the distance between 2 station is 5400 m find the time taken by a train to cover this distance, if the train travels with speed 60m/s

Answer:

The width of the river is  [tex]z = 60.62 \ m[/tex]

Explanation:

From the question we are told that

     The distance of the base line is D = 130 m

       The angle is  [tex]\theta = 25^o[/tex]

A diagram illustration the question is shown on the first uploaded image

    Applying Trigonometric Rules for Right-angled Triangle,

            [tex]tan 25 = \frac{z}{130}[/tex]

Now making  z the subject

           [tex]z = 130 * tan (25)[/tex]

          [tex]z = 60.62 \ m[/tex]

Answer:

2.95m

Explanation:

The farthest distance the object can move is the radius of the circle of which the Simple harmonic motion is assumed to be a part

But V = w× r; where V is velocity,

w is angular velocity and r is radius.

Also,

a= w2r; where a is linear acceleration

but a = v× r ; by comparing both equations

Hence r = a/v =8.6/2.45 =3.51m

But the horizontal distance of the motion is given by:

X = rcosx ; where x is the angle

X is the distance covered.

We know that the maximum value of cos x is 1 which is 0°

When the object moves in a fashion directly parallel to an horizontal distance, maximum distance would be reached and hence:

X = r=3.51m

Meaning the object needs to travel 3.51-0.56=2.95m further.

Note: the acceleration of the motion is constant whether it is swinging towards the left or right.

When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left and the amplitude of motion A = 0.732 m.

The distance between the central and extreme points for a moving particle is known as the amplitude of motion.

The given data to find the amplitude of motion,

Object displaced = 0.560 m

Velocity = 2.45 m/s

Acceleration = 8.60 m/s²

Starting with sine:

x(t) = Asin(ωt)

so that t = 0, x = 0

x(t) = 0.56 m = Asin(ωt)

v(t) = x(t)'= 2.45 m/s = Aωcos(ωt)

a(t) = v(t)'= -8.60 m/s² = -Aω²sin(ωt)

x(t) / a(t) = Asin(ωt) / -Aω²sin(ωt)

0.56m / -8.60 m/s² = -1 / ω²

ω² = 15.3571 rad^2/s^2

ω = 3.91881 rad/s  

x(t) / v(t) = Asin(ωt) / Aωcos(ωt)

0.560m / 2.45m/s = tan(3.91t) / 3.91rad/s

0.8937= tan(3.91t)

t = 0.176 s  

x(0.176) = Asin(3.59×0.176)

0.65 m= Asin(0.631)

A = 0.732 m is the amplitude of motion.

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Answer:

a. SO2

Explanation:

Answer:

The apparent weight of the person as she pass the highest point is  [tex]N = 458.8 \ N[/tex]

Explanation:

From the question we are told that

   The radius of the Ferris wheel is [tex]r = 5.0 \ m[/tex]

    The period of revolution is [tex]T = 8.0 \ s[/tex]

     The weight of the person is  [tex]W = 670 \ N[/tex]

Generally the speed of the wheel is mathematically represented as

      [tex]v = \frac{2 \pi r}{T }[/tex]

substituting values

      [tex]v = \frac{2 * 3.142 * 5}{8 }[/tex]

       [tex]v = 3.9 3 \ m/s[/tex]

The apparent weight (the normal force exerted on her by the bench) at the highest point is mathematically evaluated as

          [tex]N = mg - \frac{mv^2}{r}[/tex]

Where m is the mass of the person which is mathematically evaluated as

     [tex]m = \frac{W}{g}[/tex]

substituting values

    [tex]m = \frac{670}{9.8}[/tex]

    [tex]m = 68.37 \ kg[/tex]

So

    [tex]N = 68.37 * 9.8 - \frac{68.37 * {3.93}^2}{5}[/tex]

    [tex]N = 458.8 \ N[/tex]

Answer:

[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]

Explanation:

In order to calculate the angular momentum of the particle you use the following formula:

[tex]\vec{L}=\vec{r}\ X\ \vec{p}[/tex]       (1)

r is the position vector respect to the point (0 , 5.0), that is:

r = 0m i + 5.0m j    (2)

p is the linear momentum vector and it is given by:

[tex]\vec{p}=m\vec{v}=(2.0kg)(3.0m/s)(\hat{i+\hat{j}})=6\frac{kgm}{s}(\hat{i}+\hat{j})[/tex]   (3)

the direction of p comes from the fat that the particle is moving along the i + j direction.

Then, you use the results of (2) and (3) in the equation (1) and solve for L:

[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]

The angular momentum is -30 kgm^2/s ^k

Answer:5.35nm

Explanation:

Consider that 1 inch is = 0.0254m

we have,

1m= 1x10^9 nm  

While:

0.0254m = 2.54x10^7nm  

1/55 (2.54x10^7) = 4.6181 x 10^5nm  

1 day= 24 hrs  

= (24x60) when calculating in min  

= (24x60x60) calculating in seconds we have:

= 8.64x10⁴sec  

In 8.64x10^4 seconds, the hair grows by 4.6181 x 10^5nm

Therefore, the amount by which the hair grows in 1 second  will be;

= (4.6181 x 10^5)/(8.64x10^4)  

= 5.35nm  

The rate of growth will be 5.35nm

Answer:

Explanation:

Assuming that the TV signal is sent in a straight line from the camera to the TV receiver, which is very far from the truth.

The reporter hears the sound is

4.1 / 343 = 0.01195 s later

The viewer hears the sound from the TV is

2.9 / 343 = 0.00845s

the difference is 0.00845 sec

the question is how far the TV signal can travel in that time.

the distance between politician and TV set is

= 0.00845 * 3*10^8 m

= 2536 km

d = 2536km

Answer:

1.92 Watt lost

Explanation:

Power rating of each toaster = 600 Watts

Current that flows = 5 Amperes

Wasted power = 1 Watt

Voltage of toaster can be gotten from P = [tex]I^{2}[/tex]R

where I = current

and R = Resistance

600 = [tex]5^{2}[/tex] x R

R = 600/25 = 24 Ohms.

According to joules loss due to heating of wire

Power loss P ∝ [tex]I^{2}[/tex]R

imputing values,

1 ∝ [tex]5^{2}[/tex] x 24

1 ∝ 600

to remove the proportionality sign, we introduce a constant k

1 = 600k

k = 1/600 = 0.00167

For the case where the current is doubled to 10 ampere, as the power doubles to 1200 W.

The resistance across the wire becomes

1200 = [tex]10^{2}[/tex]R

R = 1200/100 = 12 Ohms

power loss P = k x [tex]I^{2}[/tex]R

P = 0.0016 x [tex]10^{2}[/tex] x 12

P = 1.92 Watt lost

This question involves the concepts of power, current, and resistance.

The power wasted by the wires in the home for two units will be "4 watt".

The power wasted by the wires can be given in terms of current and resistance by the following formula:

[tex]P=I^2R\\\\\frac{P}{I^2}=R=Constant\\\\\frac{P_1}{I_1^2}=\frac{P_2}{I_2^2}[/tex]

where,

Therefore,

[tex]\frac{1\ watt}{(5\ A)^2}=\frac{P_2}{(10\ A)^2}\\\\P_2=\frac{(1\ watt)(100\ A^2)}{25\ A^2}[/tex]

P₂ = 4 watt

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Answer:

A sound level of 96 dB is 4 times as intense as one of 90 dB

Explanation:

The formula of the intensity level of sound in decibels is given as follows:

Intensity Level = 10 log₁₀(I/I₀)

where,

I = Intensity of Sound

I₀ = Reference Intensity Level = 10⁻¹² W/m²

Therefore, for 96 dB sound level:

96 = 10 log₁₀(I₁/10⁻¹²)

log₁₀(I₁/10⁻¹²) = 96/10

I₁/10⁻¹² = 10^9.6

I₁ = (10⁻¹²)(4 x 10⁹)

I₁ = 0.004 W/m²

For 90 dB sound level:

90 = 10 log₁₀(I₂/10⁻¹²)

log₁₀(I₂/10⁻¹²) = 90/10

I₂/10⁻¹² = 10^9

I₂ = (10⁻¹²)(10⁹)

I₂ = 0.001 W/m²

Therefore,

I₁/I₂ = 0.004/0.001

I₁ = 4 I₂

Hence, the sound level of 96 dB is 4 times as intense as one of 90 dB.

Answer:

About 2104m/s

Explanation:

[tex]F=ma \\\\F=\dfrac{2660kg}{2}\cdot 1614m/s=2,146,620N \\\\2,146,620N=1,269kg\cdot a \\\\a\approx 1691m/s \\\\v_f=v_o+at=413m/s+1691m/s=2104m/s[/tex]

Hope this helps!

Starting from rest, the wheel attains an angular velocity of 25 rad/s in a matter of 10 s, which means the angular acceleration [tex]\alpha[/tex] is

[tex]25\dfrac{\rm rad}{\rm s}=\alpha(10\,\mathrm s)\implies\alpha=2.5\dfrac{\rm rad}{\mathrm s^2}[/tex]

For the next 37 s, the wheel maintains a constant angular velocity of 25 rad/s, meaning the angular acceleration is zero for the duration. After this time, the wheel undergoes an angular acceleration of -1.5 rad/s/s until it stops, which would take time [tex]t[/tex],

[tex]0\dfrac{\rm rad}{\rm s}=25\dfrac{\rm rad}{\rm s}+\left(-1.5\dfrac{\rm rad}{\mathrm s^2}\right)t\implies t=16.666\ldots\,\mathrm s[/tex]

which makes B, approximately 17 s, the correct answer.

The time interval of angular deceleration is 16.667 seconds, whose closest integer is 17 seconds. (B. 17 s.)

Let suppose that the grinding wheel has uniform Acceleration and Deceleration. In this question we need to need to calculate the time taken by the grinding wheel to stop, which is found by means of the following Kinematic formula:

[tex]t = \frac{\omega - \omega_{o}}{\alpha}[/tex] (1)

Where:

[tex]\omega_{o}[/tex] - Initial angular velocity, in radians per second.

[tex]\omega[/tex] - Final angular velocity, in radians per second.

[tex]\alpha[/tex] - Angular acceleration, in radians per square second.

[tex]t[/tex] - Time, in seconds.

If we know that [tex]\omega = 0\,\frac{rad}{s}[/tex], [tex]\omega_{o} = 25\,\frac{rad}{s}[/tex] and [tex]\alpha = -1.5\,\frac{rad}{s^{2}}[/tex], then the time taken by the grinding wheel to stop:

[tex]t = \frac{0\,\frac{rad}{s}-25\,\frac{rad}{s}}{-1.5\,\frac{rad}{s^{2}} }[/tex]

[tex]t = 16.667\,s[/tex]

The time interval of angular deceleration is 16.667 seconds. (Answer: B)

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Answer:

91.5 miles

Explanation:

61 miles per hour so 61(x amount of hours)

so 61 x 1.5 hours is 91.5 miles

Answer:

For an object to be an equilibrium it must be experiencing no acceleration.

Explanation:

Hope it helps.

Answer:

[tex]\tau = 1\ ms[/tex]

Explanation:

First we need to find the capacitance of the capacitor.

The capacitance is given by:

[tex]C = \epsilon_0 * area / distance[/tex]

Where [tex]\epsilon_0[/tex] is the air permittivity, which is approximately 8.85 * 10^(-12)

The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:

[tex]Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2[/tex]

So the capacitance is:

[tex]C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}[/tex]

[tex]C = 10^{-12}\ F = 1\ pF[/tex]

The time constant of a rc-circuit is given by:

[tex]\tau = RC[/tex]

So we have that:

[tex]\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms[/tex]

Complete question

The complete question is shown on the first uploaded image  

Answer:

The velocity is  [tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]

Explanation:

From the question we are told that

           a = nb

The length of the minor axis  of  the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the  Empire's symbol, (an ellipse)

Now this length seen by the observer can be mathematically represented as

        [tex]h = t \sqrt{1 - \frac{v^2}{c^2} }[/tex]

Here t  is the actual length of the major axis of of the  Empire's symbol, (an ellipse)

So t = a = nb

and  b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the  Empire's symbol, (an ellipse)

 i.e    h = b

So

    [tex]b = nb [\sqrt{1 - \frac{v^2}{c^2} } ][/tex]  

     [tex][\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}[/tex]

      [tex]v^2 =c^2 [1- \frac{1}{n^2} ][/tex]

       [tex]v^2 =c^2 [\frac{n^2 -1}{n^2} ][/tex]

        [tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]

Answer:

Explanation:

From Newton's 2nd Law,

F = m×a

Where F is Force

m is mass

a is acceleration

Hence a= F/m

a= 16/8= 2m/s2

Answer:

a

The speed is   [tex]s = 5.857 m/s[/tex]

b

The distance is  [tex]D = 22.4 \ m[/tex]

Explanation:

From the question we are told that

     The speed of the banana is  [tex]v = 16 \ m/s[/tex]

   The distance from my  location is  [tex]d = 8.2 \ m[/tex]  

     The time taken is  [tex]t = 1.4 \ s[/tex]

The speed of the ice cream is

          [tex]s = \frac{d}{t}[/tex]

substituting values

        [tex]s = \frac{8.4}{1.4}[/tex]

        [tex]s = 5.857 m/s[/tex]

The distance of separation between i and Valdimir is the same as the distance covered by the banana

   So  

          [tex]D = v * t[/tex]

substituting values

        [tex]D = 16 * 1.4[/tex]

        [tex]D = 22.4 \ m[/tex]

Answer:

down, backwards, and outwards.

Explanation:

For a hornet that is accelerating in flight, this means that there is a net forward motion at a relatively constant vertical height above the ground.

For this flight, the wings beat downwards to counter the weight of the hornet due to gravity, keeping it at that height above the floor.

For the hornet to accelerate forward, there has to be a net backwards force by the wing on the air. This backwards force accelerates tr forward due to the absence of an equal opposing force in the opposite direction save for a little drag.

The wings also beat with forces directed outwards to provide centripetal force to keep the hornet stable. The absence of this would cause it to spiral out of control.

Answer:

[tex]52.25\times10^4W\\699.1 hp[/tex]

Explanation:

According to the energy conversation:

ΔK=[tex]-f_kd+W[/tex]

ΔK=[tex]K_f-K_i ; K=1/2 mv^2[/tex]

where,

[tex]k_i, k_f[/tex] are initial and final kinetic energy of the system.

[tex]v_i[/tex]= initial velocity of the system

[tex]v_f[/tex]=final velocity of the system

W= total work done on the system

[tex]f_k[/tex]= friction force

d= distance traveled

Given: [tex]v_f[/tex]=110m/s

d=400m

[tex]f_k[/tex]=1200N

[tex]v_i[/tex]=0m/s

t=7.3s

ΔK=[tex]-f_kd+W[/tex]

W= ΔK + [tex]f_kd[/tex]

  =[tex]K_f-K_i+f_kd\\[/tex]

  [tex]=1/2 mv_f^2-1/2 mv_i^2+f_kd\\=\frac{1}{2} \times 550\times110^2 - \frac{1}{2} \times 550\times0^2+ (1200\times400)\\=3807500[/tex]

[tex]P=\frac{W}{t} =\frac{3807500}{7.3} \\P=52.15 \times10^4w\\P=\frac{52.15 \times10^4}{746} \\P=699.1 hp[/tex]

Answer:

49.725× 10^-24J

Explanation:

The Energy associated with a Photon us defined as;

E = hf

Where h is Planck's constant = 6.63× 10^-34m2kg/s

f is the frequency= 7.5 x 10^14 Hz

Hence

E = 6.63× 10^-34 × 7.5 x 10^14 =49.725× 10^-24J

Answer:

1A,2D,3B

Explanation:

hope this helps

Answer:

a bale

Explanation:

a bale is a group of turtles

Answer:

A bale or nest

Explanation:

Answer:

B

Explanation:

Because the force has 2 components (horizontal and vertical), the horizontal component must be smaller than the total force. The Pythagorean theorem only adds positive values (because they're squared) so it makes sense. Using trigonometry, 100*cos(-20) yields a horizontal force of around 939.7N, which is less than 1000N.

The dragster's velocity v at time t with constant acceleration a is

[tex]v=at[/tex]

since it starts at rest.

After 2.1 s, it will attain a velocity of

[tex]v=\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)[/tex]

or 92.4 m/s.

Doubling the time would double the final velocity,

[tex]v=a(2t)=2at[/tex]

so the velocity would be twice the previous one, 184.8 m/s.

The dragster undergoes a displacement x after time t with acceleration a of

[tex]x=\dfrac12at^2[/tex]

if we take the starting line to be the origin.

After 2.1 s, it will have moved

[tex]x=\dfrac12\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)^2[/tex]

or 88 m.

Doubling the time has the effect of quadrupling the displacement, since

[tex]x=\dfrac12a(2t)^2=4\left(\dfrac12at^2\right)[/tex]

so after 4.2 s it will have moved 352 m.

Answer:

1.16 Hz

Explanation:

frequency, basically, is the number of wave on 1 second

so, in math we write like this

f = n/t

n = number of waves

t = time to do that (in sec)

f = 140/120 = 7/6 Hz

f = 1.16 Hz

Answer:

K = 3.9 kJ

Explanation:

The kinetic energy ([tex]K_{T}[/tex]) added is given by the difference between the final kinetic energy and the initial kinetic energy:

[tex] K_{T} = K_{f} - K_{i} [/tex]  

The initial kinetic energy is:

[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} [/tex]

Where m₁ is the mass of the object before the explosion and v₁ is its velocity

[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} = \frac{1}{2}70 kg*(21 m/s)^{2} = 1.54 \cdot 10^{4} J [/tex]

Now, the final kinetic energy is:

[tex] K_{f} = \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2} [/tex]

Where m₂ and m₃ are the masses of the 2 pieces produced by the explosion and v₁ and v₂ are the speeds of these pieces

Since m₂ is 4 times as massive as m₃ and v₃ = 0, we have:

[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}v_{2}^{2} + \frac{1}{2}*\frac{1}{5}m_{1}*0 [/tex]   (1)          

By conservation of momentum we have:

[tex] p_{i} = p_{f} [/tex]

[tex] m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3} [/tex]  

[tex] m_{1}v_{1} = \frac{4}{5}m_{1}v_{2} + \frac{1}{5}m_{1}*0 [/tex]

[tex] v_{2} = \frac{5}{4}v_{1} [/tex]     (2)

By entering (2) into (1) we have:

[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}(\frac{5}{4}v_{1})^{2} = \frac{1}{2}*\frac{4}{5}70 kg(\frac{5}{4}*21 m/s)^{2} = 1.93 \cdot 10^{4} J [/tex]  

Hence, the kinetic energy added is:

[tex] K_{T} = K_{f} - K_{i} = 1.93 \cdot 10^{4} J - 1.54 \cdot 10^{4} J = 3.9 \cdot 10^{3} J [/tex]  

Therefore, the kinetic energy added to the system during the explosion is 3.9 kJ.

I hope it helps you!

Answer:

Explanation:

From Newton's second Law of Motion,

F = ma

Ff + F = ma

Where F is the applied force, Ff is the frictional force, a is the acceleration and m is the mass of the object or box.

Magnitude of the acceleration:

a = Ff+F/m

This must act in the direction of F or the box would slide or accelerate off the negative side of the board (taking the direction of F to be positive

Answer:

WEIGHT ON MOON IS 0.2004N

Explanation:

mass of the body=120g=[tex]\frac{120}{1000}[/tex]kg=0.12kg (we will convert g into kg)

gravity on moon=1.67m/s²( to find the mass of anybody on another we should know its gravity)

as we know that (from the formula of weight)

weight=mass×gravity

w=mg

w=0.12kg²×1.67m/s²

w=0.2004N

Answer:

14.68m/s

Explanation:

As per the question, the data provided is as follows

Mass = M = 0.148 kg

Height = h = 11 m

Initial velocity = U = 0 m/s

Final velocity = V

Gravitational force = F

Mass = M

Based on the above information, the speed that hit to the ground is

As we know that

Work to be done = Change in kinetic energy

[tex]F ( S) = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]

[tex]M g h = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]

[tex]g h = (\frac{1}{2} ) ( V^2 - U^2 )[/tex]

[tex]V^2 - U^2 = 2gh[/tex]

[tex]V^2 - 0 = 2gh[/tex]

[tex]V = \sqrt{2 g h}[/tex]

[tex]= \sqrt{2\times9.8\times11}[/tex]

= 14.68m/s

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

[tex]K=qV[/tex]          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

[tex]K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J[/tex]

The kinetic energy of the particle is also:

[tex]K=\frac{1}{2}mv^2[/tex]       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}[/tex]

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

[tex]|F_B|=qvBsin(\theta)[/tex]

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

[tex]|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N[/tex]

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

[tex]r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm[/tex]

the radius of the trajectory of the electron is 10.1 cm

The speed, magnetic force and radius are respectively; 10.75 * 10⁶ m/s; 7.57 * 10⁻¹² N; 0.101 m

A) We know that the formula for kinetic energy can be expressed as;

K = qV

where;

q is charge of the particle = 2e = 2(1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C

V is potential difference = 1.2 × 10⁶ V

K = 3.2 × 10⁻¹⁹ *  1.2 × 10⁶

K = 3.84 × 10⁻¹³ J

Also, formula for kinetic energy is;

K = ¹/₂mv²

where v is speed

Thus;

v = √(2K/m)

v = √(2 * 3.84 × 10⁻¹³)/(6.64 * 10⁻²⁷)

v = 10.75 * 10⁶ m/s

B) The magnetic force is given by the formula;

F_b = qvB

F_b = (3.2 × 10⁻¹⁹ * 10.75 * 10⁶ * 2.2)

F_b = 7.57 * 10⁻¹² N

C) The formula to find the radius is;

r = mv/qB

r = (6.64 * 10⁻²⁷ * 10.75 * 10⁶)/(1.6 × 10⁻¹⁹ * 2.2)

r = 0.101 m

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