Let A be the last two digits, and let B be the last three digits, and the C be the sum of the last 4 digits of your 8-digit student ID. (Example: For 20245347, A = 47, B = 347, and C = 19) A train moves at an average speed of (23.0 + A) m/s for (250.0 + B) seconds and then at an average speed of (45.0 + C) m/s for (800.0 + B) seconds. Determine the average speed for the entire time in meters per second (m/s). Round your final answer to 3 significant figures.

Answers

Answer 1
Answer:

Answer:

66.053m/s

Explanation:

A = 47

B = 347

C = 19

Train moves at

(23 + A)m/s

= 23 + 47 = 60m/s

At (250.0+B) seconds

250.0+347 =

547 seconds

Distance d,

= 70 x 597

= 41790

It also moves at

(45.0 + c)

= 45 + 19

= 64m/s

Time = 800 + B

= 800 + 347

= 1147

Distance,

= 64 x 1147

= 73408m

Total distance,

= 73408 + 41790

= 115,198

Total time,

= 597 + 1147

= 1744

Average speed,

= Total distance / total time

= 115198/1174

= 66.053m/s

Answer 2
Answer:

Final answer:

The average speed over the entire time can be calculated by first finding the distances the train travels over both periods, then finding the total distance and the total time, and finally dividing the total distance by the total time. The value must be rounded to three significant figures.

Explanation:

You can find the average speed of the train over the full-time interval by dividing the total distance travelled by the total time. To begin with, you would have to find the distances the train covered during both periods.

  1. The distance (D1) it travelled during the first period can be found by multiplying the average speed (23.0 + A) by the time (250.0 + B).
  2. The distance (D2) it travelled during the second period can be calculated by multiplying the average speed (45.0 + C) by the time (800.0 + B).

Then you add D1 and D2 to get the total distance (TD). This will be (D1 + D2). The total time (TT) will be found by adding both time intervals, which means it equals (250.0 + B) + (800.0 + B). You then divide the total distance by the total time to get the average speed, i.e., TD/TT. Lastly, round the average speed to 3 significant figures.

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he magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.9 AA , how many turns of wire would you need? Express your answer using two significant figures.
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?
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Last night, Shirley worked on her accounting homework for one and one half hours. During that time, she completed 6 problems. What is the velocity in problems per hour? a. 6 per hour
b. 4 per hour
c. 10 per hour
d. 0.67 per hour
e. 15 per hour

Answers

Answer:

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

Explanation:

given data

worked on  homework time = 1.5 hour

completed = 6 problems

to find out

What is the velocity in problems per hour

solution

we know that Shirley solve complete 6 accounting homework problem in 1.5 hour so her velocity  in problems per hour will be as

velocity in problems per hour =  (complete\ problem)/(time\ taken)   ..................1

put here value we will get

velocity in problems per hour =  (6)/(1.5)

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

Answer:

option (b) is correct

Explanation:

time t = 1 and half hour = 1.5 hour

Number of problems, n = 6

So, the velocity in problems

v = n / t

v = 6 / 1.5

v = 4

So, the rate is 4 problems per hour.

A cylinder with a diameter of 2.0 in. and height of 3 in. solidifies in 3 minutes in a sand casting operation. What is the solidification time if the cylinder height is doubled? What is the time if the diameter is doubled?

Answers

Answer:

3 min 55 sec is the solidification time if the cylinder height is doubled

7min 40 sec if the diameter is doubled

Explanation:

see the attachment

Hercules X-1 is a pulsating X-ray source. The X-rays from this source sometimes completely disappear for 6 hours every 1.7 days because the neutron star has a 1.7-day orbital period around its companion star, and it is eclipsed for ____ hours once every orbital period.

Answers

Answer:

06 Hours

Explanation:

As per the details given in the question it self, the neutron star X-1 is revolving around its companion star. The orbital period is 1.7 years which means it will complete the revolution in 1.7 years. During the movement in the orbit we will be able to detect the x-rays except for the time when it goes behind the companion star and eclipsed by it as seen from Earth.

Since the x-rays disappear completely for around 6 hours. This clearly means that eclipse period is 06 hours.

How does an increase in cold working effect Modulus of Elasticity and why?

Answers

Answer:

There is a decrease in modulus of elasticity

Explanation:

Young's Modulus of elasticity also known as elastic modulus is the deformation of a body along a particular axis under the action of opposing forces along that axis. at atomic levels, it depends on bond energy or strength.

In cold working processes, plastic deformation a metal occurs below its re-crystallization temperature due to which crystal structure of metal gets distorted and as a result of dislocations fractures also occur resulting in hardening of metal but bonds at atomic levels defining elasticity are temporarily affected.

Thus an increase in cold working results in a decrease in modulus of elasticity.

Assume: The bullet penetrates into the block and stops due to its friction with the block.

The compound system of the block plus the

bullet rises to a height of 0.13 m along a

circular arc with a 0.23 m radius.

Assume: The entire track is frictionless.

A bullet with a m1 = 30 g mass is fired

horizontally into a block of wood with m2 =

4.2 kg mass.

The acceleration of gravity is 9.8 m/s2 .

Calculate the total energy of the composite

system at any time after the collision.

Answer in units of J.

Taking the same parameter values as those in

Part 1, determine the initial velocity of the

bullet.

Answer in units of m/s.

Answers

To solve this problem we will start considering the total energy of the system, which is given by gravitational potential energy of the total of the masses. So after the collision the system will have an energy equivalent to,

E_T = (m_1+m_2)gh

Here,

m_1= mass of bullet

m_2= Mass of Block of wood

The ascended height is 0.13m, so then we will have to

PART A)

E_T = (m_1+m_2)gh

E_T = (0.03+4.2)(9.8)(0.13)

E_T = 5.389J

PART B) At the same time the speed can be calculated through the concept provided by the conservation of momentum.

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Since the mass at the end of the impact becomes only one in the system, and the mass of the block has no initial velocity, the equation can be written as

m_1v_1 =(m_1+m_2)v_f

The final velocity can be calculated through the expression of kinetic energy, so

E_T = KE = (1)/(2) (m_1+m_2)v_f^2

v_f = \sqrt{(2E_T)/(m_1+m_2)}

v_f = \sqrt{(2*5.389J)/(0.03+4.2)}

v_f = 1.5962m/s

Using this value at the first equation we have that,

m_1v_1 =  (m_1+m_2)v_f

v_1 =((m_1+m_2)v_f)/(m_1)

v_1 = ((0.03+4.2)(1.5962))/(0.03)

v_1 = 225.06m/s

The particle with charge q is now released and given a quick push; as a result, it acquires speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m , what was its initial speed v ?.

Answers

The given situation is illustrated below. A particle is released and given a quick push. As a result, it acquires a speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m, the initial speed of the particle is √((2qV/m).)


To solve the problem, we need to apply the law of conservation of energy, which states that energy can neither be created nor destroyed; it can only be transformed from one form to another.
Initial potential energy = Final kinetic energy
The initial potential energy of the particle is given by
U = qV
where V is the potential difference between the corner and the center of the square.
At the center of the square, the potential energy is zero.
The final kinetic energy of the particle is given by
K = (1/2) mv^2
where m is the mass of the particle and v is its final velocity.
Since the particle is momentarily at rest at the center of the square, its final kinetic energy is zero.
Therefore, we have
qV = (1/2) mv^2
Solving for v, we get
v = √((2qV/m).)

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