A
$5\text{.}\text{00}\times {\text{10}}^{5}\text{-kg}$ rocket is accelerating straight up. Its engines produce
$1\text{.}\text{250}\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{N}$ of thrust, and air resistance is
$4\text{.}\text{50}\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{N}$ . What is the rocket’s acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.
The wheels of a midsize car exert a force of 2100 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 250 N and the acceleration of the car is
$1\text{.}{\text{80 m/s}}^{2}$ , what is the mass of the car plus its occupants? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. For this situation, draw a free-body diagram and write the net force equation.
Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.
Use Newton’s laws of motion.
Given :
$a=4.00g=(4.00)(9.{\text{80 m/s}}^{2})=\text{39.2}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\text{;}\phantom{\rule{0.25em}{0ex}}$$m=\text{70}\text{.}\text{0 kg}$ ,
Find:
$F$ .
$$$$$\sum F\text{=+}F-w=\text{ma}\text{,}$ so that
$F=\text{ma}+w=\text{ma}+\text{mg}=m(a+g)$ .
$F=(\text{70.0 kg})[(\text{39}\text{.}{\text{2 m/s}}^{2})+(9\text{.}{\text{80 m/s}}^{2})]$$=3.\text{43}\times {\text{10}}^{3}\text{N}$ . The force exerted by the high-jumper is actually down on the ground, but
$F$ is up from the ground and makes him jump.
This result is reasonable, since it is quite possible for a person to exert a force of the magnitude of
${\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{N}$ .
When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.
A freight train consists of two
$8.00\times {10}^{4}\text{-kg}$ engines and 45 cars with average masses of
$5.50\times {10}^{4}\phantom{\rule{0.25em}{0ex}}\text{kg}$ . (a) What force must each engine exert backward on the track to accelerate the train at a rate of
$5.00\times {\text{10}}^{\text{\u20132}}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$ if the force of friction is
$7\text{.}\text{50}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}$ , assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?
Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor. (a) An 1800-kg tractor exerts a force of
$1\text{.}\text{75}\times {\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N}$ backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is
$0\text{.}{\text{150 m/s}}^{2}$ , what is the mass of the airplane? (b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane. (c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each.
Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
static fiction is friction between two surfaces in contact an none of sliding over on another, while Kinetic friction is friction between sliding surfaces in contact.
MINDERIUM
I don't get it,if it's static then there will be no friction.
author
It means that static friction is that friction that most be overcome before a body can move
kingsley
static friction is a force that keeps an object from moving, and it's the opposite of kinetic friction.
author
It is a force a body must overcome in order for the body to move.
Eboh
If a particle accelerator explodes what happens
Eboh
why we see the edge effect in case of the field lines of capacitor?
Force equals mass time acceleration. Weight is a force and it can replace force in the equation. The acceleration would be gravity, which is an acceleration. To change from weight to mass divide by gravity (9.8 m/s^2).
the write question should be " How many Topics are in O- Level Physics, or other branches of physics.
effiom
how many topic are in physics
Praise
Praise what level are you
yusuf
If u are doing a levels in your first year you do AS topics therefore you do 5 big topic i.e particles radiation, waves and optics, mechanics,materials, electricity. After that you do A level topics like Specific Harmonic motion circular motion astrophysics depends really
Anya
Yeah basics of physics prin8
yusuf
Heat nd Co for a level
yusuf
yh I need someone to explain something im tryna solve . I'll send the question if u down for it
a ripple tank experiment a vibrating plane is used to generate wrinkles in the water .if the distance between two successive point is 3.5cm and the wave travel a distance of 31.5cm find the frequency of the vibration
Tamdy
hallow
Boniface
please send the answer
Boniface
the range of objects and phenomena studied in physics is