Explanation:
ehb-pynw-ayo
joi n fast
Ocean waves of wavelength 26 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.
Required:
How far from the wall is she? What is the period of her up-and-down motion?
Answer:
a) the distance between her and the wall is 13 m
b) the period of her up-and-down motion is 6.5 s
Explanation:
Given the data in the question;
wavelength λ = 26 m
velocity v = 4.0 m/s
a) How far from the wall is she?
Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;
x = λ/2
we substitute
x = 26 m / 2
x = 13 m
Therefore, the distance between her and the wall is 13 m
b) What is the period of her up-and-down motion?
we know that the relationship between frequency, wavelength and wave speed is;
v = fλ
hence, f = v/λ
we also know that frequency is expressed as the reciprocal of the time period;
f = 1/T
Hence
1/T = v/λ
solve for T
Tv = λ
T = λ/v
we substitute
T = 26 m / 4 m/s
T = 6.5 s
Therefore, the period of her up-and-down motion is 6.5 s
Một điện tích q = 6.10-6 C đặt tại tâm của một hình lập phương. Tính thông lượng điện trường gửi qua mỗi mặt hình lập phương.
b. Một điện tích q = 4.10-8 C đặt tại tâm của một hình vuông. Tính thông lượng điện trường gửi qua mỗi mặt của hình vuông.
Answer:
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A caterpillar climbs up a one-meter a wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the
caterpillar to climb to the top. What is the distance traveled? (Round the number to the nearest hundred.)
cm
Answer:
1.5 m is ur answer
Explanation:
A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?
16s
300s
15s
23s
The boy pushed the sled for 16 seconds.
We have a boy who pushes his little brother on a sled.
We have to determine for how long time does boy push the sled.
State Work - Energy Theorem.The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
According to the question -
The sled is initially at rest → initial velocity (u) = 0.
Final velocity (v) = 4 m/s
Mass of boy and sled (M) = 40 kg
Power developed (P) = 20 W = 20 Joules/sec
According to work - energy theorem -
Work done (W) = Δ E(K) = E(f) - E(i)
Therefore -
W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule
Now, Power is defined as the rate of doing work -
P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]
20 = [tex]\frac{320}{t}[/tex]
t = 16 seconds
Hence, the boy pushed the sled for 16 seconds.
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Kirchhoff's junction rule is a statement of: Group of answer choices the law of conservation of momentum the law of conservation of charge the law of conservation of energy. the law of conservation of angular momentum. Newton's second law
Answer:
the law of conservation of energy.
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
In Physics, a point where at least three circuit paths (wires) meet is referred to as a junction.
The Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845.
Fundamentally, they address the conservation of energy and charge in the context of electrical circuits.
One of the laws known as Kirchoff's Current Law (KCL) deals with the principle of application of conserved energy in electrical circuits.
Kirchoff's Current Law (KCL) states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.
This simply means that the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero.
The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.
This ultimately implies that, Kirchhoff's junction rule is a statement of the law of conservation of energy.
What is used to represent the magnitude of the force in an FBD?
Answer:
See explanation below
Explanation:
The length of the arrow represents the magnitude/size of the force. The longer it is, the higher the force's magnitude is.
What will be resultant force ?
Answer:
It will go to the right way 500N.
Explanation:
Because, up and down is 300 and 300 so it minus each other and get zero, that made the resultant force will never go up or down. And now we are focusing on the left and right. 900 is more than 400, so it will go to the right. But we also pull to the left too!! That make the resultant force is 500 to the right.
. A 79 g sample of water at 21oC is heated until it becomes steam with a temperature of 143oC. Find the change in heat content of the system.
Answer:
40479.6 J
Explanation:
Applying,
q = cm(t₂-t₁).................... Equation 1
Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.
From the question,
Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C
Constant: c = 4200 J/kg.°C
Substitute these values into equation 1
q = 4200(0.079)(143-21)
q = 331.8(122)
q = 40479.6 J
A electron gains electric potential energy as it moves from point 1 to point 2. Which of the following is true regarding the electric potential at points 1 and 2?
a. V1 = V2
b. V1 > V2
c. V1 < V2
Answer:
We know that the change in electric potential energy is defined as:
q*ΔV = ΔP
So, the change in the electric potential energy is the charge times the change in the electric potential.
For the case of an electron gas, we have:
q = -e
where -e is the charge of an electron (remember that is negative).
So, if the electron gains electric potential then:
ΔP > 0
this means that the final potential energy is larger than the initial one, then we have:
-e*ΔV > 0
This means that ΔV must be negative.
V₂ = electric potential at point 2, so it is the final electric potential
V₁ = electric potential at point 1, so it is the final electric potential
Then we should get:
ΔV = V₂ - V₁ < 0.
This means that:
V₂ < V₁
The correct option is b.
Can someone help me
Btw the last one say current
How much time will it take for a person to walk the length of a football field (100 yards)
at a constant speed of 5 ft/s ?
The speed is in feet per seconds so change the length of the field from yards to feet.
1 yard = 3 feet
100 yards x 3 = 300 feet
The field is 300 feet long
Time = distance / speed
Time = 300 feet / 5 feet per second = 60 seconds = 1 minute
It will take 1 minute
Answer:
A person will take 60 Seconds to walk the distance of 100 yards.
Explanation:
Data Given ;
Distance ( d ) = 100 yards = 300 Ft
Speed ( v ) = 5 Ft/s
Time ( t ) = ?
What is speed ?The distance travelled in unit time is called speed.
formula ; [tex]v = \frac{d}{t} \\[/tex]
On putting values,
[tex]5 = \frac{300}{t}[/tex]
[tex]t =\frac{300}{5}[/tex]
[tex]t = 60 sec[/tex]
Hence the time taken by the person is 60 sec.
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Please help, only answer if your 1000% correct im in summer school and need to pass this class
Action reaction forces never cancel each other out because .............................? *
1 point
they do not act on objects.
they act on the same objects
they act on different objects.
none of the above
A force can be described as? *
1 point
push or pull action
pulling direction
direction of change
push action only
Using EquatIO calculate the resultant force for the horizontal (x-axis) component. Show your calculation step. *
2 points
Captionless Image
A player hits a baseball with a bat. The action force is the impact of the bat against the ball. The reaction force is .......................? *
1 point
Captionless Image
the grip of the player's hands on the ball
the weight of the ball
the air resistance of the ball
the force of the ball against the bat
Which law is this? "An object at rest tends to stay at rest and an object in motion tends to stay in motion unless acted upon by an unbalanced force." *
1 point
Newtons third law
Newtons second law
Newtons first law
None of the above
Which law is this? "If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A." *
1 point
None of the above
Newtons second law
Newtons first law
Newtons third law
What can be said about a force on an object that is not stationary? *
1 point
Captionless Image
there are no forces acting on the object
there are unequal forces acting on the object
there is only one force acting on the object
there are equal and opposite acting on the object
Using EquatIO calculate the resultant force for the vertical (y-axis) component. Show your calculation step. *
2 points
Captionless Image
Identify the different force acting on a moving vehicle shown in the mage below. *
5 points
Captionless Image
A B C D E
Reaction force
Weight
Friction
Air resistance
Thrust
Reaction force
Weight
Friction
Air resistance
Thrust
Which law is this? "The acceleration of an object is directly related to the net force and inversely related to its mass." *
1 point
None of the above
Newtons third law
Newtons first law
Newtons second law
If forces acting on an object are unbalanced, what can occur to the object? *
1 point
acceleration
all of the above
change of shape
deceleration
change in direction
If a player hits a baseball with a force of 870N Calculate the acceleration (state the units) of the ball . Show your calculation step using only EquatIO. *
2 points
Captionless Image
Action reaction forces never cancel each other out because .............................? *
1 point
they act on different objects.
none of the above
they act on the same objects
they do not act on objects.
because they work on the same object
Explanation:
A push and pull factor
Consider a 92.0 kg ice skater who is spinning on the ice. What is the moment of inertia of the skater, if the skater is approximated to be a solid cylinder that has a 0.140 m radius and is rotating about the center axis of the cylinder.
Answer:
[tex]I=0.902kg*m^2[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=92.0kg[/tex]
Radius [tex]r=0.140m[/tex]
Generally the equation for moment of Inertia is mathematically given by
[tex]I = 0.5*m*r^2[/tex]
[tex]I=0.5*(92)*0.14^2[/tex]
[tex]I=0.902kg*m^2[/tex]
Monica pulls her daughter Jessie in a bike trailer. The trailer and Jessie together have a mass of 25 kg. Monica starts up a 100-m-long slope that's 4.0 m high. On the slope, Monica's bike pulls on the trailer with a constant force of 9.0 N. They start out at the bottom of the slope with a speed of 5.0 m/s.
What is their speed at the top of the slope?
Answer:
Explanation:
From the given information:
total mass = 25 kg
distance d = 100 m
height = 4.0 m
Force F = 9.0 N
The speed at (bottom) u = 5.0 m/s
Using the concept of energy conservation;
[tex]\dfrac{1}{2}mu^2 + W = \dfrac{1}{2}mv^2 + mgh[/tex]
divide both sides by m
[tex]\dfrac{1}{2}u^2 + \dfrac{W }{m}= \dfrac{1}{2}v^2 + gh[/tex]
multiply both sides by 2
[tex]\dfrac{1}{2}u^2\times 2 + \dfrac{W }{m}\times 2= \dfrac{1}{2}v^2\times 2 + gh\times 2[/tex]
[tex]u^2 +2 \dfrac{W }{m}=v^2 + 2gh[/tex]
[tex]v^2 =u^2 - 2gh+ 2 \dfrac{W }{m}[/tex]
Recall that:
W = Fd
∴
[tex]v^2 =u^2 - 2gh+ 2 \dfrac{Fd }{m}[/tex]
[tex]v^2 =(5.0 \ m/s)^2 - 2(9.81 \ m/s)(4.0 \ m)+ 2 \dfrac{( 9.0 \ N \times 100 \ m) }{25 \ kg}[/tex]
[tex]v^2 =(25.0 ) - 78.48 +72[/tex]
[tex]v^2 = 18.52 \ m^2/s^2[/tex]
[tex]v = \sqrt{18.52 \ m^2/s^2}[/tex]
v = 4.30 m/s
A skateboarder is inside of a half pipe, shown here. Explain her energy transformations as she jumps off at point A, slides to point B, and finally reaches point C.
A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that is perpendicular to the platform at its center. The angular speed of the merry-go-round is constant, and a child at a distance of 1.4 m from the axis has a tangential speed of 2.2 m/s. What is the tangential speed of another child, who is located at a distance of 2.1 m from the axis?
(a) 1.5 m/s
(b) 3.3 m/s
(c) 2.2 m/s
(d) 5.0 m/s
(e) 0.98 m/s
Answer:
[tex]V_2=3.3m/s[/tex]
Explanation:
From the question we are told that:
Distance [tex]d_1=1.4m[/tex]
Tangential speed [tex]V=2.2m/s[/tex]
Distance 2 [tex]d_2=2.1m[/tex]
Generally the equation for Angular velocity is mathematically given by
[tex]w=\frac{v}{r}[/tex]
Therefore
[tex]\frac{v_1}{r_1}=\frac{v_2}{r_2}[/tex]
[tex]V_2=\frac{2.2*2.1}{1.4}[/tex]
[tex]V_2=3.3m/s[/tex]
Which one is the dependent variable in distance, force, or work
Answer:
Distance
Explanation:
Work can be defined as the energy transferred to a physical object by exertion of a force on the object to cause a displacement of the object. Thus, work is typically done when a person or simple machine move an object over a distance through the application of a force.
Mathematically, work done is given by the formula;
[tex] W = F * d[/tex]
Where,
W is the work done
F represents the force acting on a body.
d represents the distance covered by the body.
A dependent variable is the event expected to change when an independent variable is manipulated.
Hence, distance is the dependent variable because its value changes with respect to the amount of force exerted on an object.
A balloon is filled with 80 liters of gas on a day where the temperature was 34 degrees at sea level which is 101.3 kPa and released. As the balloon rises to a certain altitude, the temperature drops to 0 degrees celsius and the balloon doubles in volume. What is the atmospheric pressure at that altitude?
Answer:
0.444atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (
P2 = final pressure (
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question,
P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm
P2 = ?
V1 = 80L
V2 = 160L (double of V1)
T1 = 34°C = 34 + 273 = 307K
T2 = 0°C = 0 + 273 = 273K
Using P1V1/T1 = P2V2/T2
0.999 × 80/307 = P2 × 160/273
79.92/307 = 160P2/273
Cross multiply
307 × 160P2 = 79.92 × 273
49120P2 = 21818.16
P2 = 21818.16 ÷ 49120
P2 = 0.444
P2 = 0.444atm
In 1913 Niels Bohr formulated a method of calculating the different energy levels of the hydrogen atom.
a. True
b. False
Match each planet to an accurate characteristic
Answer:
venus - 2
earth - 3
mars - 4
mercury - 1
Three resistors (16 ohm, 16 ohm and 8 ohm) are connected in parallel. The equivalent
resistance (Re)
Answer:
4 ohm
Explanation:
The equivalent resistance (Re) of three resistors in parallel is given by;
1/Re = 1/R1 + 1/R2 + 1/R3
Where; R1 = 16 ohm, R2 = 16 ohm, R3 = 8 ohm
1/Re= 1/16 + 1/16 + 1/8
1/Re= (0.0625) + (0.0625) + (0.125)
Re= 4 ohm
A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.255 m while bringing a 830 kg car to rest from an initial speed of 1.4 m/s.
Answer:
the magnitude of the average force on the bumper is 3189.8 N
Explanation:
Given the data in the question;
In terms of force and displacement, work done is;
W =[tex]F^>[/tex] × [tex]x^>[/tex]
W = [tex]Fxcos\theta[/tex] ------- let this be equation 1
where F is force applied, x is displacement and θ is angle between force and displacement.
Now, since the displacement of the bumper and force acting on it is in the same direction,
hence, θ = 0°
we substitute into equation 1
W = [tex]Fxcos([/tex] 0° [tex])[/tex]
W = [tex]Fx[/tex] ------- let this be equation 2
Now, using work energy theorem,
total work done on the system is equal to the change in kinetic energy of the system.
[tex]W_{net[/tex] = ΔKE
= [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu² --------- let this be equation 3
where m is mass of object, v is final velocity, u is initial velocity.
from equation 2 and 3
[tex]Fx[/tex] = [tex]\frac{1}{2}[/tex]mv² - [tex]\frac{1}{2}[/tex]mu²
we make F, the subject of formula
F = [tex]\frac{m}{2x}[/tex]( v² - u² )
given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s
so we substitute
F = [tex]\frac{830}{(2)(0.255)}[/tex]( (0)² - (1.4)² )
F = 1627.45098 ( 0 - 1.96 )
F = 1627.45098 ( - 1.96 )
F = -3189.8 N
The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.
Therefore, the magnitude of the average force on the bumper is 3189.8 N
Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very few collisions with other molecules. Express your answer using two significant figures.
Answer:
The right solution is "24.39 per sec".
Explanation:
According to the question,
⇒ [tex]v=\frac{502.1}{\sqrt{3} }[/tex]
[tex]=289.9 \ m/s[/tex]
The time will be:
⇒ [tex]t=\frac{d}{v}[/tex]
[tex]=\frac{2\times 6}{289.9}[/tex]
[tex]=\frac{12}{289.9}[/tex]
[tex]=0.041 \ sec[/tex]
hence,
⇒ [tex]N=\frac{1}{t}[/tex]
[tex]=\frac{1}{0.041}[/tex]
[tex]=24.39 \ per \ sec[/tex]
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down? Assume the speed of sound is 343 m/s.
Answer:
The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
Explanation:
We can calculate the speed of the train using the Doppler equation:
[tex] f = f_{0}\frac{v + v_{o}}{v - v_{s}} [/tex]
Where:
f₀: is the emitted frequency
f: is the frequency heard by the observer
v: is the speed of the sound = 343 m/s
[tex] v_{o}[/tex]: is the speed of the observer = 0 (it is heard in the town)
[tex] v_{s}[/tex]: is the speed of the source =?
The frequency of the train before slowing down is given by:
[tex] f_{b} = f_{0}\frac{v}{v - v_{s_{b}}} [/tex] (1)
Now, the frequency of the train after slowing down is:
[tex] f_{a} = f_{0}\frac{v}{v - v_{s_{a}}} [/tex] (2)
Dividing equation (1) by (2) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}} [/tex]
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}} [/tex] (3)
Also, we know that the speed of the train when it is slowing down is half the initial speed so:
[tex] v_{s_{b}} = 2v_{s_{a}} [/tex] (4)
Now, by entering equation (4) into (3) we have:
[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}} [/tex]
[tex] \frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}} [/tex]
By solving the above equation for [tex]v_{s_{a}}[/tex] we can find the speed of the train after slowing down:
[tex] v_{s_{a}} = 11.06 m/s [/tex]
Finally, the speed of the train before slowing down is:
[tex] v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s [/tex]
Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.
I hope it helps you!
If it were to snow in Phoenix in July, which type of Earth scientist would be most
surprised?
Answer:
Economists, I guess.
Explanation:
Water from a fire hose knocks over a wooden shed. Compared with the pressure within
the water, the pressure exerted against the shed is
a) less.
b) the same.
c) more.
d) nonexistent
Which type of balance is key to sitting?
dynamic
static
bosu
level
Explanation:
bosu
here is your answer
Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 24 N, and F3 = 13 N. The radial distances are R2 = 0.22 m and R3 = 0.10 m. The forces F2 and F3 are perpendicular to the radial lines labeled R2 and R3. The moment of inertia of the cylinder is 1/2 MR 2/2. Find the magnitude of the angular acceleration of the cylinder about the axis of rotation.
Answer:
α = 13.7 rad / s²
Explanation:
Let's use Newton's second law for rotational motion
∑ τ = I α
we will assume that the counterclockwise turns are positive
F₁ 0 + F₂ R₂ - F₃ R₃ = I α
give us the cylinder moment of inertia
I = ½ M R₂²
α = (F₂ R₂ - F₃ R₃) [tex]\frac{2}{M R_2^2}[/tex]
let's calculate
α = (24 0.22 - 13 0.10) [tex]\frac{2}{12 \ 0.22^2}[/tex]2/12 0.22²
α = 13.7 rad / s²
Alice's friends Bob and Charlie are having a race to a distant star 10 light years away. Alice is the race official who stays on Earth, and her friend Darien is stationed on the star where the race ends. Bob is in a rocket that can travel at 0.7c; whereas Charlie's rocket can reach a speed of 0.866c. Bob and Charlie start at the same time. Draw space- diagrams from each perspective.
Required:
Estimate how long it takes Bob and Charlie to finish the race from each perspective.
Solution :
The distance between the starting point and the end point, [tex]L_0[/tex] = 10 light years
But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.
For Bob,
Speed of Bob's rocket with respect to Alice, [tex]L_b = 0.7 \ c[/tex]
So the distance appeared to Bob due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.49} \ Ly$[/tex]
[tex]$=7.1 \ Ly$[/tex]
Therefore, the time required to finish the race by Bob is
[tex]$t_b = \frac{L_b}{V_b}$[/tex]
[tex]$=\frac{7.1 \ c}{0.7 \ c}$[/tex]
= 10.143 year
For Charlie,
Speed of Charlie's rocket with respect to Alice, [tex]L_c = 0.866 \ c[/tex]
So the distance appeared to Charlie due to the length contraction,
[tex]$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$[/tex]
[tex]$L_b=10\times \sqrt{1-0.75} \ Ly$[/tex]
[tex]$=5 \ Ly$[/tex]
The time required to finish the race by Charlie is
[tex]$t_b = \frac{L_c}{V_c}$[/tex]
[tex]$=\frac{5 \ c}{0.866 \ c}$[/tex]
= 5.77 year
A 75.0 kg diver falls from rest into a swimming pool from a height of 5.10 m. It takes 1.34 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
Answer:
559.5 N
Explanation:
Applying,
v² = u²+2gs............. Equation 1
Where v = final velocity,
From the question,
Given: s = 5.10 m, u = 0 m/s ( from rest)
Constant: 9.8 m/s²
Therefore,
v² = 0²+2×9.8×5.1
v² = 99.96
v = √(99.96)
v = 9.99 m/s
As the diver eneters the water,
u = 9.99 m/s, v = 0 m/s
Given: t = 1.34 s
Apply
a = (v-u)/t
a = 9.99/1.34
a = -7.46 m/s²
F = ma.............. Equation 2
Where F = force, m = mass
Given: m = 75 kg, a = -7.46 m/s²,
F = 75(-7.46)
F = -559.5 N
Hence the average force exerted on the diver is 559.5 N