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Ipatiy [6.2K]
3 years ago
5

A straight wire in a magnetic field experiences a force of 0.023 N when the current in the wire is 2.5 A. The current in the wir

e is changed, and the wire experiences a force of 0.058 N as a result. What is the new current
Physics
1 answer:
Tatiana [17]3 years ago
8 0

Answer:

The new current in the wire is 6.3 A.

Explanation:

Given that,

Force in a straight wire, F = 0.023 N

Current in the wire, I = 2.5 A

If new force is, F' = 0.058 N, we need to find the new current.

We know that, magnetic force is given by :

F=ilB\sin\theta

It is clear that force is directly proportional to electric current. So, we can say:

\dfrac{F}{F'}=\dfrac{I}{I'}

I' is new current

I'=\dfrac{IF'}{F}\\\\I'=\dfrac{2.5\times 0.058}{0.023}\\\\I'=6.3\ A

So, the new current in the wire is 6.3 A. Hence, this is the required solution.

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Where are globular clusters located in the milky way?.
mel-nik [20]

Answer:

Globular star clusters are located in the great spherical halo.

Explanation:

Hope this helps! :)

8 0
2 years ago
A ball is shot from the ground into the air. At a height of 8.8 m, the velocity is observed to be
Mariulka [41]

Answer:

h = 10.4 m

R = 22.48 m

v= 16,2 m/s , α = 61.7°, below the horizontal

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

Explanation:

The ball describes a parabolic path, and the equations of the movement are:

Equation of the uniform rectilinear motion (horizontal ) :

x = vx*t  :

Equations of the uniformly accelerated rectilinear motion of upward   (vertical ).

y = (v₀y)*t - (1/2)*g*t² Equation (2)

vfy² = v₀y² -2gy Equation (3)

vfy = v₀y -gt Equation (4)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m  

y: vertical position in meters (m)  

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Known data

y= 8.8 m

v = ( (7.7)i + (5.7)j  ) m/s : vx= 7.7 m/s , vy= 5.7 m/s

g = 9.8 m/s²

Calculation of the  initial  vertical velocity ( v₀y)

We apply Equation (3) with the known data

(vfy)² = (v₀y)² -2*g*y

(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)

(5.7)²+ 172.48 =  (v₀y)²

v_{oy} = \sqrt{(5.7)^{2}+ 172.48 }

v₀y = 14.3 m/s

Calculation of the maximum height  the ball rise (h)

In the maximum height vfy=0

We apply the Equation (3) :

(vfy)² = (v₀y)² -2*g*y

0 = (14.3)² - 2*98*h

h = (14.3)² / 19.6

h = 10.4 m

Calculation of the time it takes for the ball to the maximum height

We apply the Equation (4) :

vfy = v₀y -gt

0 = v₀y -gt

gt = v₀y

t = v₀y/g

t = 14.3/9.8

t= 1.46 s

Flight time = 2t = 2.92 s

Total horizontal distance traveled by the ball  (R)

We replace data in the equation (1)

x =vx*t    vx= 7.7 m/s , t =2.92 s  (Flight time)

R = (7.7)* (2.92) = 22.48 m

Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground

vx = 7.7 m/s

vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s

v= \sqrt{v_{x}^{2}+v_{y}^{2}  }

v= \sqrt{(7.7)^{2}+ (-14.3)^{2}  }

v= 16,2 m/s

\alpha = tan^{-1} (\frac{v_{y} }{v_{x} })

\alpha = tan^{-1} (\frac{-14.3 }{7.7 })

α = -61.7°

α = 61.7°, below the horizontal

i- j components of the v

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

5 0
3 years ago
Part complete Sound with frequency 1240 Hz leaves a room through a doorway with a width of 1.11 m . At what minimum angle relati
Sedbober [7]

Answer:

14.43° or 0.25184 rad

Explanation:

v = Speed of sound in air = 343 m/s

f = Frequency = 1240 Hz

d = Width in doorway = 1.11 m

Wavelength is given by

\lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343}{1240}\\\Rightarrow \lambda=0.2766\ m

In the case of Fraunhofer diffraction we have the relation

dsin\theta=\lambda\\\Rightarrow \theta=sin^{-1}\frac{\lambda}{d}\\\Rightarrow \theta=sin^{-1}\frac{0.2766}{1.11}\\\Rightarrow \theta=14.43^{\circ}\ or\ 0.25184\ rad

The minimum angle relative to the center line perpendicular to the doorway will someone outside the room hear no sound is 14.43° or 0.25184 rad

6 0
3 years ago
What is the bouncing of light when it changes media
matrenka [14]

When light passes from one medium to another, part of it continues on
into the new medium, while the rest of it bounces away from the boundary,
back into the first medium.

The part of the light that continues on into the new medium is <em>transmitted</em>
light.  Its forward progress at any point in its journey is <em>transmission</em>. 

Its direction usually changes as it crosses the boundary.  The bending is <em>
refraction</em>
.

The part of the light that bounces away from the boundary and heads back
into the first medium is <em>reflected</em> light.  The process of bouncing is <em>reflection</em>.


5 0
3 years ago
Is AG on the periodic table
Taya2010 [7]
Ag on the periodic table is silver. 
The atomic number for silver is 47, atomic symbol is Ag, and the atomic weight is 107.9.
8 0
3 years ago
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