Answer:
Explanation:
Height covered = 12m
time to fall by 12 m
s = 1/2 gt²
12 = 1/2 g t²
t = 1.565 s
Horizontal distance of throw
= 8.5 x 1.565
= 13.3 m
This distance is to be covered by dog during the time ball falls ie 1.565 s
Speed of dog required = 13.3 / 1.565
= 8.5 m /s
b ) dog will catch the ball at a distance of 13.3 m .
Answer:
A.) 27000 kgm/s
18000 kgm/s
B.) Va = 22 m/s
C.) 19800 kgm/s
25200 kgm/s
Explanation: Given that the velocity of A and B are 30 m/s and 20 m/s. And of the same mass M = 9 × 10^5g
M = 9×10^5/1000 = 900 kg
A.) Initial momentum of A
Mu = 900 × 30 = 27000 kgm/s
Initial momentum of B
Mu = 900 × 20 = 18000 kgm/s
B.) if they have an accident and then the velocity of the B is 28 m/s, find out velocity of A.
Momentum before impact = momentum after impact
Given that Vb = 28 m/s
27000 + 18000 = 900Va + 900 × 28
45000 = 900Va + 25200
900Va = 45000 - 25200
900Va = 19800
Va = 19800/900
Va = 22 m/s
C.) Momentum of A after impact
MV = 900 × 22 = 19800 kgm/s
Momentum of B after impact
MV = 900 × 28 = 25200 kgm/s
Answer:
Explanation:
I can conclude that this means that the law can be broken under certain coditions as long as its not focused on a natural phenomonon
and a pheononmono is a fact or situation that is observed to exist or happen, especially one whose cause or explanation is in question.
The time elapsed is 9 seconds
Explanation:
The motion of the ball is a uniformly accelerated motion (a motion with constant acceleration), so we can use the following suvat equation:
where
:
v is the final velocity of the ball
u is the initial velocity
a is the acceleration
t is the time elapsed
For the ball in this problem, we have:
u = 3 m/s is the initial velocity
v = 34.5 m/s is the final velocity
is the acceleration
Solving for t, we find the time taken for this change in velocity:

Learn more about acceleration:
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Answer:
The high of the ramp is 2.81[m]
Explanation:
This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.
If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.
We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.
![E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D0.5%2Am%2Av%5E%7B2%7D%5C%5C%5C%5Cwhere%3A%5C%5CE_%7Bk%7D%3D3.8%5BJ%5D%5C%5Cv%20%3D%202.8%5Bm%2Fs%5D%5C%5Cm%3D%5Cfrac%7BE_%7Bk%7D%7D%7B0.5%2Av%5E%7B2%7D%20%7D%20%5C%5Cm%3D%5Cfrac%7B3.8%7D%7B0.5%2A2.8%5E%7B2%7D%20%7D%20%5C%5Cm%3D0.969%5Bkg%5D)
Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.
![E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\](https://tex.z-dn.net/?f=E_%7Bp%7D%2BW_%7Bf%7D%3DE_%7Bk%7D%5C%5Cwhere%3A%5C%5CE_%7Bp%7D%3D%20potential%20energy%20%5BJ%5D%5C%5CW_%7Bf%7D%3D23%5BJ%5D%5C%5CE_%7Bk%7D%3D3.8%5BJ%5D%5C%5C)
And therefore
![m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]](https://tex.z-dn.net/?f=m%2Ag%2Ah%20%2B%20W_%7Bf%7D%3D3.8%5C%5C%200.969%2A9.81%2Ah%20-%2023%3D%203.8%5C%5Ch%20%3D%20%5Cfrac%7B23%2B3.8%7D%7B0.969%2A9.81%7D%5C%5C%20h%20%3D%202.81%5Bm%5D)