At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 55 kg, 90 kg, and 42.5 kg respectively. Car A is moving to the right with a velocity vA = 2 m/s and car C has a velocity vC = 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8. Determine the final velocity of each car, after all impacts, assuming car A hits car B before car C does. Assume positive sign denoting forward motion and negative sign denoting backward motion.

Answer:

R1=3.333 Ohms

R2=10 Ohms

R3=16.666 Ohms

Explanation:

30 total

30=R1+R2R3

30=1(x):3(x):5(x)

x=3.3333

R1=3.333 Ohms

R2=10 Ohms

R3=16.666 Ohms

Answer:

1) v= 90km/h  d = 70 m,  2)  x₁ = v t_r,  x₁ = 6.25 m, 3) x₁=6.25 no change

4) x = 22 m

Explanation:

1) for the first part, you are asked to find the minimum safety distance with the vehicle in front

The internet is searched for the stopping distance for two typical speeds on the highway

v (km/ h)      v (m/s)      d (m)

90                  25           70

100                 27.78      84

the safe distance is this distance plus the distance traveled during the person's reaction time, which can be calculated with infirm movement

              v = x / t_r

               x₁ = v t_r

the average reaction time is t_r = 0.25s for a visual stimulus and t_r 0.17 for an auditory stimulus

therefore the safe distance is

              x_total = x₁ + d

2) The distance is the sum of the distance traveled in the reaction

              x₁ = v t_r

for v = 90 km / h

              x₁ = 25 0.25

              x₁ = 6.25 m

for v = 100 km / h

              x₁ = 27.78 0.25

              x₁ = 6.95 m

the total distance is

               x_total = x₁ + d

for v = 90 km / h

             x_total = 25 0.25 + 70

             x_total = 76.25 m

this is the distance until the cars stop and do not collide

3) the stopping distance of a truck is

   v = 90 km / h       d = 100 m

in this case we see that the braking distance is much higher,

the safe distance is given by the distance traveled during the reaction, as the truck brakes slower than the car this distance does not change

4) let's analyze the empirical rule: maintain the length of a car for each increase in speed of v = 10 m / h = 4.47 m / s

for the car case at v = 90km / h = 25 m / s

according to this rule we must this to

           x = 25 / 4.47 = 5.6 cars

each modern car is about 4 m long so the distance is

           x = 22 m

we see that this distance is much greater than the reaction  distance so it does not make much sense

Answer:

a=0.5g ms^-2

Explanation:

Let tension be T

acceleration be a

g be gravitational acceleration 9.81ms^-2

They share same T and a

by force diagram the net force on m:

T-2.1g=2.1a

net force on M:

6.3g-T=6.3a

solve:

(T-2.1g)/(6.3g-T)=2.1a/6.3a

3T+T=12.6g

T=3.15g newton

3.15g-2.1g=2.1a

a=0.5g ms^-2

Answer:

The other force is the weight of the student.

Explanation:

With respect to Newton's third law of motion, for the student to sit and balance on the chair, there must be two equal and opposite forces involved. The student applies his/ her weight on the chair which acts downwards, while the chair applies an equal but opposite force to the weight of the student.

The force applied by the chair on the student's body is counter balanced by the student's weight. Note that, if the weight of the student is greater than the opposing force from the chair, the chair would collapse.

Answer:

Option C. 2, 2, 3

Explanation:

__KClO₃ —> __ KCl + __O₂

The above equation can be balance as illustrated below:

KClO₃ —> KCl + O₂

There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by writing 2 before KClO₃ and 3 before O₂ as shown below:

2KClO₃ —> KCl + 3O₂

There are 2 atoms of K on the left side and 1 atom on the right side. It can be balance by writing 2 before KCl as shown below:

2KClO₃ —> 2KCl + 3O₂

Thus, the equation is balanced. The coefficients are: 2, 2, 3

Answer:

The magnitude of the induced current is 4.73 x 10⁻³ A.

Explanation:

Given;

number of turns, N = 1

cross sectional area of the loop, A = 8.8  cm² = 8.8 x 10⁻⁴ m²

change in magnetic field strength, ΔB = 1.8 T -  0.5 T = 1.3 T

change in time, Δt = 1.10 s

resistance of the loop, R = 2.2 ohm

The magnitude of the induced emf is calculated as;

[tex]emf = \frac{NA \Delta B}{\Delta t} \\\\emf = \frac{1 \times 8.8\times 10^{-4} \times 1.3}{1.10} \\\\emf = 1.04 \times 10^{-3} \ V[/tex]

The induced current in the loop is calculated as;

[tex]I = \frac{emf}{R} \\\\I = \frac{1.04 \times 10^{-3}}{2.2} \\\\I= 4.73 \times 10^{-4} \ A[/tex]

Therefore, the magnitude of the induced current is 4.73 x 10⁻³ A

Answer:

1) the required horizontal force F is 1095.6 N

2) W = 0 J { work done by rope will be 0 since tension perpendicular }

3) work is done by the worker is 1029.4 J

Explanation:

Given that;

mass of bag m = 125 kg

length of rope [tex]l[/tex] = 3.3 m

displacement of bag d = 2.2 m

1) What horizontal force is necessary to hold the bag in the new position?

from the figure below; ( triangle )

SOH CAH TOA

sin = opp / hyp

sin[tex]\theta[/tex] = d / [tex]l[/tex]

sin[tex]\theta[/tex] = 2.2/ 3.3

sin[tex]\theta[/tex] = 0.6666

[tex]\theta[/tex] = sin⁻¹ ( 0.6666 )

[tex]\theta[/tex]  = 41.81°

Now, tension in the string is resolved into components as illustrated in the image below;

Tsin[tex]\theta[/tex] = F  

Tcos[tex]\theta[/tex] = mg

so

Tsin[tex]\theta[/tex] / Tcos[tex]\theta[/tex] = F / mg

sin[tex]\theta[/tex] / cos[tex]\theta[/tex] = F / mg

we know that; tangent = sine/cosine

so

tan[tex]\theta[/tex] = F / mg

F = mg tan[tex]\theta[/tex]

we substitute

Horizontal force F = (125kg)( 9.8 m/s²) tan( 41.81° )

F = 1225 × 0.8944

F = 1095.6 N

Therefore, the required horizontal force F is 1095.6 N

2)  As the bag is moved to this position, how much work is done by the rope?

Tension in the rope and displacement of mass are perpendicular,

so, work done will be;

W = Tdcos90°

W = Td × 0

W = 0 J { work done by rope will be 0 since tension perpendicular }

3) As the bag is moved to this position, how much work is done by the worker

from the diagram in the image below;

SOH CAH TOA

cos = adj / hyp

cos[tex]\theta[/tex]  = ([tex]l[/tex] - h) / [tex]l[/tex]

we substitute

cos[tex]\theta[/tex]  = ([tex]l[/tex] - h) / [tex]l[/tex]  = 1 - h/[tex]l[/tex]

cos[tex]\theta[/tex] = 1 - h/[tex]l[/tex]

h/[tex]l[/tex] = 1 - cos[tex]\theta[/tex]

h = [tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

now, work done by the worker against gravity will be;

W = mgh = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

W = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

we substitute

W = (125 kg)((9.8 m/s²)(3.3 m)( 1 - cos41.81° )

W = 4042.5 × ( 1 - 0.745359 )

W = 4042.5 × 0.254641

W = 1029.4 J

Therefore,  work is done by the worker is 1029.4 J

Answer:

gravitational potential energy

Answer:

coefficient of viscosity of 8.4 poison denotes that the tangential frictional force acting per unit area when divided by the velocity gradient as a result of streamline flow conditions gives 8.4.

Explanation:

Viscosity is defined as the extent to which a fluid can resist flow when a force is applied to it.

Now, coefficient of viscosity is the term in which viscosity is calculated. It is basically the tangential frictional force acting per unit area which is divided by the velocity gradient as a result of streamline flow conditions.

Thus, coefficient of viscosity of 8.4 poison denotes that the tangential frictional force acting per unit area when divided by the velocity gradient as a result of streamline flow conditions gives 8.4.

The Work-Energy Theorem states that Work is equal to the change in kinetic energy, which is the first option . This theorem is an essential principle in physics and mechanics, so first option is correct.

Work (W) is a measure of the energy transferred to or from an object by a force acting on it. It is defined as the product of the force applied to the object and the displacement of the object in the direction of the force. When a force does positive work on an object, it transfers energy to the object, increasing its kinetic energy. Conversely, when a force does negative work on an object (opposite to its direction of motion), it takes energy away from the object, decreasing its kinetic energy. So, first option is correct.

Learn more about the work-energy theorem here.

https://brainly.com/question/30560150

#SPJ6

Answer:

towards the north.

Explanation:

Answer: A

Explanation:

Answer:

I'm gonna say d

Explanation:

bc they all seem very important

hope this helped

Answer:

5.56 A

Explanation:

From the question,

Q = it.............. Equation 1

Where Q = charges, i = current, t = time.

Make i the subject of the equation

i = Q/t.............. Equation 2

Given: Q = 200 coulombs, t = 0.6 minutes = (0.6×60) seconds

Substitite these values into equation 2

i = 200/(0.6×60)

i = 5.56 A

Hence the magnitude of the current flowing through the circuit is 5.56 A

Explanation:

Video games are developed around a structure that is unique from your usual media, where there is a sense purpose in mind when playing a game, as decided by the player, and it allows them to explore creative options in order to solve scenarios, depending on the genre. In the use of phones, internet, or computers, this structure is more rare and diverse from video games, which does not lean more toward a creative purpose to build from. That idea gives people the inspiration and discover skills they never knew they even had.

When an object is raised vertically, work is done against _gravitational force _________

The force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface is called gravitational force .

On every body there is a force acting named  gravitational force which attracts the body downward .

If a object / body is thrown up it always comes downward if only gravitational force is acting on it which act due to earth's gravity .

When an object is raised vertically, work is done against _gravitational force _________

learn more about  gravitational force :

https://brainly.com/question/12528243?referrer=searchResults

#SPJ2

Answer:

3.9 × 10^7 J

Explanation:

Given that a tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J

Solution

Since the tank is half full, the height = 2.5m

Pressure = density × gravity × height

Pressure = 900 × 9.8 × 2.5

Pressure = 22050 Pascal

The cross sectional area of the pump will be area of a circle.

A = πr^2

A = π × 15^2

A = 706.858 m^2

Using the formula

Density = mass/volume

Mass = density × volume

Mass = 900 × 706.86 × 2.5

Mass = 1590.435

Energy = mgh

Energy = 1590.435 × 9.8 × 2.5

Energy = 38965657.8 J

Since the work done = energy

Therefore, the work done = 3.9 × 10^7 J

Answer:

the magnitude of the electric field intensity is 20 N/C

Explanation:

 Given the data in the question;

mass m = 1 micro gram = 1 × 10⁻⁹ kg

time duration t = 10 sec

distance s = 1 m

the charge on the particle q = 10⁻¹² Coulomb

force applied on a charged particle due to electric field E is;

F = Eq  ------ equ 1

where q is the charge on the particle.

Also, force on a particle with mass m  will be;

F = ma  ------ equ

where a is acceleration

so F = ma = Eq

ma = Eq -------- equ 3

using kinetic equation

Distance = 1/2×at²

where a is acceleration and t is the time period

now lets consider that initial velocity is zero (0)

Here;

1 m = 1/2 × a × ( 10 s )²

1 m = a × 50 s²

a = 1 m / 50 s²

a = 0.02 m/s²

so, from equation 3

ma = Eq

E = ma / q

we substitute

E = (1 × 10⁻⁹ kg × 0.02) / 10⁻¹² Coulomb

E = 2 × 10⁻¹¹ / 10⁻¹²

E = 20 N/C

Therefore, the magnitude of the electric field intensity is 20 N/C

Answer:

24.5 m/s

Explanation:

KE=1/2mv^2

15000=1/2(50)v^2

30000=(50)v^2

600=v^2

sqrt600=v

v=24.5 m/s!!

Answer:

Momentum is given by

p

=

m

v

. Impulse is the change of momentum,

I

=

Δ

p

and is also equal to force times time:

I

=

F

t

. Rearranging,

F

=

I

t

=

Δ

p

t

=

0

−

20

,

000

5

=

−

4000

N

.

Explanation:

Momentum before the collision is

p

=

m

v

=

2000

⋅

10

=

20

,

000

k

g

m

s

−

1

.

Assuming the truck comes to a complete halt, the momentum after the collision is

0

k

g

m

s

−

1

.

The change in momentum,

Δ

p

, is initial minus final

→

0

−

20

,

000

=

−

20

,

000

This is called the impulse:

I

=

Δ

p

. Impulse is also equal (check the units) to force times time:

I

=

F

t

.

We can rearrange this expression to make

F

the subject:

F

=

I

t

=

Δ

p

t

=

−

20

,

000

5

=

−

4000

N

The negative sign just means the force acting is in the opposite direction to the initial momentum.

(This will be the average force acting during the collision: collisions are chaotic so the force is unlikely to be constant.)

Answer:

An object with no net forces acting on it which is initially at rest will remain at rest. If it is moving, it will continue to move in a straight line with constant velocity. Forces are "pushes" or "pulls" on the object, and forces, like velocity and acceleration are vector quantities.

Answer:

1.matter is any substance that has mass and takes up space by having volume.

2.The three fundamental phases of matter are solid, liquid, and gas (vapour),

3.In gases the particles move rapidly in all directions, frequently colliding with each other and the side of the container. With an increase in temperature, the particles gain kinetic energy and move faster. ... In liquids, particles are quite close together and move with random motion throughout the container.

3.In gases the particles move rapidly in all directions, frequently colliding with each other and the side of the container. With an increase in temperature, the particles gain kinetic energy and move faster. ... In liquids, particles are quite close together and move with random motion throughout the container.

Answer:

Explanation:

answer C looks good

there isn't really a "centrifugal " force. :/    when we are pushed "back" in a car seat.. it's not because there is a force pushing us backwards... but a force pushing us forwards.. just like when turning too,   a force pushes us into the corner,  not a force pushing out of the corner.  :)