Physics, 01.08.2020 01:01 skyvargasov9cad

A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 20.5 meters?

Answers: 1

Show answers

Answers

Answer from: Frenchfries13

The final speed of the crate after being pulled these 20.5 meters is 13.82 m/s

Explanation:

I'll assume that the correct question is

A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240 N. For the first 12.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.21. What is the final speed of the crate after being pulled these 22.5 meters?

mass of box = 51 kg

for the first 12 m, it is pulled with a constant force of 240 N

The acceleration of the box for this first 12 m will be

from F = ma

a = F/m

where F is the pulling force

m is the mass of the box

a is the acceleration of the box

a = 240/51 = 4.71 m/s^2

Since the body started from rest, the initial velocity u = 0

applying Newton's equation of motion to find the final velocity at the end of the first 12 m, we have

$v^{2}= u^{2}+2as$

where v is the final velocity

u is the initial velocity which is zero

a is the acceleration of 4.71 m/s^2

s is the distance covered which is 12 m

substituting value, we have

$v^{2}$ = 0 + 2(4.71 x 12)

$v^{2}$ = 113.04

$v = \sqrt{113.04}$ = 10.63 m/s

For the final 10.5 m, coefficient of friction is 0.21

from f = μF

where f is the frictional force,

μ is the coefficient of friction = 0.21

and F is the pulling force of the box 240 N

f = 0.21 x 240 = 50.4 N

Net force on the box = 240 - 50.4 = 189.6 N

acceleration = F/m = 189.6/51 = 3.72 m/s^2

Applying newton's equation of motion

$v^{2}= u^{2}+2as$

u is initial velocity, which in this case = 10.63 m/s

a = 3.72 m/s^2

s = 10.5 m

v = ?

substituting values, we have

$v^{2}$ = $10.63^{2}$ + 2(3.72 x 10.5)

$v^{2}$ = 112.9 + 78.12

v = $\sqrt{191.02}$ = 13.82 m/s

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Another question on Physics

Physics, 22.06.2019 05:10

Which diagram correctly demonstrates the various forces acting on a ball moving horizontally with some speed?

Answers: 2

Answer

Physics, 22.06.2019 06:00

Suppose water is leaking from a tank through a circular hole of area ah at its bottom. when water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to cah 2gh , where c (0 < c < 1) is an empirical constant. a tank in the form of a right-circular cone standing on end, vertex down, is leaking water through a circular hole in its bottom. (assume the removed apex of the cone is of negligible height and volume.) (a) suppose the tank is 20 feet high and has radius 8 feet and the circular hole has radius 2 inches. the differential equation governing the height h in feet of water leaking from a tank after t seconds is dh dt = − 5 6h3/2 . if the height of the water is initially 8 feet, how long will it take the tank to empty? (round your answer to two decimal places.)

Answers: 2

Answer

Physics, 22.06.2019 12:30

Which levels of government establish and implement educational requirements for minors

Answers: 1

Answer

Physics, 22.06.2019 22:00

Awhite-blue star is hotter than a red star. select the best answer from the choices provided true or false

Answers: 2

Answer

You know the right answer?

A 51.0 kg box, starting from rest, is pulled across a floor with a constant horizontal force of 240...