$0.9433=m_2$

∴ $m_2=0.9433$ kg

Mass of the box = 0.9433 kg (Answer)

4 0

3 years ago

A force of 45 N is applied tangentially to the rim of a solid disk of radius 0.12 m. The disk rotates about an axis through its

bearhunter [10]

Answer:

Mass of the disk will be 2.976 kg

Explanation:

We have given force F = 45 N

Radius of the disk r = 0.12 m

Angular acceleration $\alpha =140rad/sec^2$

We know that torque $\tau =I\alpha$

And $\tau =Fr$

So $Fr=I\alpha$ , here I is moment of inertia

So $50\times 0.12=I\times 140$

$I=0.0428kgm^2$

We know that moment of inertia $I=\frac{1}{2}mr^2$

So $0.0428=\frac{1}{2}\times m\times 0.12^2$

m = 2.976 kg

6 0

3 years ago