Answer:

**Explanation:**

The given data is as follows.

height (h) = 4.70 m, mass = 81.0 kg

t = 1.84 s

As formula to calculate the velocity is as follows.

= 2gh

=

= 92.12

As relation between force, time and velocity is as follows.

F =

Hence, putting the given values into the above formula as follows.

F =

=

= 4055.28 N

Thus, we can conclude that **the magnitude of the average force exerted on the diver during that time is 4055.28 N.**

How can scientific method solve real world problems examples

A wheel starts at rest, and has an angular acceleration of 4 rad/s2. through what angle does it turn in 3.0 s?

Two 3.7 microCoulomb charges are 0.8 m apart. How much energy (in milliJoule) went into assembling these two charges? Enter a number with one digit behind the decimal point.

You have a grindstone (a disk) that is 105.00 kg, has a 0.297-m radius, and is turning at 71.150 rpm, and you press a steel axe against it with a radial force of 46.650 N. Assuming the kinetic coefficient of friction between steel and stone is 0.451. How many turns will the stone make before coming to rest?

A diverging lens has a focal length of 23.9 cm. An object 2.1 cm in height is placed 100 cm in front of the lens. Locate the position of the image. Answer in units of cm. 007 (part 2 of 3) 10.0 points What is the magnification? 008 (part 3 of 3) 10.0 points Find the height of the image. Answer in units of cm.

A wheel starts at rest, and has an angular acceleration of 4 rad/s2. through what angle does it turn in 3.0 s?

Two 3.7 microCoulomb charges are 0.8 m apart. How much energy (in milliJoule) went into assembling these two charges? Enter a number with one digit behind the decimal point.

You have a grindstone (a disk) that is 105.00 kg, has a 0.297-m radius, and is turning at 71.150 rpm, and you press a steel axe against it with a radial force of 46.650 N. Assuming the kinetic coefficient of friction between steel and stone is 0.451. How many turns will the stone make before coming to rest?

A diverging lens has a focal length of 23.9 cm. An object 2.1 cm in height is placed 100 cm in front of the lens. Locate the position of the image. Answer in units of cm. 007 (part 2 of 3) 10.0 points What is the magnification? 008 (part 3 of 3) 10.0 points Find the height of the image. Answer in units of cm.

**Answer:**

2613.3 pa

**Explanation:**

p=F/A

p=ma/A

p=200×9.8/0.75

p=2613.3

The formula for the rotational kinetic energy is

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

**Answer:**

4833J

**Explanation:**

Length=0.777

mass=2.67

# rods= 5

ω=573 rpm--> rad/s

I=kgm^2

K=1/2(number of rods)(I)(ω)=J

I know it's very late, but hope this helps anyone else trying to find the answer.

**Answer:**

the googles are 5.3 m from the edge

**Explanation:**

**Given that**

depth of pool , d = 3.2 m

Now, let i be the angle of incidence

a laser pointer 0.90 m above the edge of the pool and laser beam enters the water 2.2 m from the edge

⇒tan i = 2.2/0.9

solving we get

i = 67.8°

Using snell's law ,

n1 ×sin(i) = n2 ×sin(r)

n1= refractive index of 1st medium= 1

n2= refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

r = 44.1°

Now,

distance of googles = 2.2 + d×tan(r)

distance of googles = 2.2 + 3.2×tan(44.1)

distance of googles = 5.3 m

the googles are 5.3 m from the edge

Answer:

Option A is true

Explanation:

For option A, it's true because a map that has a scale model with the proportion of one centimeter to one kilometer is known as verbal scale which is a type of scale.

For option B, it's not true because though a scale model is most times always smaller than the object it represents, there are sometimes when the scale model is an enlarged view/representation of a small object.

For option C, it's true because there is no one scale that is confined to just a model. A model can use different scales to depict an object.

0.67 m/s2

0.075 m/s2

54 m/s2

4.0 m/s2

it's 9 squared divided by 6

it's 9 squared divided by 6

the answer is B 4.0 m/s2

**Answer:**

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

**Explanation:**

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

Here h is the Planck's constant, c is the speed of light, is the wavelength of the light and W the work function of the element:

Now, we calculate the wavelength for the new maximum kinetic energy:

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.