The local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.



Calculate the time required for bumper car A to travel the 10 m to collide with bumper car B.
Calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s.
Is the direction of motion for bumper car A following the collision with bumper car B to the right, to the left, or is bumper car A at rest?
Is the collision elastic? Justify your answer.

Answers

Answer 1

Answer:

a. 20 s

b. 0 m/s  

c. right

d.no its inelastic because when the car b was at rest and a was coming in at it, since b had no force what so ever car a swept it away with it moving to the right

Explanation:

im not sure though

Answer 2

By applying conservation of linear momentum, the answers are:

1. Time = 2 s

2. 3 m/s

3. same direction

4. Inelastic collision

COLLISION

There are for types of collision. They are;

Elastic CollisionPerfectly elastic collisionInelastic collisionPerfectly Inelastic collision

Given that a local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.

1. The time required for bumper car A to travel the 10 m to collide with bumper car B can be calculated by using first equation of linear motion.

V = U + at

Where

V = 4 m/s

U = 2 m/s

a = 1 m/[tex]s^{2}[/tex]

Substitute all the parameters into the formula

4 = 2 + t

t = 4 - 2

t = 2s

2. To calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s, we will apply conservation of linear momentum

[tex]m_{1}u_{1}[/tex] = [tex]m_{1}v_{1}[/tex] + [tex]m_{2}v_{2}[/tex]

300 x 4 = 300V + 200 x 3

1200 = 300V + 300

300V = 1200 - 300

300V = 900

V = 900/300

V = 3 m/s

3. Since the final velocity of car A is positive, the direction of motion for bumper car A follows the collision with bumper car B to the right.

4. Since the both move at the same velocity, the collision inelastic.

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Related Questions

it takes a cheetah just 3s to reach its top speed of 30m/s what is it's acceleration​

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The acceleration will be "10 m/s²".

Given values are:

Speed,

v = 30 m/s

Time,

t = 3 s

As we know,

→ [tex]Acceleration = \frac{Speed}{Time}[/tex]

or,

→                   [tex]a = \frac{v}{t}[/tex]

By substituting the values,

                       [tex]= \frac{30}{3}[/tex]

                       [tex]= 10 \ m/s^2[/tex]

Thus the above solution is correct.    

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A roller-coaster car has a mass of 1240 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 22 m, its speed is not changing. (a) At the top of the hill, what is the normal force (using the negative sign for the downward direction) FN on the car from the track if the car's speed is v = 8.7 m/s? (b) What is FN if v = 20 m/s? Use g=9.81 m/s2.

Answers

Answer:

a)      N = 7.90 10³ N,  b) N = -1.04 10⁴ N

Explanation:

a) For this exercise we can use Newton's second law

             N -W = m (-a)

The relationship is centripetal, the negative sign of the acceleration is because it points towards the center of the circle

              a = v² / r

we substitute

             N = mg -m v² /r  

             N = m (g - v² /r)

let's calculate

v = 8.7 m / s

             N = 1240 (9.81 - 8.7²/22)

             N = 7.90 10³ N

b) v = 20 m / s

             N = 1240 (9.81 - 20²/22)

             N = -1.04 10⁴ N

A storage tank has the shape of an inverted circular cone with height 15 m and base radius of 5 m. It is filled with water to a height of 11 m. Find the work required to empty the tank by pumping all of the water to the top of the tank. (The density of water is 1000 kg/m3. Assume g = 9.8 m/s2 is the acceleration due to gravity. Round your answer to the nearest integer.) g

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One of the characteristics of a wide-angle lens is the ability to capture the expanse of a scene in an image with its wide angle of view. Also, it is able to create an intriguing deformed effect when you photograph the subject from a close distance. These characteristics can be utilised to bring out the shape of a subject or the depth of a landscape. That being said, care is needed as distortion is more likely to occur toward the edges of the image. To make effective use of a wide-angle lens, it is important to control both the angle of view as well as distortions in the image at the same time. By doing so, you will be able to add a dynamic touch to your photographic expression.

when is an object considered to be in motion​

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Answer:

An object is considered to be in motion only when its position changes over time with reference to a point which will be known as orgin

True or False
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Answers

Answer:

True

Explanation:

"a process by which someone is deprived of normal external stimuli such as sight and sound for an extended period of time, especially as an experimental technique in psychology."

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An alumium rod when measured with a steel scale both being at 25 degree celcius appears to be 1 meter long. If the scale is correct at 0 degree celcius, what is the true length of the rod at 25 degree celcius? What wull be the length of rod at 0 degree celcius?​

Answers

Given:
temperature at which the rod is measured= 25°
length of aluminium rod measured at 25°C= 1 meter
temperature at which the scale is correct= 0°C
Linear expansivity of steel= 12* 10^-6
linear expansivity of aluminum= 26* 10^-6
To find:
True length of rod at 25°C
Length of rod at 0°C
Solution:
Let the length of the aluminum and steel rod at 0°C be L0A and L0S respectively.
Length of the steel scale at 25°C , thermal expansion is:
LA25 = LA0(1+ (αS - αA)25)
LA0 = LA25/ (1+ (αS-αA)t)
LA0 = 100/ (1+ (12*10^-6 - 26*10^-6)25)
LA0 ≈ 100.05 cm at 0°C
2. Let the length at 25°C of the aluminium trod be LA25
The actual measure of aluminum rod at 25°C, will be
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LA25 ≈ 100.06 cm

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During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of 39.0 kg and its horizontal component of velocity is 6.50 m/s just before the 65.0 kg performer catches it. If the performer is initially motionless on nearly frictionless roller skates, what is his speed immediately after catching the cannon ball

Answers

Answer:

[tex]2.4375\ \text{m/s}[/tex]

Explanation:

[tex]m_1[/tex] = Mass of cannon ball = 39 kg

[tex]m_2[/tex] = Mass of performer = 65 kg

[tex]u_1[/tex] = Initial horizontal component of cannon ball's velocity = 6.5 m/s

[tex]u_2[/tex] = Initial horizontal component of performer's velocity = 0

v = Velocity of combined mass

As the momentum of the system is conserved we have

[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{39\times 6.5+0}{39+65}\\\Rightarrow v=2.4375\ \text{m/s}[/tex]

The speed of the performer immediately after catching the cannon ball is [tex]2.4375\ \text{m/s}[/tex].

it takes 840s to walk completely around a circular track, moving at a speed of 1.20m/s? what is the radius of the track?

Answers

Answer:

160.43 meters

Explanation:

T=(2*pi*r)/v

840=(2*pi*r)/1.2

1008=2*pi*r

504=pi*r

160.43=radius

The distance around the circular track is equal to its circumference. If it takes840 s  to complete the path with a velocity of 1.20 m/s then the radius of the path will be 160 m.

What is velocity?

Velocity is a physical quantity measuring the distance traveled by unit time. It is a vector quantity thus, characterized by its magnitude and direction. Velocity is usually expressed in m/s.

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Given that the time take to complete the circular path = 840 s

velocity = 1.20 m/s

distance = circumference of the circular path = 2πr.

distance = time × velocity

2πr = 840 s × 1.20 m/s

      = 1008 m.

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If a 15 N box is lifted a distance of 3 m, how much work is done?

0 J

45 J

5 J

5 N

Answers

Answer:

W=45J

Explanation:

W=Fd

W=15(3)=45

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An official major league baseball has a mass of 0.14 kg. A pitcher throws a 40 m/s fastball which is hit by the batter straight back up the middle at a speed of 46 m/s.
a) What is the change in momentum of the ball during the collision with the bat?
b) If this collision occurs during a time of 0.012 seconds, what is the average force exerted by the bat on the ball?

Answers

Answer:

(a) The change in momentum is 12.04 kg-m/s

(b) The force exerted by the bat is 1003.33 N

Explanation:

Given that,

The mass of a ball, m = 0.14 kg

Initial speed of the ball, u = 40 m/s

Final speed of the ball, v = -46 m/s

(a) The change in momentum of the ball during the collision with the bat is given by :

[tex]\Delta p=m(v-u)\\\\=0.14(-46-40)\\\\=-12.04\ kg-m/s[/tex]

(b) Time for collision, t = 0.012 s

Now the force can be calculated as follows :

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Answer:

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Answers

Answer:
C
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a. protons.ekgfsbdbsi

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A. Protons

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Atomic number is an indicator of how many protons an element possesses in its nucleus. We use atomic number to identify an element. For example, carbon has 6 protons and an atomic number of 6.

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Infrared light is the part of the electromagnetic spectrum, and is an electromagnetic form of radiation. It comes from the heat and thermal radiation, and it is not visible to the naked eyes. In infrared transmission, senders can be simple light emitting diodes (LEDs) or laser diodes. Photodiodes act as receivers.
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Netwtons 1st law of motion

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Answer:

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Explanation:

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Suppose that a uniform rope of length L resting on a frictionless horizontal surface, is accelerated along the direction of its length by means of a force F, pulling it at one end. A mass M is accelerated by the rope. Assuming the mass of the rope to be m and the acceleration is a. Stated in terms of the product ma, what is the tension in the rope at the position 0.3 L from the end where the force F is applied if the mass M is 1.5 times the mass of the rope m?

Answers

Answer:

2.2 ma

Explanation:

Given :

Length of the rope = L

Mass of the rope = m

Mass of the object pulled by the rope = M

M = 1.5 m

So, L [tex]$\rightarrow$[/tex] m

For unit length [tex]$\rightarrow \frac{m}{L}$[/tex]

∴ 0.3 L = [tex]$0.3 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.3 m

And for remaining 0.7 L =  [tex]$0.7 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.7 m

By Newtons law of motion,

F - T = ( 0.3 m) a .........(1)

T = ( M + 0.7 m) a

T = ( 1.5 m + 0.7 m) a

T = ( 2.2 m ) a  ..............(2)

So from equation (1) and (2), we have

Tension on the rope

T = 2.2 ma

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Explanation:

Answer:

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If a spring extends by 3 cm when a 4 N weight is suspended from it, find the extension
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Answers

Answer:

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λ = 5.434 x 10⁻⁷ m = 543.4 nm

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Answers

Answer:

0.204 kg

Explanation:

6/(9.8*3)

6/29.4=0.204

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