<span>The diver is heading downwards at 12 m/s
Ignoring air resistance, the formula for the distance under constant acceleration is
d = VT - 0.5AT^2
where
V = initial velocity
T = time
A = acceleration (9.8 m/s^2 on Earth)
In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m)
So let's substitute the known values and solve for T
d = VT - 0.5AT^2
-7 = 2.5T - 0.5*9.8T^2
-7 = 2.5T - 4.9T^2
0 = 2.5T - 4.9T^2 + 7
We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164.
Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So
V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s
V = 2.5 m/s - 14.47706141 m/s
V = -11.97706141 m/s
So the diver is going down at a velocity of 11.98 m/s
Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point.
V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s)
V = 2.5 m/s - (-9.477061409 m/s)
V = 2.5 m/s + 9.477061409 m/s
V = 11.97706141 m/s
And you get the exact same velocity, except it's the opposite sign.
In any case, the result needs to be rounded to 2 significant figures which is -12 m/s</span>
Force=mass*acceleration. So 88kg*10 m/s^2=880 newtons
Answer:
a
The total distance is 
b
The displacement is
Explanation:
From the question we are told that
Distance traveled by the ball for first player 
to the right
Distance traveled by the ball for second player 
to the left
The total distance traveled by the ball is mathematically represented as
Substituting values
The displacement is mathematically represented as
This is because displacement deal with direction and from the question we are told that right is positive and left is negative
Substituting values
Answer:
maximum speed 56 km/h
Explanation:
To apply Newton's second law to this system we create a reference system with the horizontal x-axis and the Vertical y-axis. In this system, normal is the only force that we must decompose
sin 10 = Nx / N
cos 10 = Ny / N
Ny = N cos 10
Nx = N sin 10
Let's develop Newton's equations on each axis
X axis
We include the force of friction towards the center of the curve because the high-speed car has to get out of the curve
Nx + fr = m a
a = v2 / r
fr = mu N
N sin10 + mu N = m v² / r
N (sin10 + mu) = m v² / r
Y Axis
Ny -W = 0
N cos 10 = mg
Let's solve these two equations,
(mg / cos 10) (sin 10 + mu) = m v² / r
g (tan 10 + μ / cos 10) = v² / r
v² = r g (tan 10 + μ / cos 10)
They ask us for the maximum speed
v² = 30.0 9.8 (tan 10+ 0.65 / cos 10)
v² = 294 (0.8364)
v = √(245.9)
v = 15.68 m / s
Let's reduce this to km / h
v = 15.68 m / s (1 km / 1000m) (3600s / 1h)
v = 56.45 km / h
This is the maximum speed so you don't skid
