In 1271, Marco Polo departed Venice and traveled to Kublai Khan's court near Beijing, approximately 7900 km away in a direction we will call positive. Assume that the Earth is flat (as some did at the time) and that the trip took him 4.0 years, with 365 days in a year. (a) What was his average velocity for the trip, in meters per second? (b) A 767 could make the same trip in about 9.0 hours. What is the average velocity of the plane in meters per second?

Answer:

The distance of c is 56.57

Explanation:

Given that,

Horizontal distance b = 50.071

Angle C = 90.286°

Angle B = 62.253°

We need to calculate the distance of c

Using sine rule

[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]

[tex]\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]

Put the value into the formula

[tex]\dfrac{50.071}{\sin62.253 }=\dfrac{c}{\sin90.286}[/tex]

[tex]c= \dfrac{50.071\times\sin90.286}{\sin62.253}[/tex]

[tex]c=56.575[/tex]

Hence, The distance of c is 56.575.

Answer:

221754385964.9123

Explanation:

Convert miles to nanometer

1 mile = 1.6 km

1 km = 1×10³×10³×10³×10³ nm

1 mile = 1.6×10¹² nm

So,

158 miles = 158×1.6×10¹² = 252.8×10¹² nm

Length of each molecule = 1140 nm

Number of molecules = Total length / Length of each molecule

[tex]\text{Number of molecules}=\frac{252.8\times 10^{12}}{1140}\\\Rightarrow \text{Number of molecules}=221754385964.9123[/tex]

There are 221754385964.9123 number of molecules in a stretch of 158 miles

Answer:

Force between two equal charges will be 608.4 N

Explanation:

We have given charges [tex]q_1=0.65C\ and\ q_2=0.65C[/tex]

Distance between the charges = 2.5 km = 2500 m

According to coulombs law force between two charges is given by

[tex]F=\frac{1}{4\pi \varepsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}[/tex], here K is constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]

So force [tex]F=\frac{9\times 10^9\times 0.65\times 0.65}{2500^2}=608.4N[/tex]

Answer:

[tex]E_{k}=1589.5ftlb[/tex]  

Explanation:

[tex]E_{k}=E_{movement}+E_{rotational}\\[/tex]    

[tex]E_{k}=\frac{1}{2}mv^{2}+\frac{1}{2}Iw^{2}[/tex]     (1)

For this wheel:

[tex]w=\frac{v}{r}[/tex]

[tex]I=mr^{2}[/tex]:    inertia of a ring

We replace (2) and (3) in (1):

[tex]E_{k}=\frac{1}{2}mv^{2}+\frac{1}{2}(mr^{2})(\frac{v}{r})^{2}=mv^{2}=5.5*17^{2}=1589.5ftlb[/tex]  

Answer:

The two joggers will pass each other after 1 minute and 4 seconds at 3:01:04 PM.

Explanation:

The situation is analogous to two joggers running in opposite direction in a straight line where one jogger starts at the beginning of the line and the other starts at the other end, 300 m ahead.

The equation for the position of the joggers will be:

x = x0 + v · t

Where:

x = position of the jogger at time t

x0 = initial position

v = velocity

t = time

When the joggers pass each other, their position will be the same. Let´s find at which time both joggers pass each other:

x jogger 1 = x jogger 2

0 m + 2.15 m/s · t = 300 m - 2.55 m/s · t

(notice that the velocity of the joggers has to be of opposite sign because they are running in opposite directions).

2.15 m/s · t + 2.55 m/s · t = 300 m

4.70 m/s · t = 300 m

t = 300 m / 4.70 m/s = 63.8 s

The two joggers will pass each other after 1 minute and 4 seconds at 3:01:04 PM.

Final answer:

The magnitude of the magnetic force on an electron moving with a speed of 5.0 × [tex]10^4[/tex] m/s perpendicular to a magnetic field of 0.20 Tesla is calculated using the formula F = qvB, resulting in a force of 1.6 × [tex]10^{-15}[/tex] Newtons.

Explanation:

The magnetic force on an electron moving perpendicular to a magnetic field can be calculated using the formula F = qvB, where F is the magnetic force, q is the charge of the electron (-1.6 × [tex]10^{-19}[/tex] C), v is the velocity of the electron, and B is the magnetic field strength. Given that the electron moves with a speed of  5.0 × [tex]10^4[/tex] m/s perpendicular to a uniform magnetic field of 0.20 T, we use the formula to find the magnitude of the force:

F = (1.6 ×[tex]10^{-19}[/tex]C)( 5.0 × [tex]10^4[/tex] m/s)(0.20 T) =1.6 ×[tex]10^{-19}[/tex]C× 104 m/s × 2 × [tex]10^{-1}[/tex] T

F = 1.6 ×[tex]10^{-15}[/tex] N

The magnitude of the magnetic force on the electron is  1.6 ×[tex]10^{-15}[/tex] Newtons.

Answer:

If the potential is zero , the electric field could be different to zero

Explanation:

The relation between the electric field and the potential is:

=−∇

∇: gradient operator

If the electric potential, , is zero at one point but changes in the neighbourhood of this point, then the Electric field, , at that point is different from zero.

Answer:

[tex]Q=4.0*10^{-9}C[/tex]

Explanation:

Electric field of a charge:

[tex]E=k*\frac{Q}{R^{2}}[/tex]

[tex]Q=\frac{E*R^{2}}{K}=1.7*4.6^{2}/(9*10^{9})=4.0*10^{-9}C[/tex]

Answer:

V=37.5miles/h

Explanation:

For convenience, let's convert the average speed to miles per minute:

V=7miles/h * 1h/60min = 0.1167 miles/min.

The distance traveled during the first 15 min was:

D = V*t = 1.75 miles   So, the remaining distance is 6.25miles.

Since you only have 10min left:

Vr = Dr / tr = 6.25 / 10 = 0.625 miles / min. If we take that to miles per hour we get final answer:

Vr = 0.625 miles/min * 60min/1h = 37.5 miles/h

Answer:

Explanation:

We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector

D₁ = - 255 cos 49 i  + 255 sin49 j

= - 167.29 i + 192.45 j

Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is

D = 125 i

So

D₁ + D₂ = D

- 167.29 i + 192.45 j + D₂ = 125 i

D₂ = 125 i + 167.29 i - 192.45 j

= 292.29 i - 192.45 j

Angle of D₂ with x axes θ

tan θ = -192.45 / 292.29

= - 0.658

θ = 33.33 south of east

Magnitude of D₂

D₂² = ( 192.45)² + ( 292.29)²

D₂ = 350 km approx

Tan

Answer:

The magnetic field at the center of a circular loop is [tex]3.14\times10^{-5}\ T[/tex].

Explanation:

Given that,

Radius = 4.0 cm

Current = 2.0 A

We need to calculate the magnetic field at the center of a circular loop

Using formula of magnetic field

[tex]B = \dfrac{I\mu_{0}}{2r}[/tex]

Where, I = current

r = radius

Put the value into the formula

[tex]B =\dfrac{2.0\times4\pi\times10^{-7}}{2\times4.0\times10^{-2}}[/tex]

[tex]B =0.00003141\ T[/tex]

[tex]B=3.14\times10^{-5}\ T[/tex]

Hence, The magnetic field at the center of a circular loop is [tex]3.14\times10^{-5}\ T[/tex].

Answer:

0.5 rad / s

Explanation:

Moment of inertia of the disk I₁ = 1/2 MR²

M is mass of the disc and R is radius

Putting the values in the formula

Moment of inertia of the disc  I₁  = 1/2 x 100 x 2 x 2

= 200 kgm²

Moment of inertia of man about the axis of rotation of disc

mass x( distance from axis )²

I₂  = 40 x 1.25²

= 62.5 kgm²

Let ω₁ and ω₂ be the angular speed of disc and man about the axis

ω₂ = tangential speed / radius of circular path

= 2 /1.25 rad / s

= 1.6 rad /s

ω₁ = ?

Applying conservation of angular moment ( no external torque is acting on the disc )

I₁ω₁ = I₂ω₂

200 X ω₁ = 62.5 X 1.6

ω₁ =  0.5 rad / s

Answer:

The distance traveled by Tina before passing David is 900 m

Given:

Initial speed of David, [tex]u_{D} = 30 m/s[/tex]

Acceleration of Tina, [tex]a_{T} = 2.0 m/s^{2}[/tex]

Solution:

Now, as per the question, we use 2nd eqn of motion for the position of David after time t:

[tex]s = u_{D}t + \frac{1}{2}at^{2}[/tex]

where

s = distance covered by David after time 't'

a = acceleration of David = 0

Thus

[tex]s = 30t[/tex]

Now, Tina's position, s' after time 't':

[tex]s' = u_{T}t + \frac{1}{2}a_{T}t^{2}[/tex]

where

[tex]u_{T} = 0[/tex], initially at rest

[tex]s' = 0.t + \frac{1}{2}\times 2t^{2}[/tex]

[tex]s' = t^{2}[/tex]                     (1)

At the instant, when Tina passes David, their distances are same, thus:

s = s'

[tex]30t = t^{2}[/tex]

[tex]t(t - 30) = 0[/tex]

t = 30 s

Now,

The distance covered by Tina before she passes David can be calculated by substituting the value t = 30 s in eqn (1):

[tex]s' = 30^{2}[/tex] = 900 m

The distance covered by Tina before passing David at an acceleration rate of  2 m/s² is 900 meters.

Given to us

Velocity of David, v = 30 m/s

Acceleration of Tina, a = 2 m/s²

Let the time taken by Tina pass David is t.

According to the given information, the distance traveled by Tina will be the same as the distance traveled by David between Tina when she was at rest and when Tina passes her.

Distance traveled by Tina = Distance traveled by David

Distance traveled by David,

[tex]s = v \times t\\\\ = 30 \times t =30t[/tex]

Using the second equation of Motion

[tex]s= ut +\dfrac{1}{2}at^2[/tex]

Substitute,

[tex]30t= (0)t +\dfrac{1}{2}(2)t^2[/tex]

[tex]t = 30\rm\ sec[/tex]

Thus, the time taken by Tina to pass David is 30 seconds.

We have already discussed,

Distance traveled by Tina = Distance traveled by David,

therefore,

[tex]s = v \times t\\\\ = 30 \times 30 =900\rm\ meters[/tex]

Hence, the distance covered by Tina before passing David at an acceleration rate of  2 m/s² is 900 meters.

Learn more about the Equation of motion:

https://brainly.com/question/5955789

Answer:

21.75 m

Explanation:

t = Time taken for the car to slow down = 0.75 s

u = Initial velocity = 50 m/s

v = Final velocity = 8 m/s

s = Displacement

a = Acceleration

Equation of motion

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{8-50}{0.75}\\\Rightarrow a=-56\ m/s^2[/tex]

Acceleration is -56 m/s²

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{8^2-50^2}{2\times -56}\\\Rightarrow s=21.75\ m[/tex]

The distance covered in the 0.75 seconds is 21.75 m

Answer:

[tex]6.70\times 10^{-13}\ m[/tex]

Explanation:

Given:

Assumptions:

WE know that the radius of the [tex]n^{th}[/tex] orbit of an atom is given by:

[tex]r = \dfrac{n^2h^2}{4\pi^2kZe^2m}\\[/tex]

Let us find out the radius of the 1st orbit of the gold atom for which n = 1 and Z = 79.

[tex]r = \dfrac{n^2h^2}{4\pi^2kZe^2m}\\\Rightarrow r = \dfrac{(1)^2(6.62\times 10^{-34})^2}{4\pi^2\times 9\times 10^9\times 79\times (1.6\times 10^{-19})^2\times 9.1\times 10^{-31}}\\\Rightarrow r =6.70\times 10^{-13}\ m[/tex]

Answer:

[tex]\mu=180\times 10^{-3}A-m^2[/tex]      

Explanation:

Given that,

Area of the loop, [tex]A=10^{-3}\ m^2[/tex]

Current flowing in the wire, I = 180 A

We need to find the magnetic moment of the current loop. It is given by :

[tex]\mu=I\times A[/tex]

[tex]\mu=180\times 10^{-3}[/tex]      

[tex]\mu=180\times 10^{-3}A-m^2[/tex]      

So, the magnetic moment of the current loop is [tex]180\times 10^{-3}A-m^2[/tex]. Hence, this is the required solution.

Answer:0.931 s

Explanation:

Given

initial speed=1.1 m/s

height(h)=28 m

after 0.5 sec blue ball is thrown upward

Velocity of blue ball is 24.4 m/s

height with which blue ball is launched is 0.9 m

Total distance between two balls is 28-0.9=27.1 m

Let in t time red ball travels a distance of x m

[tex]x=1.1t+\frac{gt^2}{2}[/tex] --------1

for blue ball

[tex]27.1-x=24.4t-\frac{g(t-0.5)^2}{2}[/tex] -----2

Add 1 & 2

we get

[tex]27.1=24.4t+1.1t+\frac{g(2t-0.5)(0.5)}{2}[/tex]

[tex]27.1=25.5t+g\frac{4t-1}{8}[/tex]

t=0.931 s

after 0.931 sec two ball will be at same height

Answer:

Explanation: This is done using the equation:

[tex]\frac{4}{3} π R^{3}[/tex]

Because the Radius is a know value. We have the following.

[tex]\frac{4}{3} π (10mm)^{3}[/tex]

Which is:

4188.7902 mm

Answer:

The rock will rise 3.3 m above the top of the window.

Explanation:

The equations used to find the height and velocity of the rock at any given time are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the rock at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

v = velocity of the rock at time t

If we place the frame of reference at the bottom of the window, we can say that at time t = 0.19 s the height of the rock is 1.7 m. That will allow us to find the initial velocity needed to find the time at which the rock is at its maximum height.

y = y0 + v0 · t + 1/2 · g · t²

1.7 m = 0 m + v0 · 0.19 s - 1/2 · 9.8 m/s² · (0.19 s)²

1.7 m + 1/2 · 9.8 m/s² · (0.19 s)²  = v0 · 0.19 s

(1.7 m + 1/2 · 9.8 m/s² · (0.19 s)²) / 0.19 s = v0

v0 = 9.9 m/s

With the initial velocity, we can find at which time the rock reaches its max- height. We know that at maximum height, the velocity of the rock is 0. Then, using the equation of velocity:

v = v0 + g · t

0 = 9.9 m/s - 9.8 m/s² · t

-9.9 m/s / -9.8 m/s² = t

t = 1.0 s

Now calculating the position at time t = 1.0 s, we will find the maximum heigth:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 9.9 m/s · 1.0 s - 1/2 · 9.8 m/s² · (1.0 s)²

y = 5.0 m  (this is the max-height meassured from the bottom of the window)

Then, the rock will rise (5.0 m - 1.7 m) 3.3 m above the top of the window.

Answer: A little more that 5 Kg for a healthy person

Explanation: First, we know the following:

The regular adult has from 9 to 12 pints of blood. This is around 5 liters for a healthy male adult.

The human body is composed mostly on water, around 80%.

Blood is mostly composed on plasma, which makes blood thicker than water.

Knowing that, almost all the body is compose of water, it is safe to think that blood density should be near to that of water but higher.

The density on water is a know value. Which makes the following true:

1 Liter of Water weights 1 Kg

It could be said then, that the total mass of blood for a healthy person should be a little more that 5 kgs.

Answer:

Force, F = −229.72 N

Explanation:

Given that,

First charge particle, [tex]q_1=-6.29\times 10^{-6}\ C[/tex]

Second charged particle, [tex]q_2=5.23\times 10^{-6}\ C[/tex]

Distance between charges, d = 0.0359 m

The electric force between the two charged particles is given by :

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{-6.29\times 10^{-6}\times 5.23\times 10^{-6}}{(0.0359)^2}[/tex]

F = −229.72 N

So, the magnitude of force that one particle exerts on the other is 229.72 N. Hence, this is the required solution.

Answer:

[tex]B=2.01 \times 10^{-5}\ T[/tex]

Explanation:

Distance between plates = 0.8 cm

Distance from one plate = 1 mm

Current density (J)= 16 A/m

Currents are flowing in opposite direction.

[tex]\mu _o=4\pi \times 10^{-7}[/tex]

When current is flowing in opposite direction then magnetic field given as

[tex]B=\dfrac{\mu _oJ}{2}+\dfrac{\mu _oJ}{2}[/tex]

[tex]B=\mu _oJ[/tex]

Now by putting the values we get

[tex]B=4\pi \times 10^{-7}\times 16[/tex]

[tex]B=2.01 \times 10^{-5}\ T[/tex]

The magnitude of the magnetic field between the plates at the given point is 2.011 x 10⁻⁵ T.

The magnitude of magnetic field between the plates due to the current flowing in opposite directions is determined by using the following formula;

B = μ₀J/2 + μ₀J/2

B = μ₀J

where;

Substitute the given parameters and solve for the magnetic field as follows;

B = (4π x 10⁻⁷) x (16)

B = 2.011 x 10⁻⁵ T

Thus, the magnitude of the magnetic field between the plates at the given point is 2.011 x 10⁻⁵ T.

Learn more about magnetic field here: https://brainly.com/question/7802337

Answer:

a) [tex]v_{1}=14.29m/s\\v_{2}=9.25m/s\\v_{3}=6.36m/s[/tex]

b) [tex]v=+9.97m/s[/tex]

Explanation:

From the exercise we know that

[tex]x_{1} =15m, t_{1}=3s[/tex]

[tex]x_{2} =-3m, t_{1}=1.74s[/tex]

[tex]x_{3} =29m, t_{3}=5.20s[/tex]

From dynamics we know that the formula for average velocity is:

[tex]v=\frac{x_{2}-x_{1}  }{t_{2}-x_{1}  }[/tex]

a) For the three intervals:

[tex]v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s[/tex]

[tex]v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s[/tex]

[tex]v_{3}=\frac{x_{3}-x_{1}  }{t_{3}-t_{1}  }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s[/tex]

b) The average velocity for the entire motion can be calculate by the following formula:

[tex]v=\frac{v_{1}+v_{2}+v_{3}   }{n} =\frac{(14.29+9.25+6.36)m/s}{3}=+9.97m/s[/tex]

Answer:

Vf = 10.76 m/s

Explanation:

Train kinematics

The train moves with uniformly accelerated movement

[tex]V_f = V_o + a*t[/tex] Formula (1)

Vf: Final speed (m/s)

V₀: Inital speed (m/s)

t: time in seconds (s)

a: acceleration (m/s²)

Movement from t = 0 to t = 5.2s

We replace in formula (1)

4.6 = 0 + a*5.2

a = 4.6/5.2 = 0.88 m/s²

Movement from t = 5.2s to t = 5.2s + 7s = 12.2s

We replace in formula (1)

[tex]V_f = 4.6 + 0.88*7[/tex]

Vf = 10.76 m/s

Answer:

25 times greater

Explanation:

Let the mass of the car is m

Initial speed, u = 20 km/h = 5.56 m/s

Final speed, v = 100 km/h = 27.78 m/s

The formula for the kinetic energy is given by

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

So, initial kinetic energy

[tex]K_{i} = \frac{1}{2}m(5.56)^{2}[/tex]

Ki = 15.466 m

final kinetic energy

[tex]K_{f} = \frac{1}{2}m(27.78)^{2}[/tex]

Kf = 385.86 m

Increase in kinetic energy is given by

= [tex]\left ( \frac{K_{f}}{K_{i}} \right )[/tex]

= 385.86 / 15.466 = 25

So, the kinetic energy is 25 times greater.

The kinetic energy of the car increases by a factor of 25.

The increase in kinetic energy of the car can be determined by comparing the initial kinetic energy to the final kinetic energy. Kinetic energy is directly proportional to the square of velocity.

In this case, the velocity is increased from 20 km/hr to 100 km/hr. Let's calculate the ratio of the final kinetic energy to the initial kinetic energy.

The initial kinetic energy is given by 1/2 * (mass of the car) * (initial velocity)^2, and the final kinetic energy is given by 1/2 * (mass of the car) * (final velocity)^2.

Let's substitute the values and calculate the ratio:

Ratio = (1/2 * (mass) * (final velocity)^2) / (1/2 * (mass) * (initial velocity)^2) = (final velocity)^2 / (initial velocity)^2.

Substituting the numbers, Ratio = (100 km/hr)^2 / (20 km/hr)^2 = 10000 / 400 = 25.

Therefore, the factor by which the kinetic energy of the car increases is 25 times greater.

Answer:

370.6 nm

Explanation:

wavelength in vacuum = 494 nm

refractive index of water with respect to air = 1.333

Let the wavelength of light in water is λ.

The frequency of the light remains same but the speed and the wavelength is changed as the light passes from one medium to another.

By using the definition of refractive index

[tex]n = \frac{wavelength in air}{wavelength in water}[/tex]

where, n be the refractive index of water with respect to air

By substituting the values, we get

[tex]1.333 = \frac{494}{\lambda }[/tex]

λ = 370.6 nm

Thus, the wavelength of light in water is 370.6 nm.

Answer:

Part a)

[tex]W = 1.38 eV[/tex]

Part b)

[tex]\lambda = 901.22 nm[/tex]

Part c)

[tex]KE = 1.2 eV[/tex]

Explanation:

As we know by Einstein's equation of energy that

incident energy of photons = work function of metal + kinetic energy of electrons

here we know that incident energy of photons is given as

[tex]E = \frac{hc}{\lambda}[/tex]

[tex]E = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{480 \times 10^{-9}}[/tex]

now we have

[tex]E = 4.125 \times 10^{-19} J[/tex]

[tex]E = 2.58 eV[/tex]

kinetic energy of ejected electrons = qV

so we have

[tex]KE = e(1.2 V) = 1.2 eV[/tex]

Part a)

now we have

[tex]E = KE + W[/tex]

[tex]2.58 = 1.2 + W[/tex]

[tex]W = 1.38 eV[/tex]

Part b)

in order to find cut off wavelength we know that

[tex]W = \frac{hc}{\lambda}[/tex]

[tex]1.38 eV = \frac{1242 eV-nm}{\lambda}[/tex]

[tex]\lambda = 901.22 nm[/tex]

Part c)

Maximum energy of ejected electrons is the kinetic energy that we are getting

the kinetic energy of electrons will be obtained from stopping potential

so it is given as

[tex]KE = 1.2 eV[/tex]

Answer:

The moment is 81.102 k N-m in clockwise.

Explanation:

Given that,

Force = 260 N

Side = 580 mm

Distance h = 370 mm

According to figure,

Position of each point

[tex]O=(0,0)[/tex]

[tex]A=(0,-b)[/tex]

[tex]B=(h,0)[/tex]

We need to calculate the position vector of AB

[tex]\bar{AB}=(h-0)i+(0-(-b))j[/tex]

[tex]\bar{AB}=hi+bj[/tex]

We need to calculate the unit vector along AB

[tex]u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}[/tex]

[tex]u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}[/tex]

We need to calculate the force acting along the edge

[tex]\hat{F}=F(u_{AB})[/tex]

[tex]\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})[/tex]

We need to calculate the net moment

[tex]\hat{M}=\hat{F}\times OA[/tex]

Put the value into the formula

[tex]\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})[/tex]

[tex]\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))[/tex]

[tex]\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})[/tex]

[tex]\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}[/tex]

Put the value into the formula

[tex]\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}[/tex]

[tex]\hat{M}=-81.102\ \hat{k}\ N-m[/tex]

Negative sign shows the moment is in clockwise.

Hence, The moment is 81.102 k N-m in clockwise.

The moment will be  81.102 k N-m in a clockwise direction. The moment is used to rotate or twist the object.

The moment is defined as the product of the force and the perpendicular distance from the pivot point. Its unit is KN-m.

The given data in the problem is;

F is the Force = 260 N

b is the Side = 580 mm

h is the distance = 370 mm

Position of the points is found by;

O(0,0)

A(0,-b)

B(h,0)

The position vector for the AB will be;

[tex]\vec AB = (h-0)+ (0-(-b))j \\\\ \vec AB =h \vec i + b \vec j[/tex]

The unit vector along with AB

[tex]\rm u_AB = \frac{\vec AB }{|\vec AB|} \\\\ \rm u_AB = \frac{h \vec i + b \vec j}{\sqrt{h^2+b^2} }[/tex]

The net moment is found by;

[tex]\hat M = \hat F \times OA \\\\ \hat M =F \frac{h \vec i + b \vec j}{\sqrt{h^2+b^2} }\times (b \vec j) \\\\ \hat M =\frac{F}{\sqrt{h^2+b^2}} \times (bh \hat k) \\\\ \hat M =- \ \frac{bhf}{\sqrt{h^2+b^2}}[/tex]

[tex]\hat M =- \ \frac{bhf}{\sqrt{h^2+b^2}} \\\\ \hat M =- \ \frac{580 \times 10^-3 \times 370 \times 10^-3 \times 260 }{\sqrt{(370\times 10^-3)^2+(580\times 10^-3)^2}} \\\\ \hat M =- 81.102 \ KNm[/tex]

-ve sign shows that moment is clockwise.

Hence the moment will be  81.102 k N-m in a clockwise direction.

To learn more about the moment refer to the link;

https://brainly.com/question/6278006

Answer:

The angle θ is 6.1° below the horizontal.

Explanation:

Please, see the figure for a description of the situation.

The vector "r" gives the position of the ball and can be expressed as the sum of the vectors rx + ry (see figure).

We know the magnitude of these vectors:

magnitude rx = 11.6 m

magnitude ry = 2.23 m - 0.99 m = 1.24 m

Then:

rx = (11. 6 m, 0)

ry = (0, -1.24 m)

r = (11.6 m + 0 m, 0 m - 1.24 m) = (11.6 m, -1.24 m)

Using trigonometry of right triangles:

magnitude rx = r * cos θ = 11. 6 m

magnitude ry = r * sin θ = 2.23 m - 0.99 = 1.24 m

where r is the magnitude of the vector r

magnitude of vector r:

[tex]r = \sqrt{(11.6m)^{2} + (1.24m)^{2}} = 11.667m[/tex]

Then:

cos θ = 11.6 m / 11.667 m

θ = 6.1°

Using ry, we should obtain the same value of θ:

sin θ = 1.24 m/ 11.667 m

θ = 6.1°

( the exact value is obtained if we do not round the module of r)

Answer: in air 33.33 ps and in the glass 50 ps: so the difference 16.67 ps

Explanation: In order to calculate the time for a pulse travellin in air  and in a glass we have to consider the expresion of the speed given by:

v= d/t  v the speed in a medium is given by c/n where c and n are the speed of light and refractive index respectively.

so the time is:

t=d/v=d*n/c

in air

t=0.01 m*1/3*10^8 m/s= 33.33 ps

while for the glass

t=0.01 m*1.5* 3* 10^8 m/s= 50 ps

Finally the difference is (50-33.33)ps = 16.67 ps