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3 years ago

Why do the compass needles change direction when the electric current is Flowing

Physics

2 answers:

zhenek [66]3 years ago

6 0

Explanation:

Since compasses work by pointing along magnetic field lines, this means that there must be a magnetic field near the wire through which the current is flowing.

REY [17]3 years ago

3 0

Answer:

i have noo idea

Explanation:

idk im slow

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Which statements describe intensity? Check all that apply.

REY [17]

Answer:

it's A, D, E

Explanation:

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3 years ago

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A 0.80-kg soccer ball experinces an impulse of 25 N x s . Determine the momentum change of the soccer ball.

borishaifa [10]

If the impulse is 25 N-s, then so is the change in momentum.
The mass of the ball is extra, unneeded information.

Just to make sure, we can check out the units:

Momentum = (mass) x (speed) = kg-meter / sec

Impulse = (force) x (time) = (kg-meter / sec²) x (sec) = kg-meter / sec

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3 years ago

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Somebody please explain how to solve this. Thanks in advance!

Jobisdone [24]

work done is product of force and displacement of point of application of force

so here we have to check the product of force and displacement both

Now we will put the least to maximum work in the following order

1. -A man exerts strenuous effort in pushing a stationary wall

2. -A flea pushes a speck of dirt 1 cm

3. -A farmer pushes a 2 kg wheelbarrow 20 m

4. -A cannon launches a 3 kg cannonball a distance of 200

5. -A 2000 kg car travels 400 m down a road

6. -Space shuttle Atlantis launches from the ground into near-Earth orbit

5 0

2 years ago

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

Given:

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

Assume:

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0

2 years ago

a small table has a mass of 4kg, stands on four legs, each leg having an area of 0.001 m2. what is the pressure exerted by the t

Gemiola [76]

Answer:

P = 10 kPa

Explanation:

Given that,

The mass of a small table, m = 4 kg

The area of each leg = 0.001 m²

We need to find the pressure exerted by the table on the floor. Pressure is equal to the force per unit area. So

$P=\dfrac{mg}{4\times A}\\\\P=\dfrac{4\times 10}{4\times 0.001}\\P=10000\ Pa\\\\or\\\\P=10\ kPa$

So, the required pressure is 10 kPa.

3 0

2 years ago