Answer:
the tension in the string is 5.59 N
Explanation:
Here ,
m_1 = 0.385 Kg
m_2 = 0.710 Kg
Using second law of motion ,
a = F_net / effective mass
a = (0.710- 0.385)×9.8/(0.710 + 0.385 + 0.0125/0.15^2)
a = 1.93 m/s^2
Now , let tension be T ,
then,
mg-T=ma
0.710×g - T = 0.710×1.93
T = 5.59 N
the tension in the string is 5.59 N
Answer: 592.37m
Explanation:
Person D is the blue line.
The total displacement is equal to the difference between the final position and the initial position, if the initial position is (0,0) we have that he first goes down two blocks, then right 6 blocks. then up 4 blocks, then left 1 block.
Now i will considerate that the positive x-axis is to the right and the positive y-axis is upwards.
Then the new position will be, if B is a block:
P =(6*B - 1*B, -2*B + 4*B) = (5*B, 2*B)
And we know that B = 110m
P = (550m, 220m)
Now, then the displacement will be equal to the magnitude of our vector, (because the difference between P and the initial position is equal to P, as the initial position is (0,0)) this is:
P = √(550^2 + 220^2) = 592.37m
Answer:
We know that the speed of sound is 343 m/s in air
we are also given the distance of the boat from the shore
From the provided data, we can easily find the time taken by the sound to reach the shore using the second equation of motion
s = ut + 1/2 at²
since the acceleration of sound is 0:
s = ut + 1/2 (0)t²
s = ut <em>(here, u is the speed of sound , s is the distance travelled and t is the time taken)</em>
Replacing the variables in the equation with the values we know
1200 = 343 * t
t = 1200 / 343
t = 3.5 seconds (approx)
Therefore, the sound of the gun will be heard at the shore, 3.5 seconds after being fired
Answer:
19.6 m
Explanation:
The total motion of the golf ball lasts 4.0 seconds: since the motion is symmetrical, it takes 2.0 s for the ball to reach the highest point and then another 2.0 s to land back on the tee.
Therefore, we can just analyze the second half of the motion that lasts
t = 2.0 s
During this time, the vertical distance covered by the ball is given by the equation:

where
u = 0 is the initial velocity (zero because the ball starts from its highest point, where the velocity is zero)
t = 2.0 s
g = 9.8 m/s^2 is the acceleration of gravity
Solving for d, we find:

So, the ball reaches a maximum height of 19.6 m.