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Elanso [62]
1 year ago
5

What is the direction of the magnetic field b⃗ net at point a? Recall that the currents in the two wires have equal magnitudes.

Physics
1 answer:
andrew11 [14]1 year ago
6 0

Answer:

Explanation:

The direction of a magnetic field indicates where the magnetic inluence on the electric charges are directed to.

From the given  question, we are to determine the direction of the magnetic field bnet at a point A.

Also, having the notion that  the currents in the two wires have equal magnitudes, Then:

\bar{B_{net}} = \bar{B_1} + \bar{B_2}

\bar{B_{net}} = \frac{\mu_oI}{2 \pi r } \bar {k}+ \frac{\mu_oI}{2 \pi r } \bar {k}

\bar{B_{net}} = \frac{2 \mu_oI}{2 \pi r } \bar {k} \ out

Thus; \bar{B_{net}} points out of the screen at A.

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A bicyclist is riding at a tangential speed of 13.2 m/s around a circular track. The magnitude of the centripetal force is 377 N
MA_775_DIABLO [31]

Answer:

40m approximately

Explanation:

Given

Force =377N

Mass =86.5kg

Velocity =13.2m/s

Required

Radius of the track

The expression for the centripetal force acting on the cyclist is

F=mv²/r

Make r subject of the formula

r= mv²/F

Substitute

r=86.5*13.2²/377

r= 15,071.76/377

r=39.97

r=40m approximately

4 0
2 years ago
A steel cable 1.25 in. in diameter and 50 ft long is to lift a 20-ton load without permanently deforming. What is the length of
Over [174]

Answer:

50.0543248872 ft

Explanation:

F = Load = 20 ton = 20\times 2000\ lb

d = Diameter = 1.25 in

L_1 = Initial length = 50 ft

L_2 = Final length

A = Area = \dfrac{\pi}{4}d^2

Y = Young's modulus = 30\times 10^6\ psi

Young's modulus is given by

Y=\dfrac{FL}{A\Delta L}\\\Rightarrow Y=\dfrac{FL_1}{\dfrac{\pi}{4}d^2(L_2-L_1)}\\\Rightarrow L_2=\dfrac{4FL_1}{Y\pi d^2}+L_1\\\Rightarrow L_2=\dfrac{4\times 40000\times 50}{30\times 10^6\times \pi\times 1.25^2}+50\\\Rightarrow L_2=50.0543248872\ ft

The length during the lift is 50.0543248872 ft

6 0
2 years ago
A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma
Nesterboy [21]

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a    ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

=  g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x  .50 x ( 10+6 )

= 78.4 N

5 0
2 years ago
A rubber ball with a mass 0.20 kg is dropped vertically from a height of 1.5 m above the floor. The ball bounces off of the floo
Digiron [165]
Potential Energy = mass * Hight * acceleration of gravity
PE=hmg
PE = 1.5 * .2 * 9.81
PE = 2.943
it lost .6 so 2.943 - .6 = 2.343
now your new energy is 2.343 so solve for height
2.343 = mhg
2.334 = .2 * h * 9.81
h = 1.194
the ball after the bounce only went up 1.194m
8 0
2 years ago
Backpackers often use canisters of white gas to fuel a cooking stove's burner. If one canister contains 1.45 L of white gas, and
lorasvet [3.4K]

Answer: One canister contains 1.03 Kg of fuel,

Explanation:

The density is defined as the relation between the mass and the volume p=\frac{m}{v}

First of all you need to have the same units for volume and density, so:

For the volume,

1.45 L * \frac{1.10*10^{-3}m^{3}  }{1 L} = 1.45 *10^{-3} m^{3}

For the density,

p = 0.710\frac{g}{cm^{3} } *\frac{1cm^{3} }{1*10^{-6}m^{3}  }*\frac{1 Kg}{1*10^{3}g } = 710\frac{Kg}{m^{3} }

From the density equation we have m=p*v, so:

710\frac{Kg}{m^{3} } *1.45*10^{-3} m^{3} = 1.03Kg

4 0
2 years ago
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