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3 years ago

An electric current in a wire flows to the West in a magnetic field directed

Physics

1 answer:

vredina [299]3 years ago

7 0

Answer: Current in a wire

We can use the same right-hand rule as we did for the moving charges—pointer finger in the direction the current is flowing, middle finger in the direction of the magnetic field, and thumb in the direction the wire is pushed.

Explanation:

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20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What is the new vol

Setler79 [48]

The correct answer is C.

We will use Boyle's law that states that for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional.

P1 V1 = P2 V2

Where

P1 is initial pressure = 5 psi

V1 is initial volume = 20 cubic inch

P2 is final pressure = 10 psi

V2 is final volume = unknown

V2 = P1,V1 / P2

V2 = 20 × 5 / 10

V2 = 100/10

V2 = 10 cubic inches

3 0

3 years ago

Read 2 more answers

An unwary football player collides with a padded goalpost while running at a velocity of 9.50 m/s and comes to a full stop after

Amanda [17]

Answer:

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

Explanation:

Given:

Initial velocity of the player (u) = 9.50 m/s

Final velocity of the player (v) = 0 m/s (Comes to a stop)

Displacement of the player (S) = 0.250 m

We know that, using equation of motion relating displacement (S), acceleration (a), initial velocity (u) and final velocity (v), we have:

$v^2=u^2+2aS$

Expressing in terms of 'a', we get:

$a=\frac{v^2-u^2}{2S}$

Plug in the given values and solve for 'a'. This gives,

$a=\frac{0-9.50^2}{2\times 0.250}\\\\a=\frac{-90.25}{0.5}=-180.5\ m/s^2$

Therefore, the acceleration of the player is -180.5 m/s². So, the deceleration is 180.5 m/s².

Now, using the first equation of motion, we have:

$v=u+at\\\\t=\frac{v-u}{a}$

Plug in the given values and solve for 't'. This gives,

$t=\frac{0-9.5}{-180.5}\\\\t=0.053\ s$

Therefore, the the collision will last for 0.053 s.

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

8 0

3 years ago

Which three metal do you think might be present in alnico explain ? ans which one are magnetic

erma4kov [3.2K]

Answer:

Aluminium, Nickel, Cobalt and Iron plus varying levels of Copper, Titanium and Niobium.

MAGNETIC: Alni, Alcomax, Hycomax, Columax, and Ticonal.

8 0

3 years ago

46 points :)

IgorC [24]

Answer:

Its is dividing by 2

Explanation:

It starts with 100 them it goes to 50, 25, 12.5 so its a cycle of dividing by 2

3 0

3 years ago

Read 2 more answers

Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are

Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

$d_{12} = \sqrt{3^{2}+3^{2} } = \sqrt{18}$ m : distance from q₁ to q₂

(d₁₂)² = 18 m²

$d_{13} =\sqrt{1^{2}+3^{2} } = \sqrt{10}$ m : distance from q₁ to q₃

(d₁₃)² = 10 m²

$d_{14} =\sqrt{3^{2}+2^{2} } = \sqrt{13}$ m : distance from q₁ to q₄

(d₁₄)² = 13 m²

K= 8.98755 × 10⁹ N *m²/C²

q₁= 7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

$F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }$

$F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }$

Fn₁= 23.95*10⁻⁹ N

8 0

3 years ago