PHYSICS
HIGH SCHOOL

b) the time at which the ball reaches maximum height

c) the time required for the entire trip

d) the ball's velocity just before it hits the ground upon returning

Answer:

A. The maximum height reached by the ball is 8.62 m

B. The time taken to reach the maximum height is 1.33 s

C. The time required for the entire trip is 2.66 s

D. The ball's velocity just before it hits the ground upon returning is 13.034 m/s

A. How to determine the maximum height- Initial velocity (u) = 13 m/s
- Final velocity (v) = 0 m/s (at maximum height)
- Acceleration due to gravity (g) = 9.8 m/s²

- Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 13² – (2 × 9.8 × h)

0 = 169 – 19.6h

Collect like terms

0 – 169 = –19.6h

–169 = –19.6h

Divide both side by –19.6

h = –169 / –19.6

h = 8.62 m

B. How to determine the timeThe time taken to reach the maximum height can be obtained as follow:

- Initial velocity (u) = 13 m/s
- Final velocity (v) = 0 m/s (at maximum height)
- Acceleration due to gravity (g) = 9.8 m/s²

- Time to reach maximum height (t) =?

v = u – gt (since the ball is going against gravity)

0 = 13 – (9.8 × t)

0 = 13 – 9.8t

Collect like terms

0 – 13 = –9.8t

–13 = –9.8t

Divide both side by –9.8

t = –13 / –9.8

t = 1.33 s

C. How to determine the time for the entire trip- Time to reach maximum height (t) = 1.33 s
- Time for the entire trip (T) =?

T = 2t

T = 2 × 1.33

T = 2.66 s

D. How to determine the velocityThe velocity of the ball before hitting the ground can be obtained as follow:

- Time to reach ground from maximum height (t) = 1.33
- Initial velocity (u) = 0 m/s
- Acceleration due to gravity (g) = 9.8 m/s²

- Final velocity (v) =?

v = u + gt

v = 0 + (9.8 × 1.33)

v = 0 + 13.034

v = 13.034 m/s

Learn more about motion under gravity:

brainly.com/question/22719691

Answer: 1. the max height is solved from v = square root of 2 gy where g is equal to 9.8 m/s2/ hence y is equal to 8.62 meters.

2. time at max height is solve when velocity = 0. ( 0 = -9.81*t +13); t = 1.325 seconds

3. time required for the entire trip is 2*1.325 equal to 2.65 seconds.

4. without air resistance, the velocity just before the ball hits the ground is the same as its initial velocity, 13 m/s.

2. time at max height is solve when velocity = 0. ( 0 = -9.81*t +13); t = 1.325 seconds

3. time required for the entire trip is 2*1.325 equal to 2.65 seconds.

4. without air resistance, the velocity just before the ball hits the ground is the same as its initial velocity, 13 m/s.

HIGH SCHOOL

It is estimated that 26 large pizzas are about right to serve 66 students of a physics club meeting. How many pizzas would be required if the group, because of a conflicting math club meeting, was only going to have 38 students in attendance?

Answer:

The answer to your question is: 15 pizzas

Explanation:

data

26 large pizzas ------ 66 students

? large pizzas -------- 38 students

Rule of three

x = 38 (26) / 66 = 14.96 ≈ 15 pizzas

HIGH SCHOOL

A 1530 kg car moving south at 12.1 m/s collides with a 2560 kg car moving north. The cars stick together and move as a unit after the collision at a velocity of 5.48 m/s to the north. Find the velocity of the 2560 kg car before the collision. Assume that North is positive. Answer in units of m/s.

Answer:

16 m/s north

Explanation:

According to the law of momentum conservation, total momentum before and after must be the same

Total momentum after the collision as both cars stick together as 1 unit traveling at speed of 5.48m/s to the north

(m + M)V = (1530 + 2560)*5.48 = 22413 kgm/s

So the total momentum before must also be 22413, which is made of

north

COLLEGE

Ch 31 HW Exercise 31.10 7 of 15 Constants You want the current amplitude through a inductor with an inductance of 4.90 mH (part of the circuitry for a radio receiver) to be 3.00 mA when a sinusoidal voltage with an amplitude of 12.0 V is applied across the inductor. Part A What frequency is required? ff = nothing Hz Request Answer Provide Feedback

Answer:

f = 130 Khz

Explanation:

In a circuit driven by a sinusoidal voltage source, there exists a fixed relationship between the amplitudes of the current and the voltage through any circuit element, at any time.

For an inductor, this relationship can be expressed as follows:

VL = IL * XL (1) , which is a generalized form of Ohm's Law.

XL is called the inductive reactance, and is defined as follows:

XL = ω*L = 2*π*f*L, where f is the frequency of the sinusoidal source (in Hz) and L is the value of the inductance, in H.

Replacing in (1), by the values given of VL, IL, and L, we can solve for f, as follows:

f = VL / 2*π*IL*L = 12 V / 2*π*(3.00*10⁻³) A* (4.9*10⁻³) H = 130 Khz

HIGH SCHOOL

What pressure difference is required between the ends of a 2.0-m-long, 1.0-mm-diameter horizontal tube for 40 C water to flow through it at an average speed of 4.0 m/s?

Answer : The pressure difference required is,

Explanation : Given,

Length of tube = 2.0 m

Diameter of tube = 1.0 mm = 1.0 × 10⁻³

Average speed of water = 4.0 m/s

Viscosity of water at 40°C = 0.7 × 10⁻³

The expression used for pressure difference is:

where,

= pressure difference

= viscosity of water = 0.7 × 10⁻³

L = length of tube = 2.0 m

A = area of tube =

= average speed of water = 4.0 m/s

Now put all the given values in the above expression, we get:

Thus, the pressure difference required is,

COLLEGE

calculate the workdone to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it

The amount of work that would be required to stretch this elastic string by 40 cm is 20 Joules.

Given the following data:

- Force = 10 Newton

- Extension = 4 cm to meters = = 0.04 meters.

To find how much work will be required to stretch an elastic string by 40 cm:

First of all, we would determine the spring constant by using the formula:

K = 250 N/m

Mathematically, the work done by a string is given by the formula:

Where:

- K is the spring constant.

- e is the extension.

Substituting the given parameters into the formula, we have;

Work done = 20 Joules.

Therefore, the amount of work that would be required to stretch this elastic string by 40 cm is 20 Joules.

Read more: brainly.com/question/14621920

The force of F=10 N produces an extension of

on the string, so the spring constant is equal to

Then the string is stretched by . The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:

on the string, so the spring constant is equal to

Then the string is stretched by . The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:

HIGH SCHOOL

If anybody has the rest of the answers it would be SO appreciated!! will make brainliest!!!!! A vector of components (6, 2) is multiplied by the scalar value of 3. What is the magnitude and direction of the resultant vector?

magnitude: 19.0; direction: 18.4°

magnitude: 8.49; direction: 45.0°

magnitude: 2.11; direction: 12.6°

magnitude: 18.1; direction: 16.7°

A :magnitude: 19.0; direction: 18.4°

MIDDLE SCHOOL

If a forklift raises a 76-kg load a distance of 2.5 m, about how much work has it done?

I don't know sorry !!!

COLLEGE

A 5.36 kg object falls freely (ignore air resistance), after being dropped from rest. Determine the initial kinetic energy (in J), the final kinetic energy (in J), and the change in kinetic energy (in J) for the following. (a) first meter of fallinitial kinetic energy? Jfinal kinetic energy? Jchange in kinetic energy? J(b) second meter of fallinitial kinetic energy? Jfinal kinetic energy? Jchange in kinetic energy

Answer: 52.53

Explanation:

m*g = 5.36k * 9.8N/kg = 52.53 N. = Wt.

of object.

a. KE = 0 J. = Initial KE

V^2 = Vo^2 + 2g*h

V^2 = 0 + 19.6*1 = 19.6

V = 4.43 m/s.

KE = 0.5m*V^2 = 2.68*4.43^2 = 52.59 J.

b. V^2 = Vo^2 + 2g*h

V^2 = 4.43^2 + 19.6*1 = 39.22

V = 6.26 m/s.

KEo = 2.68*4.43^2 = 52.59 J.

KE = 2.68*6.26^2 = 105 J.

HIGH SCHOOL

What did Thomson's and Rutherford's experiments have in common?

They both used charged particles in their experiments.

Answer:

the answer is A

Explanation: