How do similar (S-S or N-N) magnetic poles interact?

A 3.0kg mass tied to a string

dem82 [27]

Answer:

$\boxed{\sf Tension \ in \ the \ string \ (T) = 3 \ kN}$

Given:

Mass (m) = 3.0 kg

Uniform speed (v) = 20 m/s

Length of string (r) = 40 cm = 0.4 m

To Find:

Tension in the string (T)

Explanation:

Tension (T) is the string will be equal to centripetal force ( $\sf F_c$ ).

$\boxed{ \bold{ T = F_c = \frac{m {v}^{2} }{r} }}$

Substituting value of m, v & r in the equation:

$\sf \implies T = \frac{3 \times {20}^{2} }{0.4} \\ \\ \sf \implies T = \frac{3 \times 400}{0.4} \\ \\ \sf \implies T =3 \times 1000 \\ \\ \sf \implies T =3000 \: N \\ \\ \sf \implies T =3 \: kN$

$\therefore$

Tension in the string (T) = 3 kN

5 0

3 years ago

If a car goes down Lake at 30 miles per hour how far will it go in 0.25 hours?

kati45 [8]

Answer:

7.5 miles

Explanation:

0.25 hours=15minutes.

So you do

30 miles=60minutes

x =15minutes

cross multiply

Ans-7.5

7 0

3 years ago

Explain how an intrusive igneous rock could become a metamorphic rock and then an extrusive igneous rock.

sleet_krkn [62]

1.igneous rocks are formed when magma or lava cools and hardens
2. the cooling rate of the rock

5 0

3 years ago

1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant

Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction $(v_j)$ = 550 mph

Velocity of jet stream from west to east direction $(v_s)=80\ mph$

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

$\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph$

Similarly, the velocity of the stream is, $\vec{v_s}=80\vec{i}$

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

$\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}$

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

$|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph$

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle $\theta$ with the x axis in the third quadrant.

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)$

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

5 0

3 years ago

We know for newton's 2nd law of motion that the force on an object is equal to the mass of the object times the acceleration of

Dmitrij [34]

   Force = (mass) x (acceleration)

   5 N = (9 kg) x (acceleration)

Divide each side
by 9 kg : 5 N / 9 kg = acceleration

   Acceleration = (5/9) kg-meter/sec²-kg

   = 0.555... m/s²   .

3 0

3 years ago