A cat jumps from a balcony. About how fast is the cat moving after 1.5 s?

Two concentric, coplanar, circular loops of wire of different diameter carry currents in the same direction. Describe the nature

Ierofanga [76]

Answer:

Both will be attractive in nature.

Explanation:

In the given case, the direction of the magnetic field is same in both loops as the direction of the current is same in both loops. When two parallel straight wire carrying current in the same direction are brought close to each other, the force between them is attractive in nature. In the same way, when two coplanar, circular and concentric loops of wire are carrying current in the same direction, the force between them is attractive in nature. It can be checked by using right hand thumb rule.

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4 0

2 years ago

A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the loc

sattari [20]

Answer:

Explanation:

velocity of ship with respect to water = 6.5 m/s due north

$\overrightarrow{v}_{s,w}=6.5 \widehat{j}$

velocity of water with respect to earth = 1.5 m/s at 40° north of east

$\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)$

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth

$\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}$

$\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j} \right )$

$\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}$

The magnitude of the velocity of ship relative to earth is $\sqrt{1.15^{2}+5.54^{2}}$ = 5.66 m/s

5 0

3 years ago

What phase difference between two otherwise identical harmonic waves, moving in the same direction along a stretched string, wil

lora16 [44]

Answer:

$\theta=145$

Explanation:

The amplitude of he combined wave is:

$B=2Acos(\theta/2)\\$

A, is the amplitude from the identical harmonic waves

B, is the amplitude of the resultant wave

θ, is the phase, between the waves

The amplitude of the combined wave must be 0.6A:

$0.6A=2Acos(\theta/2)\\ cos(\theta/2)=0.3\\\theta/2=72.5\\\theta=145$

5 0

3 years ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So