PLZZZ HELP
Question 2 Multiple Choice Worth 2 points)
(04.05 LC)
Which of the following scenarios describe an object being acted on by a non-conservative force?
O Apuck hovers and glides across an air track at constant velocity.
A dishwasher was pulled up to a window by a rope using a pulley system.
O A feather falls from one end of a tube to the other inside a vacuum.
O A charged particle was pushed away by another charged particle.

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

[tex] d = v*t \rightarrow v = \frac{d}{t} [/tex]    

Where:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

[tex] t = \sqrt{\frac{2d}{g}} [/tex]

Where:

g: is gravity = 9.8 m/s²

[tex] t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s [/tex]

Now, the horizontal velocity of the rock is:

[tex] v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s [/tex]      

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.          

2. To calculate the distance at which the projectile will land, first, we need to find the time:

[tex] t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s [/tex]

So, the distance is:

[tex] d = v*t = 6.67 m/s*1.43 s = 9.54 m [/tex]    

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!        

Answer:

a) 10.6V

b) E = 4.9V/m, +x direction

c) E = 4.9V/m, +x direction

Explanation:

You have the following function:

[tex]V=a+bx[/tex]  (1)

for the potential in a region between x=0 and x=6.00 m

a = 10.6 V

b = -4.90V/m

[tex]V=10.6V-4.90\frac{V}{m}x[/tex]

a) for x=0 you obtain for V:

[tex]V=10.6V-4.90\frac{V}{m}(0m)=10.6V[/tex]

b) The relation between the potential difference and the electric field can be written as:

[tex]E=-\frac{dV}{dx}=-b=-(-4.9V/m)=4.9V/m[/tex]  (2)

the direction is +x

c) The electric field is the same for any value of x between x=0 and x=6m.

Hence,

E = 4.9V/m, +x direction

Answer:

i think its 1-10 ´ 10-15 m

Explanation:

It is found that nuclear radii range from 1-10 ´ 10-15 m. This radius is much smaller than that of the atom, which is typically 10-10 m. Thus, the nucleus occupies an extremely small volume inside the atom. The nuclei of some atoms are spherical, while others are stretched or flattened into deformed shapes.

Answer:

Explanation:

Initially skier is at a height of 351 m . Her kinetic enery will be nil because she is at rest . Her potential energy will be calculated as follows

potential energy = mgh where m is mass , h is height and g is acceleration due to gravity

potential energy = 58 x 9.8 x 351

= 199508 .4 J

Total mehanical energy = potential energy + kinetic energy

= 199508.4 J

According to conservation of mechanical energy , at the height of 142 m also total mechanical energy will be same . At this height some potential energy will be converted into kinetic energy but total of potential and kinetic energy will be same.

A natural force of attraction exerted by the earth upon objects, that pulls objects towards earth's center is called Gravitational force .

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

[tex]v_{avg}=\frac{x_{all}}{t}[/tex]

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

[tex]v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}[/tex]

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

Answer:

The average velocity of the object is 0.1cm/s

Explanation:

Given that the object travels from point 15cm to 165cm and back to 25cm within 100 seconds

The average velocity is calculated as thus.

Average Velocity = ∆D/t

Where ∆D represent the displacement.

The displacement is calculated as follows.

∆D = End point - Start Point.

From the question, the end and start point are 25cm and 15cm respectively.

Hence,

∆D = 25cm - 15cm

∆D = 10cm.

t = 100 seconds

So, Average Velocity = 10cm/100s

Average Velocity = 0.1cm/s

Hence, the average velocity of the object is 0.1cm/s

Answer:

a) h_max = 400ft

b) v = 1024 ft/s

Explanation:

The general equation of a motion is:

[tex]s(t)=v_ot-\frac{1}{2}gt^2[/tex]  (1)

You have the following equation of motion is given by:

[tex]s(t)=160t-16t^2[/tex]  (2)

You compare both equations (1) and (2) and you obtain:

vo: initial velocity = 160ft/s

g: gravitational acceleration = 32ft/s

a) The maxim height reached by the ball is given by:

[tex]h_{max}=\frac{v_o^2}{2g}=\frac{(160ft/s)^2}{64ft/s^2}=400ft[/tex]

b) The velocity is given by:

[tex]v^2=v_o^2-2gh\\\\v=\sqrt{(160ft/s)^2-2(32ft/s^2)(384ft)}=1024ft/s[/tex]

Answer:

a) Smax = 400 ft

b) v = 32 ft/s

c) v = - 32 ft/s

Explanation:

(a)

The function given for the height of ball is:

s = 160 t - 16 t²

Therefore, in order to find the time to reach maximum height (in-flexion point), we must take the derivative with respect to t and set it equal to zero:

Therefore,

160 - 32 t = 0

32 t = 160

t = 160/32

t  = 5 sec

Therefore, maximum distance is covered at a time interval of 5 sec.

Smax = (160)(5) - (16)(5)²

Smax = 800 - 400

Smax = 400 ft

b)

First we calculate the time at which ball covers 384 ft

384 = 160 t - 16 t²

16 t² - 160 t + 384 = 0

Solving Quadratic Equation:

Either:

t = 6 sec

Or:

t = 4 sec

Since, the time to reach maximum height is 5 sec. Therefore, t < 5 sec

Therefore,

t = 4 sec

Now, we find velocity at 4 sec by taking derivative od s with respect to t at 4 sec:

v = 160 - 32 t

v = 160 - (32)(4)

v = 32 ft/s

c)

Since, t = 6 s > 5 s

The second value of t = 6 sec must correspond to the instant when the ball is 384 ft above the ground while traveling downward.

Hence, velocity at that time will be:

v = 160 - (32)(6)

v = -32 ft/s

Negative sign due to downward motion.

Answer:

171.5Hz,514.5Hz and 857.5Hz

Explanation:

We are given that

Distance between two loudspeaker,d=2.65 m

Distance of listener from one end=19.1 m

Distance of listener from other end=20.1 m

[tex]\Delta L=20.1-19.1=1[/tex]m

Speed of sound,v=343m/s

For destructive interference

[tex]f_{min,n}=\frac{(n-0.5)v}{\Delta L}[/tex]

Using the formula  and substitute n=1

[tex]f_{min,1}=\frac{(1-0.5)\times 343}{1}=171.5Hz[/tex]

For n=2

[tex]f_{min,2}=\frac{(2-0.5)\times 343}{1}=514.5Hz[/tex]

For n=3

[tex]f_{min,3}=(3-0.5)\times 343=857.5Hz[/tex]

Hence, the three lowest frequencies that give minimum signal (destructive interference) at the listener's location is given by

171.5Hz,514.5Hz and 857.5Hz

Answer:

B. less

Explanation:

acceleration due to gravity on Earth, g = 9.8 m/s²

acceleration due to gravity on Moon, g = 1.6 m/s²

Given mass of the object as, m = 5 kg

Weight of an object is given as, W = mg

Weight of the object on Earth, W = 5 x 9.8 = 49 N

Weight of the object on Moon, W = 5 x 1.6 = 8 N

Therefore, the object weighs less on the moon compared to its weight on Earth.

The correct option is "B. less"

The so-called "force" of friction always acts exactly opposite to the direction in which an object is moving.  So for this question, since the box is moving to the right, the "force" of friction is acting toward the left.

There's not enough information given in the question to determine its magnitude.

Answer:

The wheels turn 101.27 radians in that time.

Explanation:

First, we need to find the angular velocity of the tires. We use the following formula for this purpose:

Δv = rω

ω = Δv/r

where,

ω = Angular Velocity of Tires = ?

Δv = change in linear velocity = 5.33 m/s - 0 m/s = 5.33 m/s

r = radius of tire = 0.33 m

Therefore,

ω = (5.33 m/s)/(0.33 m)

ω = 16.15 rad/s

Now, the angular displacement covered by tires can be found out by using the general formula of angular velocity:

ω = θ/t

θ = ωt

where,

θ = angular displacement = ?

t = time = 6.27 s

Therefore,

θ = (16.15 rad/s)(6.27 s)

θ = 101.27 radians

Answer:

Explanation:

jhgfhmhvmghmm

Answer:

A

Explanation:

Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.

Answer:

i see they did already i am just supporting them

Answer:

A

Explanation:

Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.

Answer:

[tex]v = \sqrt{20h} [/tex]

Explanation:

The potential energy (PE) we are looking here is gravitational potential energy (GPE).

GPE= mgh,

where m is the mass of an object,

g is the gravitational field strength

h is the height of the object

KE= ½mv²,

where m is the mass and v is the velocity

loss in GPE= gain in KE

mgh= ½mv²

gh= ½v² (divide by m throughout)

Assuming that the object is on earth, then g= 10N/Kg

½v²= 10h (substitute g=10)

v²= 20h (×2 on both sides)

v= √20h (square root both sides)

Answer:

The field lines go out of Earth near Antarctica, enter Earth in northern Canada, and are not aligned with the geographic poles.

Explanation:

Answer: the field lines go out of earth near the north pole, enter earth in the south pole, and are not aligned with the geographic poles.

Explanation: just took the test on edge 2020.

Answer:

h = 31.46 m

Explanation:

We have,

Initial speed of a projectile is 37.7 m/s

It was projected at an angle of 41.2° above the horizontal on a long flat firing range.

It is required to find the maximum height reached by the projectile. The formula used to find it is given by :

[tex]h=\dfrac{u^2\sin^2\theta}{2g}[/tex]

Plugging all the known values,

[tex]h=\dfrac{(37.7)^2\times \sin^2(41.2)}{2\times 9.8}\\\\h=31.46\ m[/tex]

So, the maximum height reached by the projectile is 31.46 m.

Answer:

h = 0.288m

Explanation:

Assume

[tex]v_1[/tex] = Speed of Sue

[tex]v_2[/tex] = Speed of Chris immediately after the push

Sue's KE = [tex]\frac{1}{2} mv_1\ ^2 = 26 v_1 \ ^2[/tex]

now she swings this is converted into gravitation at PE of

mg n = 52 × 9.8 × 0.65  

= 331.24

[tex]26v_{1}\ ^2[/tex] = 331.24

So, [tex]v_1 = 3.569[/tex]

They started at rest by conservation of momentum in case of push off the magnitude of sue momentum and it is equal to the magnitude of Chris momentum in the opposite or inverse direction

[tex]m_1v_1 = m_2v_2[/tex]

[tex]52 \times 3.569 = 78 \times v_2[/tex]

[tex]v_2 = 2.380[/tex]

Chris kt = [tex]\frac{1}{2} \times 78 \times 2.380^2[/tex]

= 220.827

220.827 = mgh

So, h = 0.288m

Answer:

The normal force is the force that the floor does as a reaction of the gravitational force that an object does against the floor (is the resistance that objects have when other objects want to move trhough them, and the force comes by the 3rd Newton's law, and this is specially used in cases where the first object is fixed, like walls or the floor). With this in mind, the point in where the normal force will be greater is the point that is closer to the center of mass of the object (the point with more mass)

If the wheels are in the extremes of the object, and the center of mass is in the middle of the object, the normal force will be equal. Now if for example, you put a little mass in one end of the object, now the center of weight displaces a little bit and is not centered, and the side is where you put the weight on will receive a bigger normal force from the floor than the other side.

The force on the front and rear wheels can be determined by finding the

force acting under equilibrium conditions.

The normal reaction between the floor and the front wheel is larger. The

correct option is c) The force on the front wheels must be greater than the

force on the rear wheels.

Reasons:

The forces acting on the wheels are;

Vertical forces acting on the cart:

The weight of the cart acting on the floor

The normal reaction of the floor on the wheels

The horizontal acting on the cart:

Pushing force acting horizontally

Friction force acting in the reverse direction

At equilibrium, the cart does not tip over

Sum of moment about the rear wheel = [tex]\mathbf{W \times \dfrac{d}{2} + F \times h - N_{front} \times d} = 0[/tex]

Where;

h = The height of the cart

d = depth of the cart

[tex]N_{front}[/tex]  = The normal reaction at the front wheels

Therefore;

[tex]\displaystyle N_{front} = \frac{W \times \dfrac{d}{2} + F \times h}{d} = \mathbf{ \frac{W}{2} + \frac{F \times h}{d}}[/tex]

Sum of moment about the front wheel = [tex]\mathbf{ F \times h + N_{rear} \times d - W \times \dfrac{d}{2} } = 0[/tex]

Therefore;

[tex]\displaystyle N_{rear} = \frac{W \times \dfrac{d}{2} - F \times h}{d} = \mathbf{ \frac{W}{2} - \frac{F \times h}{d}}[/tex]

Which gives;

[tex]\displaystyle N_{front} = \mathbf{ \displaystyle N_{rear} + 2 \times \frac{F \times h}{d}}[/tex]

[tex]\displaystyle 2 \times \frac{F \times h}{d} > 0[/tex]

Therefore;

[tex]\displaystyle N_{front} > \displaystyle N_{rear}[/tex]

The correct option is c) The force on the front wheels must be greater than

the force on the rear wheels.

Learn more here:

https://brainly.com/question/17574217

Answer:

0.824m/s

Explanation:

To calculate the average velocity we have to find the total distance and the total time

We have to find the distance and time in each motions

FIRST MOTION

The values given are

Speed= 2m/s , t = 60minutes

The time has to be converted to seconds. 60×60 = 3600seconds

Distance= speed×time

= 2× 3600

= 7200m

In the first motion the distance is 7200m and the time is 3600seconds

SECOND MOTION

The values given are

Distance= 3000m

Time= 25 mins to seconds

= 25×60

= 1500 seconds

In the second motion distance is 3000m and the time is 1500 seconds

The total distance can be calculated by applying the formular ( d1-d2) since she moved in an opposite direction

Total distance= 7200-3000

= 4200m

The total time (t1+t2) = 1500+3600

= 5100 seconds

Therefore, average velocity is calculated by applying the formular

Total distance/ Total time

= 4200/5100

= 0.824m/s South

Hence the average velocity is 0.824m/s South.

Answer:

The woman's average velocity during her entire motion is 2 m/s

Explanation:

Given;

initial speed of the woman, u =  2.00 m/s

initial time taken, t₁ = 60 minutes = 3600 seconds

initial displacement of the woman, x₁ = ?

final displacement of the woman, x₂ = 3000 m north

final time taken , t₂ = 25.0 minutes = 1500 seconds

The woman's average velocity during her entire motion:

initial displacement of the woman, x₁ = u x t₁  = 2.00 m/s x 3600 seconds

                                                                           = 7200 m South

[tex]Average \ velocity = \frac{\delta X}{\delta t} = \frac{X_1-X_2}{t_1-t_2} \\\\V_{avg.} = \frac{7200-3000}{3600-1500} = \frac{4200}{2100} = 2 \ m/s[/tex]

Therefore, the woman's average velocity during her entire motion is 2 m/s

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

     ∑  F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

          cos 41 = T₁ₓ / T1

          sin 41 = T₁y / T1

          T₁ₓ = T₁  cos 41

          T₁y= T₁  sin 41

for cable gold

           cos 63 = T₂ / T₂

           sin 63 = [tex]T_{2y}[/tex] / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

              T₃ - W = 0

              T₃ = W

              T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

   - T₁ₓ + T₂ₓ = 0

  T₁ₓ = T₂ₓ

   T1 cos 41 = T2 cos 63

   T1 = T2 cos 63 / cos 41                (1)

y Axis

      [tex]T_{1y}[/tex] + T_{2y} - T3 = 0

       T₁ sin 41 + T₂ sin 63 = T₃          (2)

to solve the system we substitute equation 1 in 2

        T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

         T₂ (cos 63 tan 41 + sin 63) = W

         T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

          T₂ = 200 / (cos 63 tan 41 + sin 63)

          T₂ = 200 / 1,2856

           T₂ = 155.6 N

we substitute in 1

            T₁ = T₂ cos 63 / cos 41

             T₁ = 155.6 cos63 / cos 41

             T₁ = 93.6 N

therefore the tension in each cable is

            T₁ = 93.6 N

             T₂ = 155.6 N

             T₃ = 200 N

Answer:

The resistance of the aluminium wire is 0.541 ohms

Explanation:

Given:

Cross sectional area of the aluminum wire, A = 4.9 x 10-4 m2

Length of the wire, L = 10 km = 10,000 m

Resistance of the wire = (pL) / A

Where:

p is resistivity of aluminium wire = 2.65 x 10^-8 ohm-m

L is length of the wire

A is cross sectional area of the wire

Substitute these values and solve for resistance of the wire.

Resistance = (2.65 x 10^-8 x 10,000) / (4.9 x 10^-4)

Resistance = 0.541 ohms

Therefore, the resistance of the aluminium wire is 0.541 ohms

Answer:

Boat A wins and 50km/hr

Explanation:

To solve this problem we need to calculate the time required for both boats to make the round trip.

For Boat A.

The time required for the first trip =distance/ speed = 50/50 = 1h

For the return trip the time is same since it's the same distance on the same speed.

Hence the time taken for boat A to make the round trip is 2hrs.

For boat B,

The time required for the first trip =distance/ speed = 50/25 = 2h

The time for the return trip is;

=distance/ speed = 50/75 = 2/3h= 2/3×60= 40mins

By comparing both time.

Boat A requires less time hence Boat A wins.

2.The average velocity of the winning boat is the velocity of boat A.

Which is total distance/ total time for the trip.

100km/2hrs = 50km/hr

boat A would win the race and the average velocity of the winning boat would be 50 km/h

The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.

As given in the problem Two boats start together and race across a 50-km-wide lake and back. Boat A goes across at 50 km/h and returns at 50 km/h. Boat B goes across at 25 km/h, and its crew, realizing how far behind it is getting, returns at 75 km/h.

Turnaround times are negligible, and the boat that completes the round-trip first wins

Average velocity of the boat A = ( 50 + 50 ) / 2

                                                              = 50 km /h

Thus, boat A would win the race and the average velocity of the winning boat would be  50 km/h

Learn more about Velocity from here,

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Answer:

6.0621 m/s

Explanation:

we all know that horizontal component of the velocity in projectile motion is

= u cosα

u= initial speed of 7.00 m/s of the gun.

α = angle of the initial velocity with the horizontal = 30°

therefore, horizontal component of the ball's initial velocity = 7×cos30°

[tex]7\times\frac{\sqrt{3} }{2} \\=6.0621 \text{m/s}[/tex]

Answer:

emf = 0.02525 V

induced current with a counterclockwise direction

Explanation:

The emf is given by the following formula:

[tex]emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}[/tex][tex]\ \ =-B\frac{A_2-A_1}{t_2-t_1}[/tex]   (1)

ФB: magnetic flux =  BA

B: magnitude of the magnetic field = 1.00T

A2: final area of the loop; A1: initial area

t2: final time, t1: initial time

You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.

In t = 4 s the final circumference will be:

[tex]c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm[/tex]

To find the areas A1 and A2 you calculate the radius:

[tex]r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm[/tex]

r1 = 0.261 m

r2 = 0.190 m

Then, the areas A1 and A2 are:

[tex]A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2[/tex]

Finally, the emf induced, by using the equation (1), is:

[tex]emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV[/tex]

The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.

HEYA!!

Here's your Answer:

Energy is the power we use for transportation, for heat and light in our homes and for the manufacture of all kinds of products. There are two sources of energy: renewable and non-renewable energy

RENEWABLE ENERGY: Renewable energy, often referred to as clean energy, comes from natural sources or processes that are constantly replenished. For example, (sunlight or wind keep shining and blowing), even if their availability depends on time and weather.

NON-RENEWABLE ENERGY:Non-renewable energy comes from sources that will run out or will not be replenished in our lifetimes—or even in many, many lifetimes.

Most non-renewable energy sources are (fossil fuels: coal, petroleum, and natural gas).

HOPE IT HELPS!!!

The ball's position in the air at time t is given by the vector,

p(t) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ²

and its velocity is given by

v(t) = (60 i + 64 k) + (-6 j - 32 k) t

The ball is in the air for as long as it takes for the vertical (k) component of the position vector to reach 0, so we solve,

64 t - 32/2 t ² = 0  ==>  t = 0 OR t = 4

and so the ball is in the air for 4 s.

After this time, the ball has position vector

p(4) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ² = 240 i - 48 j

which has magnitude

||p(4)|| = √(240² + (-48)²) = 48 √26 ≈ 244.8 ft

in a direction θ in the x,y plane from the positive x axis such that

tanθ = -48/240 = -1/5  ==>  θ = -arctan(1/5) ≈ -11.3º

or 11.3º South of East.

The ball hits the ground with speed

||v(4)|| = ||60 i - 24 j - 64 k|| = √(60² + (-24)² + (-64)²) = 4 √517 ≈ 91.0 ft/s

kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is

        r = 244.8 ft, tea = 21.8º from East to South

        v = 91.0 ft / s

given parameters

to find

Kinematics allows finding the position, velocity and acceleration of the body, in this case we have a problem in three dimensions, where they establish a Cartesian coordinate system, a method to solve this exercise is to solve each component independently

a)  The acceleration of gravity acts on the z axis, so let's find the time it takes to reach the ground, if the initial vertical velocity is v_{oz} = 64 ft/s and the acceleration is a_z = g = -32 ft / s², we assume that the ball leaves the ground (z₀ = 0)

         z = z₀ + v_{oz} t + ½ a_z t²

when reaching the ground its height of zero and

        0 = 0 + v_{oz} t + ½ a_z t²

        t (v_{oz) + ½ a_z t) = 0

        t (64 - 16 t) = 0

the solusion of this squadron is

       y = 0

       t = 4 s

the first time is when it leaves and the second time is for when it reaches the ground, therefore the flight time is t = 4s

with this time we find the displacement is each exercise

X axis

    in this axis there is no acceleration, so we use the uniform motion relationships

        vₓ = x / t

        x = vₓ t

        x = 60 4

        x = 240 ft

Y Axis

on this axis there is an acceleration of a_y = -6 ft/s and an initial velocity v_{oy} = 0

we use the kinematic relation

       y = v_{oy} t + ½ a_y t²

       y = 0 - ½ 6 4²

       y = - 48 ft

let's use the Pythagoras theorem to find the position

       r² = (x -x₀) ² + (y -y₀) ² + (z -z₀) ²

       r² = (240-0) ² + (-48-0) ² + (0-0) ²

       r = 244.8 ft

We use trigonometry to find the direction

      tan θ = y / x

       θ = tan⁻¹ [tex]\frac{y}{x}[/tex]

       θ = tan⁻¹  [tex]\frac{96}{240}[/tex]

       θ = -21.8º

This angle is measured clockwise from the x axis, it can also be read

        θ = 21.8º from East to South

b)  Let's look for the speed when we hit the ground

X axis

       vₓ = v_{ox} + aₓ t

       vₓ = 60 - 0

       vₓ = 60 ft / s

Y Axis

       v_y = v_{oy} + a_y t

       v_y = 0 - 6 4

        v_y = -24 ft / s²

Z axis

       v_z = v_{oz} + a_z t

       v_z = 64 -32 4

       v_z = -64 ft / s

With the Pytagoras  theorem find the modulus of this speed is

        v² = vx² + vy² + vz²

        v² = 60² + 24 ² + 64²

        v = 91.0 ft / s

In conclusion, the kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is

       a) r = 244.8 ft, θ = 21.8º from East to South

       b) v = 91.0 ft / s

learn more about kinematics here:

https://brainly.com/question/11503629

Answer:

7.7 km 26°

Explanation:

The total x component is:

x = 2.5 cos(35°) + 5.2 cos(22°) = 6.87

The total y component is:

y = 2.5 sin(35°) + 5.2 sin(22°) = 3.38

The magnitude is:

d = √(x² + y²)

d = 7.7 km

The direction is:

θ = atan(y/x)

θ = 26°

Answer:

33 m/s

Explanation:

v = √(T / (m/L))

where v is the velocity,

T is the tension,

and m/L is the mass per length.

v = √(200 N / (0.055 kg / 0.30 m))

v = 33 m/s