Answer:
x = 1474.9 [m]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.
We must understand that when forces are applied on the body, they tend to slow the body down to stop it.
So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.
ΣF = m*a
![10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]](https://tex.z-dn.net/?f=10%20%2B12%2Asin%2860%29%3D%20-%206%2Aa%5C%5Ca%20%3D%20-%203.39%5Bm%2Fs%5E%7B2%7D%5D)
Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].
where:
Vf = final velocity = 0 (the block stops)
Vo = initial velocity = 100 [m/s]
a = - 3.39 [m/s²]
x = displacement [m]
![0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]](https://tex.z-dn.net/?f=0%20%3D%20100%5E%7B2%7D-2%2A3.39%2Ax%5C%5Cx%3D%5Cfrac%7B10000%7D%7B2%2A3.39%7D%5C%5Cx%3D1474.9%5Bm%5D)
Answer:
The answer is B
Explanation:
5/2=2.5
2.5x2=5
Hope this helps ik its kinda confusing lol
Answer:at 21.6 min they were separated by 12 km
Explanation:
We can consider the next diagram
B2------15km/h------->Dock
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B1 at 20km/h
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V
So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.
Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.
Answer:
d = 421.83 m
Explanation:
It is given that,
Height, h = 396.9 m
Horizontal speed, v = 46.87 m/s
We need to find the distance traveled by the ball horizontally. Let t is the time taken by the ball. Using second equation of motion for vertical direction. So,
Now d is the distance covered by the cannonball. So,
Hence, this is the required solution.
Answer:
141.152 miles per hour is the speed of the plane in miles per hour
Explanation:
Speed of plane = Total distance travelled/total time taken -
v = D/t
Substituting the given values in the above equation, we get
v = 467/3.3 miles /hour
v = 141.152 miles per hour
141.152 miles per hour is the speed of the plane in miles per hour
