Answer:
The load has a mass of 2636.8 kg
Explanation:
Step 1 : Data given
Mass of the truck = 7100 kg
Angle = 15°
velocity = 15m/s
Acceleration = 1.5 m/s²
Mass of truck = m1 kg
Mass of load = m2 kg
Thrust from engine = T
Step 2:
⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:
T = (m1+m2)*g*sinθ
⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .
Resultant force on truck is F = T – m1*gsinθ
F causes the acceleration of the truck: F= m*a
This gives the equation:
T – m1*gsinθ = m1*a
T = m1(a + gsinθ)
Combining both equations gives:
(m1+m2)*g*sinθ = m1*(a + gsinθ)
m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ
m2*g*sinθ = m1*a
Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:
m2*g*sinθ = (7100 – m2)*a
m2*g*sinθ = 7100a – m2a
m2*gsinθ + m2*a = 7100a
m2* (gsinθ + a) = 7100a
m2 = 7100a/(gsinθ + a)
m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)
m2 = 2636.8 kg
The load has a mass of 2636.8 kg
Answer:
F = 75[J]
Explanation:
We know that work is defined as the product of force by distance.
In this way we have two forces, the weight of the block down, and the force that bring about the block to rise.
where:
W = work = 50 [J]
d = distance = 2 [m]
Fweight = 50 [N]
Fupward [N]
Now replacing:
![50=-(50*2)+(F_{upward}*2)\\50+100=F_{upward}*2\\F_{upward}=150/2\\F_{upward}=75[J]](https://tex.z-dn.net/?f=50%3D-%2850%2A2%29%2B%28F_%7Bupward%7D%2A2%29%5C%5C50%2B100%3DF_%7Bupward%7D%2A2%5C%5CF_%7Bupward%7D%3D150%2F2%5C%5CF_%7Bupward%7D%3D75%5BJ%5D)
Answer:
The final position of the ship after the given time period is 42 km West of B.
Explanation:
Given;
average velocity of the ship, v = 35 km/h
time taken for the ship to reach point D, t = 1.2 hours
The position of the ship after the given time period is calculated as follows;
x = v x t
x = (35 km/h) x 1.2 h = 42 km
x = 42 km West of B.
Therefore, the final position of the ship after the given time period is 42 km West of B.
Answer:
a. 2.0secs
b. 20.4m
c. 4.0secs
d. 141.2m
e. 40m/s, ∅= -30°
Explanation:
The following Data are giving
Initial speed U=40m/s
angle of elevation,∅=30°
a. the expression for the time to attain the maximum height is expressed as
where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at
b. the expression for the maximum height is expressed as
c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,
Hence T=2t
T=2*2.0
T=4.0secs
d. The range of the projectile is expressed as
e. The landing speed is the same as the initial projected speed but in opposite direction
Hence the landing speed is 40m/s at angle of -30°
