A 81.0 kg diver falls from rest into a swimming pool from a height of 4.70 m. It takes 1.84 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
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.
As relation between force, time and velocity is as follows.
Hence, putting the given values into the above formula as follows.
= 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.
A piston above a liquid in a closed container has an area of 0.75m^2, and the piston carries a load of 200kg. What will be the external pressure on the upper surface of the liquid?
A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.777 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?
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.
I know it's very late, but hope this helps anyone else trying to find the answer.
It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.2m from the edge.-How far are the goggles from the edge of the pool?
the googles are 5.3 m from the edge
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
r = 44.1°
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
Which one of these are true about scale models? a. A map may have a scale model with the proportion of one centimeter to one kilometer. b. A scale model is always smaller than the object it represents, c. A model may be built to different scales.
Option A is true
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.
A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. If the radius of the ride is 9.0 meters, what is the acceleration of the ride?4.0 m/s2 0.67 m/s2 0.075 m/s2 54 m/s2
it's 9 squared divided by 6
the answer is B 4.0 m/s2
Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. What could you do to increase the maximum kinetic energy of electrons to 1.5 eV?
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.
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.