Answer:
the magnetic field of the solenoid will remain same.
Explanation:
The magnetic field of a solenoid is given by the following formula:
[tex]B = \frac{\mu_o NI}{l}[/tex]
where,
B = magnetic field
μ₀ = permeability of free space
N = total number of coils
I = current passing through the solenoid
L = length of the solenoid
The formula clearly shows that the radius of the solenoid has no effect on the magnetic field produced by it.
Therefore, the magnetic field of the solenoid will remain same.
The displacement x of a particle varies with time t as x = 4t 2 -15t + 25. Find the position,
velocity and acceleration of the particle at t = 0. When will the velocity of the particle becomes
zero? Can we call the motion of the particle as one with uniform acceleration?
Answer:
x = 4 t^2 - 15 t + 25 displacement of particle
dx / dt = 8 t - 15 velocity of particle
d^2x / dt^2 = 8 acceleration of particle
If 8 t -15 = o then t = 8 / 15
Since acceleration is a constant 8 then motion has uniform acceleratkon
HELP NEEDED FAST (last cram sessions before finals)
BRAINLIEST!
Three resistors are connected in series across a 75-V potential difference. R, is 170 and R2 is 190. The potential difference across R3 is 21 V. Find the current in the circuit.
Explanation:
The sum of the voltages of the components connected in a series circuit is equal to the voltage across the battery.
[tex]V_T = V_1 + V_2 +V_3[/tex]
From Ohm's law ([tex]V=IR[/tex]) and in a series circuit, the amount of current flowing through the components is the same for all. So we can write [tex]V_T[/tex] as
[tex]V_T= 75\:\text{V} = I(170)+I(190) + 21\:\text{V}[/tex]
[tex]I(170+190)=54\:\text{V}[/tex]
[tex]I= \dfrac{54\:\text{V}}{360\:\text{ohms}}=0.15\:\text{A}[/tex]
Consider the video tutorial you just watched. Suppose we repeat the experiment, but this time place the divider closer to one side of the tube than to the other. How will the speed of the air on the wide and narrow sides of the divider compare
Answer:
The answer is "The air will move faster on the narrow side".
Explanation:
The air on the top slows down in hypertensive. This is why light travels quicker on top. This results in air deflection downwards, required for its energy conservation to generate lift, and that is why air has to be moved quicker on the narrow side by the very same airflow per unit time as it departs.
what is the final velocity if you have an initial velocity of 5 m/s with an acceleration of 3 m/s^2 over a 4 second interval
Answer:
initial velocity (u)=5m/s
final velocity (v)=?
acceleration (a)=3m/s^2
time (t)=4s
now,
acceleration (a)=v-u/t
3=v-5/4
3×4=v-5
12=v-5
12+5=v
17=v
v=17
A boy of mass 50 kg on a motor bike is moveny coith 20m/see what is hio k.E
An electron with an initial speed of 660,000 m/s is brought to rest by an electric field.
A) What was the potential difference that stopped the electron?
B) What was the initial kinetic energy of the electron, in electron volts?
Answer:
A) ΔV = 1.237 V
B) K.E = 1.237 eV
Explanation:
B)
The initial kinetic energy of the electron is given by the following formula:
[tex]K.E = \frac{1}{2}mv^2\\\\[/tex]
where,
K.E = Kinetic Energy of electron = ?
m = mass of elctron = 9.1 x 10⁻³¹ kg
v = speed of electron = 660000 m/s
Therefore,
[tex]K.E = \frac{1}{2}(9.1\ x\ 10^{-31}\ kg)(660000\ m/s)^2[/tex]
K.E = 1.98 x 10⁻¹⁹ J
K.E = (1.98 x 10⁻¹⁹ J)([tex]\frac{1\ eV}{1.6\ x\ 10^{-19}\ J}[/tex])
K.E = 1.237 eV
A)
The energy applied by the potential difference must be equal to the kinetic energy of the electron, in order to stop it:
[tex]e\Delta V = K.E\\\\\Delta V = \frac{K.E}{e}[/tex]
where,
e = charge on electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]\Delta V = \frac{1.98\ x\ 10^{-19}\ J}{1.6\ x\ 10^{-19}\ C}[/tex]
ΔV = 1.237 V
A falcon is hovering above the ground, then suddenly pulls in its wings and begins to fall toward the ground. Air resistance is not negligible.
Identify the forces on the falcon.
a. Kinetic friction
b. Weight w
c. Static friction
d. Drag D
e. Normal force n
f. Thrust
g. Tension T
Answer:
Explanation:
When a falcon is hovering, the force of up thrust is balanced by the weight.
When it begins to fall towards the ground, the weight acts downwards, kinetic friction is upwards, drag is upwards, normal force is upwards, thrust is upwards.
What is the approximate radius of an equipotential spherical surface of 30 V about a point charge of +15 nC if the potential at an infinite distance from the surface is zero?
Answer:
V = k Q / R potential at distance R for a charge Q
R = k Q / V
R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m
Note: Our equation says that if R if infinite then V must be zero.
A free undamped spring/mass system oscillates with a period of 4 seconds. When 10 pounds are removed from the spring, the system then has a period of 2 seconds. What was the weight of the original mass on the spring? (Round your answer to one decimal place.)
Answer:
13.3 pounds.
Explanation:
For a spring of constant K, with an attached object of mass M, the period can be written as:
T = 2*π*√(M/K)
Where π = 3.14
First, we know that the period is 4 seconds, then we have:
4s = (2*π)*√(M/K)
We know that if the mass is reduced by 10lb, the period becomes 2s.
Then the new mass of the object will be: (M - 10lb)
Then the period equation becomes:
2s = (2*π)*√((M-10lb)/K)
So we have two equations:
4s = (2*π)*√(M/K)
2s = (2*π)*√((M-10lb)/K)
We want to solve this for M.
First, we need to isolate K in one of the equations.
Let's isolate K in the first one:
4s = (2*π)*√(M/K)
(4s/2*π) = √(M/K)
(2s/π)^2 = M/K
K = M/(2s/π)^2 = M*(π/2s)^2
Now we can replace it in the other equation.
2s = (2*π)*√((M-10lb)/K)
First, let's simplify the equation:
2s/(2*π) = √((M-10lb)/K)
1s/π = √((M-10lb)/K)
(1s/π)^2 = ((M-10lb)/K
K*(1s/π)^2 = M - 10lb
Now we can use the equation: K = M*(π/2s)^2
then we get:
K*(1s/π)^2 = M - 10lb
(M*(π/2s)^2)*(1s/π)^2 = M - 10lb
M/4 = M - 10lb
10lb = M - M/4
10lb = (3/4)*M
10lb*(4/3) = M
13.3 lb = M
(a) What is the efficiency of an out-of-condition professor who does 1.90 ✕ 105 J of useful work while metabolizing 500 kcal of food energy? % (b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of 25%? kcal
Answer:
a) The energy efficiency of the out-of-condition professor is 9.082 %.
b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.
Explanation:
a) The energy efficiency of the food metabolization ([tex]\eta[/tex]), no unit, is defined by following formula:
[tex]\eta = \frac{W}{E}\times 100\,\%[/tex] (1)
Where:
[tex]W[/tex] - Useful work, in joules.
[tex]E[/tex] - Food energy, in joules.
If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]E = 2.092\times 10^{6}\,J[/tex], the energy efficiency of the food metabolization is:
[tex]\eta = \frac{1.90\times 10^{5}\,J}{2.092\times 10^{6}\,J} \times 100\,\%[/tex]
[tex]\eta = 9.082\,\%[/tex]
The energy efficiency of the out-of-condition professor is 9.082 %.
b) If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]\eta = 25\,\%[/tex], then the quantity of food energy is:
[tex]E = \frac{W}{\eta}\times 100\,\%[/tex]
[tex]E = 1.90\times 10^{5}\,J\times \frac{100\,\%}{25\,\%}[/tex]
[tex]E = 7.60\times 10^{5}\,J[/tex]
[tex]E = 181.644\,kcal[/tex]
The food calories needed by the well-conditioned athlete is 181.644 kilocalories.
True or False: The forces applied by our muscles on our bones are usually several times larger than the forces we exert on the outside world with our limbs.
Answer:
True
Explanation:
This is because of the point where the forces are applied by our muscles and
the angle they have about the bones. Take for example the diagram I uploaded.
If we do a free body diagram and a sum of torques, we would get that:
[tex]F_{muscle}sin \theta r1 - mg r2 = 0[/tex]
In this case, mg is the same in magnitude as the force made by the hand to hold the ball, so:
[tex]F_{muscle}sin \theta r_{1} - F_{hand} r_{2} = 0[/tex]
If we solve the equation for the force of the muscle we would get that:
[tex]F_{muscle}=\frac{F_{hand}r_{2}}{r_{1}sin \theta}[/tex]
Since r2 is greater than r1 and the sin function can only return values that are less than 1, this means that the force of the muscle is much greater than the force used by the hand to hold the weight.
Let's use some standard values to prove this, let's say that r1=10cm, r2=35cm and theta=60 degrees. When inputing the values into the equation we get:
[tex]F_{muscle}=\frac{F_{hand}(35cm)}{(10cm)sin (60^{o})}[/tex]
which yields:
[tex]F_{muscle}=4.04 F_{hand}[/tex]
so in this example, the force made by the muscle is 4 times as big as the force exerted by the hand.
How far away should a Cliff be from a source of sound to give an echo in 5.3 second?(given speed of sound at 0°c=331 m/s
Answer:
Total distance traveled by the sound wave = 2d . Thus we can conclude that the cliff should be at a distance of 877.15 metres from the source of sound to meet the required conditions
Explanation:
this is what i found on the web
A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m. With what speed was it thrown, assuming negligible air resistance
The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.
What is the projectile motion?An item or particle that is propelled in a gravitational influence, such as from the crust of the Ground, and moves along a curved route while solely being affected by gravity is said to be in projectile motion.
A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m.
The initial velocity is zero. Then the time taken to reach the ground is given as,
h = ut + 1/2at²
- 24 = 0×t + 1/2 (-9.8)t²
24 = 4.9t²
t² = 4.8979
t = 2.213 seconds
Then the horizontal speed is given as,
v = 18 / 2.213
v = 8.133 meters per second
The velocity in the horizontal direction will not change. Then the horizontal speed will be 8.133 meters per second.
Learn more about projectile motion:
https://brainly.com/question/29761109
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The Sun is a type G2 star. Type G stars (from G0 to G9) have a range of temperatures from 5200 to 5900. What is the range of log(T) for G stars? Show your work
Answer:
log T = 3.72 to 3.77
Explanation:
Temperature range is
T = 5200 to 5900
Take the log
So,
log T = log 5200 to log 5900
log T = 3.72 to 3.77
A particle, mass 0.25 kg is at a position () m, has a velocity () m/s, and is subject to a force () N. What is the magnitude of the torque on the particle about the origin
A particle, mass 0.25 kg is at a position (-7i + 7j + 5k) m, has a velocity (6i - j + 4k) m/s, and is subject to a force (-5i + 0j - k) N. What is the magnitude of the torque on the particle about the origin?
Answer:
47.94Nm
Explanation:The torque (τ) on a particle subject to a force (represented as force vector F) at a position (represented as position vector r) about the origin is given by the cross product of the position vector r for the point of application of a force and the force F. i.e
τ = r x F
Given:
r = (-7i + 7j + 5k) m
F = (-5i + 0j - k) N
| i j k |
r x F = | -7 7 5 |
| -5 0 -1 |
r x F = i(-7 - 0) - j(7+25) + k(0+35)
r x F = i(-7) - j(32) + k(35)
r x F = -7i - 32j + 35k
Therefore the torque τ = -7i - 32j + 35k
The magnitude of the torque is therefore;
|τ| = [tex]\sqrt{(-7)^2 + (-32)^2 + (35)^2}[/tex]
|τ| = [tex]\sqrt{49 + 1024 + 1225}[/tex]
|τ| = [tex]\sqrt{2298}[/tex]
|τ| = 47.94Nm
The magnitude of the torque on the particle about the origin is 47.94Nm
you decide to work part time at a local supermarket. The job pays eight dollars and 60 per hour and you work 20 hours per week. Your employer withhold 10% of your gross pay federal taxes, 7.65% for FICA taxes, and 5% for state taxes
I guess that we want to find how much money you get each week.
We know that the job pays $8.60 per hour.
We know that you work 20 hours per week.
Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:
20*$8.60 = $172.
Now we know that your employer witholds:
10% + 7.65% + 5% = 22.65%
Then your employer withholds 22.65% of your gross pay.
if the 100% of your gross pay is $172
Then the 22.65% will be:
(22.65%/100%)*$172 = 0.2265*$172 = $38.96
This means that your employer withholds $38.96 of your weekly gross pay.
Then each week you get:
$172 - $38.96 = $133.04
If you want to learn more, you can read:
https://brainly.com/question/6692050
A 47-kg box is being pushed a distance of 8.0 m across the floor by a force whose magnitude is 188 N. The force is parallel to the displacement of the box. The coefficient of kinetic friction is 0.24. Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.
Answer:
(a) 1504 J
(b) - 884.35 J
(c) 0 J
(d) 0 J
Explanation:
Mass, m = 47 kg
displacement, s = 8 m
Force, F = 188 N
Coefficient of friction = 0.24
(a) Work done by applied force
W = F s = 188 x 8 = 1504 J
(b) Work done by the friction force
W' = - 0.24 x 47 x 9.8 x 8 = - 884.35 J
(c) Work done by the gravitational force
W''= m g s cos 90 = 0 J
(d) Work done by the normal force
W''' = m g scos 90 = 0 J
Give an example of how you could make a measurement that is, at the same time, very precise and very inaccurate
Answer:
Accuracy refers to how close a measurement is to the true or accepted value. ... Precision is independent of accuracy. That means it is possible to be very precise but not very accurate, and it is also possible to be accurate without being precise. The best quality scientific observations are both accurate and precise.
A plane flying horizontally at an altitude of 1 mi and a speed of 480 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.
Answer:
First remember that the distance between two points (a, b) and (c, d) is given by the equation:
[tex]d = \sqrt{(a - c)^2 + (b - d)^2}[/tex]
Now let's define the position of the radar as:
(0mi, 0mi)
Then we can write the position of the plane as:
(480mi/h*t, 1mi)
where t is time in hours.
Then we can write the distance equation as:
[tex]d(t) = \sqrt{(480\frac{mi}{h}*t - 0mi)^2 + (1mi -0mi)^2 } \\\\d(t) = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }[/tex]
Now we want to get:
the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station.
So first we want to find the value of t such that:
d(3) = 3mi
We will look at the positive value of t, because at this point the plane is increasing its distance to the station.
[tex]3mi = \sqrt{(480\frac{mi}{h}*t )^2 + (1mi)^2 }\\\\(3mi)^2 = (480\frac{mi}{h}*t )^2 + (1mi)^2\\\\9mi^2 - 1mi^2 = (480\frac{mi}{h}*t )^2\\\\8mi^2 = (230,400 mi^2/h^2)*t^2\\\\\\\sqrt{\frac{8mi^2}{230,400 mi^2/h^2} } = t = 0.0059 h[/tex]
The rate of change when the plane is 3 mi away from the station is:
d'(0.0059h)
remember that:
d'(t) = dd(t)/dt
We can write:
d(t) = h( g(t) )
such that:
h(x) = √x
g(t) = (480mi/h*t)^2 + (1mi)^2
then:
d'(t) = h'(g(t))*g'(t)
This is:
[tex]d'(t) = \frac{dd(t)}{dt} = \frac{1}{2}*\frac{2*t*480mi/h}{\sqrt{(480mi/h*t)^2 + (1mi)^2} }[/tex]
The rate of change at t = 0.0059h is then:
[tex]d'(0.0059h) = \frac{1}{2}*\frac{2*0.0059h*(480mi/h)^2}{\sqrt{(480mi/h*0.0059h)^2 + (1mi)^2} } =452.6 mi/h^2[/tex]
A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant angular acceleration, a of 6 rad/s If the initial angular velocity is 1.5 rad/s, determine The maximum angular velocity and linear velocity of the wheel after 1 complete revolution.
Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:
[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]
where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,
[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]
ωf = 8.8 rad/s
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
v = 2.2 m/s
A supertrain with a proper length of 100 m travels at a speed of 0.950c as it passes through a tunnel having a proper length of 50.0 m. As seen by a trackside observer, is the train ever completely within the tunnel? If so, by how much do the train’s ends clear the ends of the tunnel?
Answer:
19m
Explanation:
we have proper length L = 100m
the speed of the train v = 0.95
the speed of light is given as = 3x10⁸
length of the tunnel is given as = 50 meters
we can solve for the lenght contraction as
LX√1-v²/c²
= 100 * √1-(0.95*3x10⁸)²/(3x10⁸ )
= 31.22 metres
the train would be well seen at
50 - 31.22
= 18.78
= this is approximately 19 metres
we conclude tht the trains ends clears the ends of the tunnel by 19 meters.
thank you!
A series LR circuit contains an emf source of having no internal resistance, a resistor, a inductor having no appreciable resistance, and a switch. If the emf across the inductor is of its maximum value after the switch is closed, what is the resistance of the resistor
Answer:
b. 1.9 Ω
Explanation:
Here is the complete question
A series LR circuit contains an emf source of 14 V having no internal resistance, a resistor, a 34 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor? a. 1.5 ? b. 1.9 ? c. 5.0 ? d. 14 ?
Solution
The voltage across the inductor V is
[tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex] where V₀ = emf of source = 14 V, R = resistance, L = inductance = 34 H and t = time
Given that V = 80% of its maximum value after 4.0 s, this implies that V = 80 % of V₀ = 0.8V₀ and t = 4.0 s
Since [tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex] and V = 0.8V₀.
Since we need to find R, we make R subject of the formula, we have
[tex]V = V_{0}e^{-\frac{Rt}{L} }[/tex]
[tex]V/V_{0}= e^{-\frac{Rt}{L} }[/tex]
taking natural logarithm of both sides, we have
㏑(V/V₀) = -Rt/L
R = -L㏑(V/V₀)/t
Substituting the values of the variables into the equation, we have
R = -34㏑(0.8V₀/V₀)/4.0 s
R = -34㏑(0.8)/4.0 s
R = -34 × -0.2231/4.0 s
R = 7.587/4
R = 1.896 Ω
R ≅ 1.9 Ω
So, B is the answer
what does it mean to have an acceleration of 8m/s^2
A current of 1 mA flows through a copper wire. How many electrons will pass a point in each second?
Answer:
A current of 1ma flows through a copper wire, how many electron will pass a given point in one second? 1 Coulomb = 6.24 x 10^18 electrons (or protons)/1Sec which is also equal to 1 Amp/1 Sec. 1mA is 1/1000th of 1A so only 1/1000th of 6.24 x 10^18 electrons will pass a given point in 1 Sec.
A magnetic force acting on an electric charge in a uniform magnetic field what happend
Answer:
hgff
Explanation:
Answer:
The charge moves to equilibrium.
E.e = B.e.V
E is electric field force.
e is the charge.
B is magnetic field force.
V is acceleration voltage.
A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency is 890 Hz, the wavelength is .10m, and the amplitude is 6.5 mm. The tension in the line, in SI units, is closest to
Answer:
T = 712.9 N
Explanation:
First, we will find the speed of the wave:
v = fλ
where,
v = speed of the wave = ?
f = frequency = 890 Hz
λ = wavelength = 0.1 m
Therefore,
v = (890 Hz)(0.1 m)
v = 89 m/s
Now, we will find the linear mass density of the wire:
[tex]\mu = \frac{m}{L}[/tex]
where,
μ = linear mass density of wie = ?
m = mass of wire = 90 g = 0.09 kg
L = length of wire = 1 m
Therefore,
[tex]\mu = \frac{0.09\ kg}{1\ m}[/tex]
μ = 0.09 kg/m
Now, the tension in wire (T) will be:
T = μv² = (0.09 kg/m)(89 m/s)²
T = 712.9 N
A Man has 5o kg mass man in the earth and find his weight
Answer:
49 N
Explanation:
Given,
Mass ( m ) = 50 kg
To find : Weight ( W ) = ?
Take the value of acceleration due to gravity as 9.8 m/s^2
Formula : -
W = mg
W = 50 x 9.8
W = 49 N
friction between two flat surfaces can be divided into two categories. what are the two most common kinds of friction?
Answer:
kinetic and static
Explanation:
hope it helps! ^w^
Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen if the spacing between the first and second dark fringes is to be 6.0 mm
Answer:
The correct answer is "4.26 m".
Explanation:
Given:
Wavelength,
[tex]\lambda = 436.1 \ nm[/tex]
or,
[tex]=436.1\times 10^{-9} \ m[/tex]
Distance,
[tex]d = 0.31 \ mm[/tex]
or,
[tex]=0.31\times 10^{-3} \ m[/tex]
Distance between the 1st and 2nd dark fringes,
[tex](y_2-y_1) = 6\times 10^{-3} \ m[/tex]
As we know,
⇒ [tex]\frac{d}{L} (y_2-y_1) = \lambda[/tex]
or,
⇒ [tex]L=\frac{d(y_2-y_1)}{\lambda}[/tex]
By substituting the values, we get
[tex]=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}[/tex]
[tex]=\frac{0.31\times 6\times 10^3}{436.1}[/tex]
[tex]=\frac{1860}{436.1}[/tex]
[tex]=4.26 \ m[/tex]
plz help me with hw A bus of mass 1000 kg moving with a speed of 90km/hr stops after 6 sec by applying brakes then calculate the distance travelled and amount of force applied.
Answer:
Mass, M = 1000 kg
Speed, v = 90 km/h = 25 m/s
time, t = 6 sec.
Distance:
[tex]{ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}[/tex]
Force:
[tex]{ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}[/tex]
