<u>ALL of the following work assumes NO AIR RESISTANCE:</u>
1). an object moving under the influence of only gravity, and not in orbit; its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²
2). a parabola
3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.
4). a). the one that was thrown horizontally; b). both hit the ground at the same time; c). both hit the ground with the same vertical velocity
5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion
6). a). 1.8 seconds; b). 13.1 meters; c). 17.6 m/s down; d). 7.3 m/s; gravity has no effect on horizontal motion
7). 45 m/s
8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand
9). a). 4.49 m/s; b). 29.7 m/s
10). 7.24 meters
11). 700 meters
12). A). 103.7 meters ( ! she's in big trouble ! ); B). 17.5 meters
Answer:
The bullet's initial speed is 243.21 m/s.
Explanation:
Given that,
Mass of the bullet, 
Mass of the pendulum, 
The center of mass of the pendulum rises a vertical distance of 10 cm.
We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,
V is the speed of the bullet and pendulum at the time of collision
Now using conservation of momentum as :
Put the value of V from equation (1) in above equation as :
So, the bullet's initial speed is 243.21 m/s.
Answer:
thank for making me give up on life
Explanation:
I thought the stuff I had was hard wth is even that
Answer:
The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Explanation:
Given:-
- The diameter of the drill bit, d = 98 cm
- The power at which drill works, P = 5.85 hp
- The rotational speed of drill, N = 1900 rpm
Find:-
What Torque And Force Is Applied To The Drill Bit?
Solution:-
- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).
- The relation between these quantities is given:
T = 5252*P / N
T = 5252*5.85 / 1900
T = 16.171 Nm
- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):
T = F*r
Where, r = d / 2
F = 2T / d
F = 2*16.171 / 0.98
F = 33 N
Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
