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Leokris [45]
4 years ago
5

A particle with charge -5 C initially moves at v = (1.00 i^ + 7.00 j^ ) m/s. If it encounters a magnetic field B =80 Tkˆ, find t

he magnetic force vector on the particle.
Physics
1 answer:
algol [13]4 years ago
8 0

Answer:

The magnetic force is  \= F  =  400\r j + 2800\r i

Explanation:

From the question we are  told that

  The  charge is  q =  -5C

  The  velocity is  v  =  (1.00\ \r i +  7.00 \ \r j )\ m/s

   The  magnetic field is  B =  80 \r k \ \ T

Generally the magnetic force is mathematically represented as

        \=  F =  q  \= v  \  \  X \ \ \= B

=>     \=  F = -5  (1.0 \r i  + 7.0 \r j  ) \  \  X \ \  80 \r k

=>    \= F  =  -5.0 \r i +  35\r j  \ \ \ X \ \ 80\r k

=>  \= F  =  400\r j + 2800\r i    N/B - Applied cross - product of unit vector

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Explanation:

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A wave has a wavelength of 5 meters and a frequency of 3 hertz. What is the speed of the wave?
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8 0
3 years ago
g At some point the road makes a right turn with a radius of 117 m. If the posted speed limit along this part of the highway is
user100 [1]

Answer:

Ф = 28.9°

Explanation:

given:

radius (r) = 117m

velocity (v) = 25.1 m/s

required: angle Ф

Ф = inv tan (v² / (r * g))      we know that g = 9.8

Ф = inv tan (25.1² / (117 * 9.8))

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6 0
4 years ago
A man lifts a 25.9 kg bucket from a well and does 5.92 kJ of work.The acceleration of gravity is 9.8 m/s^2 . How deep is the wel
Jet001 [13]

Answer:

The well is 23.3 m

Explanation:

As the bucket is lifted out of the well, energy in the man is being transferred to the bucket as gravitational potential energy.

Work done against gravity =  mass * height * acceleration due to to gravity

W  =  mgh

5 920 J = 25.9 kg * h * 9.8 m/s²

h = 23.3 m

4 0
3 years ago
What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 22 km/h and the coef
Aleks [24]
First, let's put 22 km/h in m/s:

22 \frac{km}{h} \times  \frac{1000m}{1km}  \times  \frac{1h}{3600s}=6.11 \frac{m}{s}

Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by

F_{rad}=m \frac{v^2}{R}

The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:

F_{fric}= mg \mu_{s}

Notice we don't use the kinetic coefficient even though the bike is moving.  This is because when the tires meet the road they are momentarily stationary with the road surface.  Otherwise the bike is skidding.

Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:

m\frac{v^2}{R} = mg \mu_{s} \\ \\ \frac{v^2}{R} = g \mu_{s} \\ \\R= \frac{v^2}{g \mu_{s}} \\ \\ R= \frac{6.11}{9.8 \times 0.37}=1.685m

3 0
3 years ago
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