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alisha [4.7K]
3 years ago
13

A wire carries a steady current of 2.60 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uni

form magnetic field, vector B = 1.50 k T. If the current is in the positive x direction, what is the magnetic force on the section of wire?
Physics
1 answer:
ad-work [718]3 years ago
7 0

Answer:

The magnetic force on the section of wire is -2.925\hat{j}\ N.

Explanation:

Given that,

Current I = 2.60\hat{i}\ A

Length = 0.750 m

Magnetic field B = 1.50\hat{k}\ T

We need to calculate the magnetic force on the section of wire

Using formula of magnetic force

\vec{F}=l\vec{I}\times\vec{B}

\vec{F}=0.750\times2.60\hat{i}\times1.50\hat{k}

Since, \hat{i}\times\hat{k}=-\hat{j}

\vec{F}=-2.925\hat{j}\ N

Hence, The magnetic force on the section of wire is -2.925\hat{j}\ N.

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A child has a mass of 35 kg. The child is running across a fiend and has a speed of 3 m/s. What is the kinetic energy of the chi
Sladkaya [172]

Answer:

Explanation:

Given the following data;

Mass = 35 kg

Velocity = 3 m/s

To find the kinetic energy of the child;

K.E = ½mv²

4 0
3 years ago
Which of the following simple machines is not properly identified? A. Scissors: lever B. Pizza cutter: wedge C. Screw: wheel and
GrogVix [38]

Pretty sure the answer is   C. Screw: wheel and axle

8 0
3 years ago
Kevin is refinishing his rusty wheelbarrow. He moves his sandpaper back and forth 45 times over a rusty area, each time moving a
dmitriy555 [2]
W = _|....F*dx*cos(a)........With F=force, x=distance over which force acts on object,
.......0.............................and a=angle between force and direction of travel.

Since the force is constant in this case we don't need the equation to be an integral expression, and since the force in question - the force of friction - is always precisely opposite the direction of travel (which makes (a) equal to 180 deg, and cos(a) equal to -1) the equation can be rewritted like so:

W = F*x*(-1) ............ or ............. W = -F*x

The force of friction is given by the equation: Ffriction = Fnormal*(coeff of friction)

Also, note that the total work is the sum of all 45 passes by the sandpaper. So our final equation, when Ffriction is substituted, is:

W = (-45)(Fnormal)(coeff of friction)(distance)
W = (-45)...(1.8N).........(0.92).........(0.15m)
W = ................-11.178 Joules
5 0
3 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
Holly puts a box into the trunk of her car. Later, she drives around an unbanked curve that has a radius of 48 m. The speed of t
LiRa [457]

Answer:

The minimum coefficient of friction is 0.544

Solution:

As per the question:

Radius of the curve, R = 48 m

Speed of the car, v = 16 m/s

To calculate the minimum coefficient of static friction:

The centrifugal force on the box is in the outward direction and is given by:

F_{c} = \frac{mv^{2}}{R}  

f_{s} = \mu_{s}mg

where

\mu_{s} = coefficient of static friction

The net force on the box is zero, since, the box is stationary and is given by:

F_{net} = f_{s} - F_{c}  

0 = f_{s} - F_{c}  

\mu_{s}mg = \frac{mv^{2}}{R}  

\mu_{s} = \frac{v^{2}}{gR}  

\mu_{s} = \frac{16^{2}}{9.8\times 48} = 0.544  

3 0
3 years ago
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