Answer:
12+ 18 divide by 2 is the average minutes
Answer:
v₃ = 5 [m/s]
Explanation:
To solve this problem we must use the definition of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.
P = m*v
where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
We must also clarify that the momentum is preserved i.e. it is equal before the collision and after the collision
Pbeforecollision = Paftercollision
(m₁*v₁) + (m₂*v₂) = (m₁*v₃) + (m₂*v₄)
where:
m₁ = mass of the truck = 3000 [kg]
v₁ = velocity of the truck = 10 [m/s]
m₂ = mass of the car = 1000 [kg]
v₂ = velocity of the car before the collision = 0 (the car is parked)
v₃ = velocity of the truck after the collision [m/s]
v₄ = velocity of the car after the collision = 15 [m/s]
(3000*10) + (1000*0) = (3000*v₃) + (1000*15)
30000 = 3000*v₃ + 15000
3000*v₃ = 30000 - 15000
3000*v₃ = 15000
v₃ = 5 [m/s]
Answer:
865.08 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 243 m/s
Height (h) of the cliff = 62 m
Horizontal distance (s) =?
Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:
Height (h) of the cliff = 62 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
62 = ½ × 9.8 × t²
62 = 4.9 × t²
Divide both side by 4.9
t² = 62/4.9
Take the square root of both side.
t = √(62/4.9)
t = 3.56 s
Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:
Initial velocity (u) = 243 m/s
Time (t) = 3.56 s
Horizontal distance (s) =?
s = ut
s = 243 × 3.56 s
s = 865.08 m
Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.
Answer: P= mad/t or P=w/t so P= 300/6= 50 W
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m
