Two people are dragging a canoe on a beach using light ropes. Person 1 applies a
force of 60.0N [Forward] and person 2 applies a force of 90.0N [Forward]. The
friction exerted by the beach on the canoe is 85.0N. Determine the net force of the
canoe.
a) 50.0 N [forward]
Ob) 6.50 N [backward]
c) 6.50 N [forward]
d) 65.0 N [forward]

Beats can be defined as the periodic fluctuations in the frequency of sound waves. That is option D

Sound waves are those waves that are produced by the vibration of an object whose energy is usually propagated through a medium.

When two sound waves of different frequencies meets, a periodic variation that occurs is called beats.

Therefore, Beats can be defined as the periodic fluctuations in the frequency of sound waves.

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The frequency the student detect from waves omitted from the fork and waves coming from the wall is 348.1 Hz.

The observed frequency is determined by applying Doppler effect;

f' = f₀(v + v₀)/(v)

where;

f' = 342(6 + 336)/(336)

f' = 348.1 Hz

Thus, the frequency the student detect from waves omitted from the fork and waves coming from the wall is 348.1 Hz.

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(a) The maximum potential difference across the resistor is 339.41 V.

(b) The maximum current through the resistor is 0.23 A.

(c) The rms current through the resistor is 0.16 A.

(d)  The average power dissipated by the resistor is 38.4 W.

Vrms = 0.7071V₀

where;

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

I(rms) = V(rms)/R

I(rms) = (240)/(1,540)

I(rms) = 0.16 A

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

P = I(rms) x V(rms)

P = 0.16 x 240

P = 38.4 W

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(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

Torque equation for frictional force;

[tex]f = (\frac{r}{R} T) - (\frac{I}{R^2} )a[/tex]

solve (1) and (2)

[tex]a = \frac{TR(R - r)}{I + MR^2}[/tex]

since the yoyo is pulled in vertical direction, T = mg [tex]a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2[/tex]

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

W = mg

W = 0.15 x 9.8 = 1.47 N

T = mg = 1.47 N

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

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Answer:

The long pointer measures altitude in intervals of 10,000 feet (2 = 20,000 feet). The short, wide pointer measures altitude in intervals of 1,000 feet (2 = 2,000 feet). The medium, thin pointer measures altitude in intervals of 100 feet (2 = 200 feet).

The work done on  both Elastic X and Elastic Y is 25.2 J.

The work done on the Elastic X is calculated as follows;

W = ¹/₂fx

where;

W = ¹/₂(42)(1.2)

W = 25.2 J

The work done on the Elastic Y is calculated as follows;

W = ¹/₂fx

where;

W = ¹/₂(24)(2.1)

W = 25.2 J

Thus, the work done on  both Elastic X and Elastic Y is 25.2 J.

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Answer:

122.17 m/s

Explanation:

x cos 40 = horizontal velocity

1500 m / x cos 40   = time in the air = 1958.11 / x

x sin 40 = vertical velocity

 find when shell vertical velocity = 0 (this is max height....1/2 way through its flight)  , the time when it hits the ground will be twice this...  

   0 =  x sin 40 - 9.8 t

  t =  x sin40 / 9.8        time in the air is twice this = .13118 x

Equate the two times from above to solve for x

1958.11/ x  =  .13118 x

x = 122.17 m/s

The boom makes an angle of θ = 60.2° with the vertical wall and the tension in the horizontal rope is mathematically given as

[tex]T_1=730.85 \mathrm{~N}[/tex]

[tex]T_1'=\frac{1980.51 \mathrm{~N}}{}[/tex]

Generally, the equation for is  mathematically given as

1) Tension in vertical rope,

[tex]T_{1}=74.5 \times 9.81\\T_1=730.85 \mathrm{~N}[/tex]

2) Tension in horizontal rope,

[tex]\sum{Mg} =0\\Mg\frac{l}{2} \sin \theta+T_{1} \frac{l}{2} \sin \theta &=T_{2} l \cos \theta \\[/tex]

[tex]T_{2} &=\frac{M g(l / 2) \sin \theta+T_{1}(l / 2) \sin \theta}{l \cos \theta} \\\\&=\frac{M g \tan \theta}{2}+\frac{T_{1} \tan \theta}{2} \\\\T_{2} &=\frac{142 \times 9.81 \tan 61.8}{2}+\frac{730.85 \times \tan 61.8}{2} \\\\=& 1299+681.51[/tex]

[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]

In conclusion, the tension in the horizontal rope is

[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]

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The middle ear provides impedance matching between air and fluid by the reduction in area between the tympanic membrane and the stapes footplate.

Middle ear is part of ear that separates the outer ear from the inner ear.

The middle ear consists of the following bones:

The middle ear provides impedance matching between air and fluid by

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The work done in dragging an object along a straight path is  173.2 J.

Work done is equal to product of force applied and distance moved.

Given You are trying to drag an object 5.0 m along a straight path. You are pulling with a force of 40 N along a rope that is inclined 30° to the horizontal.

Work = Force x Distance x cos(angle)

W= 40 x 5 x cos 30°

W = 173.2 Joules

Thus, the work done is  173.2 Joules

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The two forces are 12N and 10N (option B)

The net force is the force that has the same effect in magnitude and direction as the two forces acting together.

This problem can be solved using a right angled triangle;

Let the forces be P and Q

sin 40 = P/16

P= 16 sin 40

P = 10 N

Cos 40 = Q/16

Q = 16 cos 40

Q = 12 N

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The distance traveled by car A when they pass each other is 3,428.56 ft.

(Va)t - (Vb)t = d

where;

80t - (-60)t = 6000

80t + 60t = 6000

140t = 6000

t = 6000/140

t = 42.857 s

s = vt

s = 80 x 42.857

s = 3,428.56 ft

Thus, the distance traveled by car A when they pass each other is 3,428.56 ft.

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Answer:

And so calculating we get the maximum speed at each proton will reach To be 1.36 Times 10 to the four m the second

The reverberation time with 800 audiences is 0.875 seconds.

R₁V₁ = R₂V₂

where;

R₂ = R₁V₁/V₂

R₂ = (1.4 x 500) / 800

R₂ = 0.875 seconds

Thus, the reverberation time with 800 audiences is 0.875 seconds.

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The height of the building in meters, given the data is 108 m

The height of the building can be obtained as illustrated below:

h = ½gt²

h = ½ × 9.8 × 4.7²

h = 108 m

Thus, the height of the building is 108 m

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Rightward and upward

Whenever any object moves in one direction only, the motion is known as one-dimensional motion.

Two-dimensional motion is the movement of the object in two directions simultaneously.
Example: A ball thrown at an angle is a two-dimensional motion.

In two-dimensional motion, there are two axes used, generally the x-axis and the y-axis.

Generally, the x-axis is positive towards the rightward direction and negative towards the leftward direction.

Similarly, the y- axis is positive towards the upward direction and negative toward the downward direction.

So, the rightward and upward directions are chosen as positive.

Therefore:

When working with two-dimensional motion, choose the direction of  , the rightward and upward to be a positive direction.

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Some household appliances and their fuse ratings are given below:

A fuse is a device which is designed to protect an appliance by cutting off the electric circuit when current greater than its rating flows through it.

Some household appliances and their fuse ratings are given below:

In conclusion, a fuse is designed to protect an electrical appliance from excess current.

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Answer:

H = L sin θ / 2    height of center of mass

E = m g H = m g L sin θ / 2

I = m L^2 / 3        moment of inertia about end of rod

1/2 I ω^2 = m g L sin θ / 2      KE of rod when it hits surface

1/2 m L^2 / 3  ω^2 = m g L sin θ / 2

ω^2 = 3 g sin θ / L

ω = (3 g sin θ / L)^1/2      angular speed of rod striking surface

P = I α       if P = torque    

α = m g L/2 / (m L^2 / 3) = 3 g / (2 L)       angular acceleration of rod

Answer:

All people with the observant trait are often a steadying force that always wants to get things done. Their energy is very elegant in the sense of working on real things in real time.

Explanation:

Hope I helped

Answer:

5

Explanation:

Use the Pythagorean theorem. 3^2 + 4^2 = 25, which is the square of 5.

Brainliest, please :)

The direction of the vector would eventually depend on the horizontal component and vertical component which are 4 and 3 respectively. So, it would be calculated as 5 according to the Pythagoras theorem.

In physics, the direction of the vector may be defined as the orientation of the vector, that is, the angle it makes with the X-axis. A vector is drawn by a line with an arrow on the top and a fixed point at the other end.

The direction in which the arrowhead of the vector is directed gives the direction of the vector. The direction of the vector measures the angle it makes with a horizontal line.

According to the question,

The horizontal component of the vector = 4

The vertical component of the vector = 3

The direction of the vector =  square of the horizontal component + square of the vertical component.

                                             = [tex]4^2+3^2[/tex] = 16 + 9 = 25 i.e. = [tex]5^2.[/tex]

Therefore, the direction of the vector would eventually depend on the horizontal component and vertical component which are 4 and 3 respectively. So, it would be calculated as 5 according to the Pythagoras theorem.

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(a) The magnitude of the static friction force is 145 N

(b) The minimum possible value of static friction is 0.44.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

it is defined as the product of the coefficient of friction and normal reaction.

Given data;

The horizontal force exerted, F =145 N

Mass of crate,m =33.2 kg.

The velocity of the crate,v =0 m/sedc

The magnitude of the static friction force, [tex]\rm f_s[/tex] =? N

The minimum possible value of the coefficient of static friction between the crate and the floor is,

Normal reaction,N = mg

a)

The forces acting on the crate are balanced since it is at rest; as a result, the horizontal force is equal to the frictional force, f:

F = [tex]\rm f_s[/tex] = 145 N

(b)

The following equations provide the greatest allowable force of friction between the floor and the crate:

[tex]\rm f_{min}= \mu_s N \\\\ \rm f_{min}= \mu_s mg \\\\[/tex]

The force applied to the box must be less than or equal to the maximum force of friction in order for it to stay at rest.

[tex]\rm f \leq f_{min} \\\\ f \leq \mu_s mg \\\\ 145 \leq \mu_s (33.2 \times 9.81 ) \\\\ \mu_s \geq 0.44[/tex]

Hence, the magnitude of the static friction force and minimum possible value of the coefficient of static friction between the crate and the floor will be 145 N and 0.44.

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A. The force that causes the water on the lettuce to come off the lettuce and go to the walls of the bowl is centrifugal force.

Centrifugal force is an inertial force that appears to act on all objects when viewed in a rotating frame of reference.

This force is directed away from the center around which the body is moving.

This is force that acts on a body moving in a circular path and is directed towards the center around which the body is moving.

While centripetal force is directed towards to the center, the centrifugal force is directed away.

Thus, the force that causes the water on the lettuce to come off the lettuce and go to the walls of the bowl is centrifugal force.

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Centripetal force causes the water on the lettuce to come off the lettuce and go to the walls of the bowl.

Centripetal force is the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation. This force can move the body in the circular direction.

So we can conclude that Centripetal force causes the water on the lettuce to come off the lettuce and go to the walls of the bowl.

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The percentage decrease in the braking distance of the car is 19.5%.

v² = u² + 2as

where;

s = (v² - u²)/(2a)

v = 100%u - 10.3%u = 89.7% u = 0.897u

s = [(0.897u)² - u²]/(2a)

s = -0.195u²/2a

percent decrease = 0.195 x 100% = 19.5%

Thus, the percentage decrease in the braking distance of the car is 19.5%.

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The mouse will be 4.61 meters distant horizontally. and the owl won't fall into the nest.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

Diameter of nest,d = 30 cm

The velocity of flying, V =  3.50 m/s toward the east

The angle of flying below the horizontal, Θ = 32°

Position of owl,12.0 m above the middle of the 30 cm-diameter nests, at 4.00 m west.

Height of owl above the ground,h = 12 m

Distance from the west direction, a = 4.00 m

The vertical component of velocity,vₐ

The horizontal component of velocity,vₓ

After covering a vertical distance of 12 m, the vertical component of the mouse's velocity, v, is given by

vₐ-u²=2gh

u = asinΘ

Substitute the above value;

vₐ²- (a sin Θ )² = 2gh

Substitute the given value;

vₐ² - (4 sin 32)² = 2 ×9.81 ×12

vₐ² - (4 × 0.53)² = 235.44

vₐ² - 4.494 = 235.44

vₐ² = 235.44 + 4.494

vₐ² = 239.934

vₐ = 15.49 m/s

The time required to descend the aforementioned 12 meters;

[tex]\rm t = \frac{v_{rel}}{g}[/tex]

t = (vₐ - u )/g

t =( vₐ - a sin Θ )/g

Substitute the given value;

t =(15.49 - 4 ×sin 32°) / 9.81

t= (15.49 - 4 * 0.53) / 9.81

t =(15.49 - 2.12) / 9.81

t = 13.37 / 9.81

t= 1.36 s

Horizontal distance traveled for the given period;

x=uₓ × t

x =a cosΘ × t

Substitute the given value;

x = 4×(cos 32°)×1.36

x= 4×0.848×1.36

x= 4.61 m

Hence, the mouse will be 4.61 meters distant in the horizontal direction. and the owl won't fall into the nest,

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The gravitational attraction between electron and proton is 10−40 whereas electrostatic force of attraction between a proton and an electron is 10-8.

The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10−40 while on the other hand, the electrostatic force of attraction between a proton and an electron in a hydrogen atom is 10- 8 which is 9 times.

The electric charge of the electron and proton are the same i.e. -1.60x10-19C whereas their gravitational force is different due to difference in mass.

So we can conclude that gravitational attraction between electron and proton is 10−40 whereas electrostatic force of attraction between a proton and an electron is 10-8.

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The tension in the string marked A is determined as 331.35 N.

tanθ = Δy/Δx

tanθ = (6 - 1)/(7 - 1) = 0.8333

θ = arc tan(0.8333) = 39.81⁰

with respect to horizontal = 90 - 39.81 = 50.19⁰

tanθ = Δy/Δx

tanθ = (9 - 6)/(7 - 5) = 1.5

θ = arc tan(1.5) = 56.31 ⁰

tanθ = Δy/Δx

tanθ = (6 - 4)/(14 - 7) = 0.2857

θ = arc tan(0.2857) = 15.95 ⁰

Bcosθ + Aosθ = Ccosθ

Bcos(56.31) + A[cos(50.19)] = 56.3cos(15.95)

0.55B + 0.64A = 54.13 ----- (1)

Bsinθ + Asinθ = Csinθ

Bsin(56.31) + A[sin(50.19)] = 56.3sin(15.95)

0.832B + 0.77A = 15.47---- (2)

Solve (1) and (2)

0.2A = 66.27

A = 66.27/0.2

A = 331.35 N

Thus, the tension in the string marked A is determined as 331.35 N.

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The average acceleration of snowboarder's is 0.19 m/s² and distance travel by snowboarder is 6.70 metres.

We have given that snowboarder start moving from rest means

Initial speed (u) = 0 m/s

Final speed (v) = 1.6 m/s

time taken (t) = 8.4 s

Applying first equation of motion

v = u + at

1.6 = 0 + a×8.4

1.6 = 8.4a

a = 1.6/8.4 m/s²

a = 0.19 m/s²

So average acceleration is 0.19 m/s².

Now Applying second equation of motion

s = ut + 1/2at²

s = (0) × (8.4) + (1/2)(0.19)(8.4)² metre

s = (1/2)(0.19)(8.4)²  metre

s = 6.70 metre

So that distance covered by snowboarder is 6.70 metre.

So we find out the average acceleration by using first equation of motion and the average acceleration came out to be 0.19 m/s² and to find out the distance covered by snowboarder we use second equation of motion and the distance came out to be 6.70 metre.

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Electric current is the vector quantity it has both magintude and direction.

The pace at which electrons travel through a conductor is known as electric current. The ampere is the SI unit for electric current. Electrons are tiny particles that reside within a substance's molecular structure.

Because it has both a magnitude and a direction, the current is a vector. However, a vector always abides by the law of vector addition.

Current is a scalar since it follows algebraic addition and doesn't obey it. Additionally, the current is represented as the dot product of the area vector and the current density vector.

Hence, electric current is the vector quantity it has both magnitude and direction.

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The moment of inertia of this set of objects with respect to the axis perpendicular to the the x-y plane passing through location x = 4.00 m and y = 3.00 m is 144.97 Kg.m^2.

The moment of inertia is the amount of rotation obtained  by an object when it is in state of motion or rest.

Three small objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 1.09 kg, mB = 2.83 kg, mC = 3.39 kg.

The coordinates  of Ball A :(2,1) Ball B :(8,2) and Ball C: (5,8) and axis is at (4,3)

Here, for the moment of inertia of each ball

For ball A

IA = 1.09 x ((4 - 2)^2 + (3-1)^2)

IA = 8.72 Kg.m^2

for the ball B

IB = 2.83 * ((4 - 8)^2 + (3 - 2)^2)

IB = 48.11 Kg.m^2

for the ball C

IC = 3.39 * ((4 - 5)^2 + (3 - 8)^2)

IC = 88.14 Kg.m^2

Total moment of inertia = 8.72 + 48.11 +  88.14

Total moment of inertia = 144.97 Kg.m^2

Thus, the moment of inertia of this set of objects with respect to the axis perpendicular to the the x-y plane passing through location x = 4.00 m and y = 3.00 m is 144.97 Kg.m^2.

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Answer:

Current = 5.45amps

Power = 654 watts

Explanation:

E =120V

I =?

R = 22.02 Ohms

I= E/R

I= 120/22.02

I = 5.45AmPs

P = ?                    P= E x R

E = 120V              P= 120x5.45

I = 5.45 AmPs      P=  654W