Answer:
Increases
Increases
Increases
Explanation:
I don't know if you answered your own question but I'll just answer this for others confused ahh
Answer:
hope this helps i think the answer is C
Hello, I see you are in a jam. Lemme help.
1.) True
2.) True
3.) True
4.) True
5.) True
LOL these are all true ;)
Answer:
1500 mph
Explanation:
Take east to be +x and north to be +y.
The x component of the velocity is:
vₓ = 889 cos 0° + 830 cos 59°
vₓ = 1316.5 mph
The y component of the velocity is:
vᵧ = 889 sin 0° + 830 sin 59°
vᵧ = 711.4 mph
The speed is found with Pythagorean theorem:
v² = vₓ² + vᵧ²
v² = (1316.5 mph)² + (711.4 mph)²
v = 1496 mph
Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s