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Scrat [10]
3 years ago
11

What is a sentence for crater

Physics
2 answers:
Sidana [21]3 years ago
6 0
Wow he stole my answer but yeah he is correct
never [62]3 years ago
3 0
Most active volcanic mountains have a crater on the top.
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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
How many planets are in our solar system
astra-53 [7]
There’s 8 planets in our solar system: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune.
5 0
3 years ago
Read 2 more answers
A ball of mass m on a string of length L is attached to a pivot. The ball is released from rest while the string is parallel to
mash [69]

Answer:

L/2

Explanation:

Neglect any air or other resistant, for the ball can wrap its string around the bar, it must rotate a full circle around the bar. This means the ball should be able to swing to the top position where it's directly above the bar. By the law of energy conservation, this happens when the ball is at the same level as where it's previously released vertically. It means the swinging radius around the bar must be at least half of the string length.

So the distance d between the bar and the pivot should be at least L/2

8 0
3 years ago
A boat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, space, start fraction, m, divided by, s,
zepelin [54]

Answer:

d=42m(-\hat{i})

Explanation:

For this problem, we need to apply the formulas of constant accelerated motion.

To obtain the boat displacement we need to calculate the displacement because of the river flow and the displacement done because of the boat motor.

for the river:

d_r=v*t\\d_r=5m/s*6s\\d_r=30m(\hat{i})

for the boat:

x=\frac{1}{2}*a*t^2\\\\x=\frac{1}{2}*4.0m/s*(6s)^2\\\\\\x=72m(-\hat{i})

So the final displacement is given by:

d=dr+x\\d=30m-72m\\d=42m(-\hat{i})

8 0
3 years ago
Read 2 more answers
Can someone help me please
stealth61 [152]
B hope this helps you!!!
4 0
3 years ago
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