Answer:
a) parallel to the ground True
c) parallel to the ground towards man True
Explanation:
To examine the possibilities, we propose the solution of the problem.
Let's use Newton's second law
F = m a
The force is exerted by the arm and the centripetal acceleration of the golf club, which in this case varies with height.
In our case, the stick is horizontal in the middle of the swing, for this point the centripetal acceleration is directed to the center of the circle or is parallel to the arm that is also parallel to the ground;
Ask the acceleration vector
a) parallel to the ground True
b) down. False
c) parallel to the ground towards True men
d) False feet
e) the head. False
Complete question:
A dog runs 3 miles east and 4 miles west in 6 hours. What’s the dogs total distance and displacement ?
Answer:
The total distance covered by the dog is 7 miles
The displacement of the dog is 1 mile west
Explanation:
Given;
initial position of the dog = 3 miles east
final position of the dog = 4 miles west
time of motion, t = 6 hours
The total distance covered by the dog is given as;
Total distance = 3 miles + 4 miles = 7 miles
The displacement of the dog is given as;
displacement = final position of the dog - initial position of the dog
displacement = 4 miles west - 3 miles east = 1 mile west
Answer:
-78.96 J
Explanation:
The workdone by the torque in stopping the wheel = rotational kinetic energy change of wheel.
So W = 1/2I(ω₁² - ω₀²) where I = rotational inertia of wheel = 0.04 kgm², r = radius of wheel = 0.02 m, ω₀ = initial rotational speed = 10 rev/s × 2π = 62.83 rad/s, ω₁ = final rotational speed = 0 rad/s (since the wheel stops)
W = 1/2I(ω₁² - ω₀²) = 1/2 0.04 kgm² (0² - (62.83 rad/s)²) = -78.96 J
Explanation:
Not enough information. It really depends on the technical details of the car ( the data provided is offering just the human factor of the reaction, not the time for getting the impulse through when using the breaks
Answer:
Explanation:
λ = wave length = 632 x 10⁻⁹
slit width a = 2 x 10⁻³ m
angular separation of central maxima
= 2 x λ /a
= 2 x 632 x 10⁻⁹ / 2 x 10⁻³
= 632 x 10⁻⁶ rad
width in m of light spot.
= 632 x 10⁻⁶ x 376000 km
= 237.632 km
