What is the Y-component of a vector A, which is of magnitude
16-12 and at a 45° angle to the horizontal?

What Is The Y-component Of A Vector A, Which Is Of Magnitude16-12 And At A 45 Angle To The Horizontal?

Answers

Answer 1

Explanation:

the answer is in the image above

What Is The Y-component Of A Vector A, Which Is Of Magnitude16-12 And At A 45 Angle To The Horizontal?
Answer 2

The Y-component of a vector A, which is of magnitude 16√2 and at a 45° angle to the horizontal would be 16

What is a vector quantity?

The quantities that contain the magnitude of the quantities along with the direction are known as the vector quantities.

Examples of vector quantities are displacement, velocity acceleration, force, etc.

As given in the problem we have to find out the  Y-component of a vector A, which is of magnitude  16√12  and at a 45° angle to the horizontal,

Y component of the vector A =  16√2 sin45°

                                                =16√2 ×1/√2

                                                =16

Thus, the Y component of vector A would be 16.

To learn more about the vector quantity here, refer to the link;

https://brainly.com/question/15516363

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Related Questions

Match each planet to an accurate characteristic

Answers

Answer:

venus - 2

earth - 3

mars - 4

mercury - 1

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.

Answers

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]

Rahul sits in a chair reading a book. Which force is equal to the force Rahul exerts on Earth?

Answers

In fact, the force Rahul exerts on Earth corresponds to the force of gravity. But Rahul's weight is, in fact, the force of gravity exerted by the Earth on Rahul, and these two forces correspond to the action-reaction pair of Newton's third law, which states that the two forces are equal.

Answer:

Rahul's weight

Explanation:

In fact, the force Rahul exerts on Earth corresponds to the force of gravity. But Rahul's weight is, in fact, the force of gravity exerted by the Earth on Rahul, and these two forces correspond to the action-reaction pair of Newton's third law, which states that the two forces are equal.

Using formulas, Rahul's weight is equal to

W=mg

where m is Rahul mass and g is the gravitational acceleration (g=9.81 m/s^2).

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?

Answers

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

Match the atmospheric energy transfer process that best fits each of the following scenarios:
Warming of the Earth's surface on a sunny day
[ Choose ] convection conduction radiation advection
On a sunny afternoon, you watch cumulus clouds forming
[ Choose ] convection conduction radiation advection
A very shallow layer of air in contact with the ground is warmed
[ Choose ] convection conduction radiation advection
A south wind carries warm air into the central United States
[ Choose ] convection conduction radiation advection

Answers

Answer:

a) RADIATION, b) CONVECTION, c)  CONDUCTION, d) CONVECTION

Explanation:

In the heating processes there can be three types: conduction, convention and radiation.

The conduction process occurs when the movement of atoms or thermal agitation of molecules creates the transfer of thermal energy.

The convention process occurs when there is a movement of matter creating the transfer of energy

The process of Radiation an electromagnetic wave indexes on a material and is absorbed, creating the process of energy transfer.

Now let's examine each situation>

a)Warming of the Earth's surface on a sunny day

in this case the sunlight heats the earth as it is absorbed, which is why it is a RADIATION process

b) On a sunny afternoon, you watch cumulus clouds forming

in this case the nines rise from the surface, since it is a moving mass, the process is CONVECTION

c) A very shallow layer of air in contact with the ground is warmed

in this case the thermal movement of the layer molecules heat the earth, for which the process of CONDUCTION

d) A south wind carries warm air into the central United States

As we have a movement of a mass of matter, the process is CONVECTION.

If it were to snow in Phoenix in July, which type of Earth scientist would be most
surprised?

Answers

Answer:

Economists, I guess.

Explanation:

In 1913 Niels Bohr formulated a method of calculating the different energy levels of the hydrogen atom.

a. True
b. False

Answers

They are right the answer is A true

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?

Answers

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

 

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.

Answers

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!                                                        

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.

Answers

Point A would be the build up and leverage to slide down the half pipe to point b she would be bending her knees to increase speed to go up the other side and meet point c

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

Answers

Answer:

1.5 m is ur answer

Explanation:

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.

Answers

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

Can someone help me
Btw the last one say current

Answers

Magnitude: Magnitude generally refers to the quantity or distance.

Attraction: A force that makes things move together and stay together.

Repulsive: The feeling of being repelled, as by the thought or presence of something; distaste, repugnance, or aversion.

Current: Current is the rate of flow of electric charge. A potential difference (voltage) across an electrical component is needed to make a current flow through it.

I hope that helps :]

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.

Answers

because they work on the same object

Explanation:

A push and pull factor

1) they act on different objects
2)push or pull action
3) (no image so i cant answer this sorry)
4)the force of the ball against the bat
5)Newton’s first law
6)newtons third law
7)there are unequal forces acting on the object
8)(no image again sorry)
9)(no image)
10)newtons second law
11)all of the above
12) f=ma
A= f/m = 870/
Mass wasnt provided so i cant solve this
13)they act on different objects

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?

Answers

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

You are helping to design a new electron microscope to investigate the structure of the HIV virus. A new device to position the electron beam consists of a charged circle of conductor. This circle is divided into two half circles separated by a thin insulator so that half of the circle can be charged positively and the other half can be charged negatively. The electron beam will go through the center of the circle. To complete the design your job is to calculate the electric field in the center of the circle as a function of the amount of positive charge on the half circle, the amount of negative charge on the other half circle, and the radius of the circle.

Answers

Answer:

The electric field in the center is  [tex]\frac{2k}{\pi R^{2}}\left ( Q_{2}-Q_{1 } \right )[/tex].

Explanation:

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.

Answers

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

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.

Answers

Answer:

mmmlbdhdjdkekkdnxnfjkckckcklclglglvkglglvkgkvkvkvkvkvkvkvkvkvkvkbkkbkbkkbbkkbkbkbkkbkbkblbkkbkb

A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and amplitude 2cm travels to the right along the string. It is observed that the displacement at x=0 at t=0 is 0.87cm.
a) What is the wave?
b) Wrote the wave function, y(x,t)
c) Find the particle velocity at position x=0 at time t=10s. What is the maximum particle velocity?​

Answers

Answer:

Explanation:wave=wavelength×frequency,

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

Answers

Answer: C. More
Pressure in the water and the pressure water can exert on something that changes its momentum.

Which type of balance is key to sitting?

dynamic

static

bosu

level

Answers

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.

Answers

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

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​

Answers

300 because the mass and weight

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.

To solve more questions on Work, Energy and Power, visit the link below -

https://brainly.com/question/208670

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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

Answers

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.

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.

Answers

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

What is used to represent the magnitude of the force in an FBD?

Answers

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.

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.

Answers

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]

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

Answers

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]

. 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.

Answers

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


Three resistors (16 ohm, 16 ohm and 8 ohm) are connected in parallel. The equivalent
resistance (Re)

Answers

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

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