When work is done, an object changes its

A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then mo

makkiz [27]

Answer:

(a) The speed of the target proton after the collision is: $V_{2f} =433(m/s)$ , and (b) the speed of the projectile proton after the collision is: $v_{1f}=250(m/s)$ .

Explanation:

We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle: $m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}$ , and y axle: $0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}$ . Now replacing the value given as: $v_{1i}=500(m/s)$ , $\beta_{1}=+60^{o}$ for the projectile proton and according to the problem $\beta_{1}and\beta_{2}$ are perpendicular so $\beta_{2}=-30^{o}$ , and assuming that $m_{1}=m_{2}$ , we get for x axle: $500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2}$ and y axle: $0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}$ , then solving for $v_{2f}$ , we get: $v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f}$ and replacing at the first equation we get: $500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}$ , now solving for $v_{1f}$ , we can find the speed of the projectile proton after the collision as: $v_{1f}=250(m/s)$ and $v_{2f}=\sqrt{3}*v_{1f}=433(m/s)$ , that is the speed of the target proton after the collision.

5 0

3 years ago

Can someone please explain to me why the net force of the ball changes in the Y direction and not the X Direction? I made little

stiks02 [169]

Whenever an object is in projectile motion, that is, it has 2-dimensional motion in the x and y axis, the resultant force on the object is in the y-direction.
This is because once the object has been projected, or the ball has been kicked in this case, there is no longer a force being applied on it in the x-direction. The air resistance is also neglected so the ball's final velocity in the x-direction is equal to its initial velocity in the x-direction.
However, the force of gravity cannot be neglected and causes the ball to come downwards. Therefore, after the ball has been projected, the net force on the ball is downwards, due to gravity.

8 0

3 years ago

If your front lawn is 21.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1350 new snowflakes every minu

ziro4ka [17]

Answer:

snow is 64.638 kg / hr

Explanation:

Given data

wide w = 21 feet

long L = 20 ft

area A = 1350 square foot

mass of snow m = 1.90 mg

to find out

snow in kilograms / hour

solution

we will find snow in kg

so we apply formula that is

snow kg / hour = w × L ×A × m × 60/10^6

put all value we get snow

snow = 21 × 20 × 1350 × 1.90 × 60/10^6

snow = 420 × 1350 × 1.90 × 60/10^6

snow = 1077300 × 60/10^6

snow = 64.638

hence snow is 64.638 kg / hr

7 0

3 years ago

If the new moon happens on January 15th, what shape will it be on February 6th?

Jobisdone [24]

-- From January 15 to February 6 is a period of 22 days.

-- The period of the full cycle of moon phases is 29.53 days.

-- So those dates represent (22/29.53) = 74.5% of a full cycle of phases.

-- That's almost exactly 3/4 of a full cycle, so on February 6, the moon would be almost exactly at Third Quarter. That's the left half of a disk (viewed from the northern hemisphere).

3 0

3 years ago

What's the minimum Out PUT WORK required to raise 14,0m3 of water 26.0m?

BartSMP [9]

Answer:

3.57 MJ

Explanation:

ASSUMING it's fresh water with density of 1000 kg/m³

W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J

Salt water would require more.

3 0

3 years ago