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A box is being pushed from the left with a force of 5 N. No force is being applied from the right.

Sergio039 [100]

Answer:

The box will move to the right with increasing speed

Explanation:

According to Newton's second law:

$F=ma$ (1)

where

F is the net force on an object

m is the mass of the object

a is its acceleration

For the box in this problem, there is only one force acting on it: a force of 5 N from the left (so, to the right). This means that according to (1) the box will experience an acceleration: the direction of the acceleration is the same as the direction of the net force (so, to the right), therefore the box will move to the right and its speed will continue increasing due to fact that it is being accelerated.

6 0

3 years ago

How do you answer this. Need help ASAP. Offering 20 points !!!

Temka [501]

in this since your volume remains at a constant you'll need to use Gay-Lussacs law, p1/t1=p2/t2.

your temp should be converted in kelvin

variables:

p1=3.0×10^6 n/m^2

t1= 270k

just add 273 to your celcius

p2= ? your solving for this

t2= 315k

then you set up the equation

(3.0×10^6)/270= (x)(315)

you then cross multiply

(3.0×10^6)315=270x

distribute the 315 to the pressure.

9.45×10^8=270x then you divide 270 o both sides to get

answer

3.5×10^6 n/m^2

7 0

3 years ago

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.