<span>What we need to first do is split the ball's velocity into vertical and horizontal components. To do that multiply by the sin or cos depending upon if you're looking for the horizontal or vertical component. If you're uncertain as to which is which, look at the angle in relationship to 45 degrees. If the angle is less than 45 degrees, the larger value will be the horizontal speed, if the angle is greater than 45 degrees, the larger value will be the vertical speed. So let's calculate the velocities
sin(35)*18 m/s = 0.573576436 * 18 m/s = 10.32437585 m/s
cos(35)*18 m/s = 0.819152044 * 18 m/s = 14.7447368 m/s
Since our angle is less than 45 degrees, the higher velocity is our horizontal velocity which is 14.7447368 m/s.
To get the x positions for each moment in time, simply multiply the time by the horizontal speed. So
0.50 s * 14.7447368 m/s = 7.372368399 m
1.00 s * 14.7447368 m/s = 14.7447368 m
1.50 s * 14.7447368 m/s = 22.1171052 m
2.00 s * 14.7447368 m/s = 29.48947359 m
Rounding the results to 1 decimal place gives
0.50 s = 7.4 m
1.00 s = 14.7 m
1.50 s = 22.1 m
2.00 s = 29.5 m</span>
Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
Answer:
v = 10 m/s
Explanation:
Let's assume the wheel does not slip as it accelerates.
Energy theory is more straightforward than kinematics in my opinion.
Work done on the wheel
W = Fd = 45(12) = 540 J
Some is converted to potential energy
PE = mgh = 4(9.8)12sin30 = 235.2 J
As there is no friction mentioned, the remainder is kinetic energy
KE = 540 - 235.2 = 304.8 J
KE = ½mv² + ½Iω²
ω = v/R
KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²
v = √(2KE / (m + I/R²))
v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6
v = 10.07968...
Explanation:
It is given that, two teams are playing tug of war.
Force applied by Team A, 
Force applied by Team B, 
We need to find the net force acting on the rope. It is equal to :
So, the net force acting on the rope is 35 N and it is acting toward right. Hence, this is the required solution.
Answer:
a. 240 N due east
b. 540 N due west
Explanation:
Let east be the reference direction
(a) if the resultant force has a magnitude of 390 N and points east, and the 1st force is 150N due East, then the additional force would also due east and has a magnitude of
390 - 150 = 240 N
(b) if the resultant force has a magnitude of 390 N and points west, it would be -390N is eastern reference, and the 1st force is 150N due East, then the additional force would also due east and has a magnitude of
-390 - 150 = -540 N
This force would point west
