<span>a. address all components of fitness
</span>According to the exercise principle of balance, a workout should address all components of fitness.
<span>NOT:
</span>b. focus on balance and coordination
<span>c. target specific areas for improvement </span>
<span>d. vary exercise activities over time</span>
Answer:
Yes. Towards the center. 8210 N.
Explanation:
Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.
In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.
The net force is equal to
Note that 95 km/h is equal to 26.3 m/s.
This is the centripetal force and equal to the x-component of the applied force.
As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.
The amount of the friction force should be
Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.
Answer:
velocity = 1527.52 ft/s
Acceleration = 80.13 ft/s²
Explanation:
We are given;
Radius of rotation; r = 32,700 ft
Radial acceleration; a_r = r¨ = 85 ft/s²
Angular velocity; ω = θ˙˙ = 0.019 rad/s
Also, angle θ reaches 66°
So, velocity of the rocket for the given position will be;
v = rθ˙˙/cos θ
so, v = 32700 × 0.019/ cos 66
v = 1527.52 ft/s
Acceleration is given by the formula ;
a = a_r/sinθ
For the given position,
a_r = r¨ - r(θ˙˙)²
Thus,
a = (r¨ - r(θ˙˙)²)/sinθ
Plugging in the relevant values, we obtain;
a = (85 - 32700(0.019)²)/sin66
a = (85 - 11.8047)/0.9135
a = 80.13 ft/s²
Answer:
3 km/h
Explanation:
Let's call the rowing speed in still water x, in km/h.
Rowing speed in upstream is: x - 2 km/h
Rowing speed in downstream is: x + 2 km/h
It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:
8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)
Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)
8*(x+2) + 8*(x-2) = (48/5)*(x² - 4)
Dividing by 8
(x+2) + (x-2) = (6/5)*(x² - 4)
2*x = (6/5)*x² - 24/5
0 = (6/5)*x² - 2*x - 24/5
Using quadratic formula
A negative result has no sense, therefore the rowing speed in still water was 3 km/h