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:
### Final answer:

### Explanation:

### Learn more about Average Speed here:

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.

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.

- 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).
- 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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According to the Heisenberg uncertainty principle, quantum mechanics differs from classical mechanics in that: Select the correct answer below: Quantum mechanics involves particles that do not move. It is impossible to calculate with accuracy both the position and momentum of particles in classical mechanics. The measurement of an observable quantity in the quantum domain inherently changes the value of that quantity. All of the above

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?

PLEASE HELP DUE BEFORE 11:30 TODAY!!!!Which of the following quantities is NOT a vector quantity? A. 926 m to the north B. 5.2 m/s to the west C. 46 m down D. 12.3 m/s faster

Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).

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?

PLEASE HELP DUE BEFORE 11:30 TODAY!!!!Which of the following quantities is NOT a vector quantity? A. 926 m to the north B. 5.2 m/s to the west C. 46 m down D. 12.3 m/s faster

Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).

b. 4 per hour

c. 10 per hour

d. 0.67 per hour

e. 15 per hour

**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 = ..................1

put here value we will get

velocity in problems per hour =

**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.

**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

**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.

**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.

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.

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,

Here,

= mass of bullet

= Mass of Block of wood

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

PART A)

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

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

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

Using this value at the first equation we have that,

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

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 =

for such more question on** speed**

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