Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is
and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:
Answer:
1) It expresses the rate (top speed) at which it can move with time.
2) P = 20 W
3) h = 18 km
Explanation:
1) Power is the rate of transfer of energy.
⇒ Power = 
i.e P = 
Thus a car's engine power is 44000W implies that the engine of the car can propel the car at this rate. This expresses the rate (top speed) at which it can move with time.
2) m = 400g = 0.4 kg
t = 20 s
h = 100m
g = 10 m/
P = 
= 
= 
P = 20 W
3) u = 600 m/s
g = 10 m/
From the third equation of free fall,
= 
- 2gh
V is the final velocity, U is the initial velocity, h is the height.
0 = 
- 2 x 10 x h
0 = 360000 - 20h
20h = 360000
h = 
= 18000
h = 18 km
The maximum height of the bullet would be 18 km.
<u>The two ways to find acceleration in non uniform motion are as follows:</u>
<u>Explanation:</u>
Non-uniform acceleration comprises the most common description of motion. Acceleration refers to the rate of changes of velocity per unit time. Basically, it implies that acceleration changes during motion. This variety can be communicated either as far as position (x) or time (t).
Accordingly, non-uniform acceleration motion can be carried out in 2 ways:
Calculus analysis is general and accurate, but limited to the availability of speed and acceleration expressions. It is not always possible to get the expression of motion attributes in the form "x" or "t". On the other hand, the graphic method is not accurate enough, but it can be used accurately if the graphic has the correct shapes.
The use of calculations involves differentiation and integration. Integration enables evaluation of the expression of acceleration of speed and expression of movement at a distance. Similarly, differentiation allows us to evaluate expression of speed position and expression speed to acceleration.
Answer:
F=5833.3 N N
Explanation:
Newton's second law applied to the car
F= m*a Formula (1)
F: Force in Newtons (N)
m : mass in kg
a: acceleration ( m/s²)
kinematics car
vf= v₀ + a*t Formula (2)
vf : final velocity (m/s)
v₀ : final velocity (m/s)
a : acceleration ( m/s²)
t : time t
Equivalences
1 km= 1000m
1 h = 3600 s
Data
m= 1000kg
v₀ = 90 km/h = 90*1000/3600 m/s = 25 m/s
vf= 0
t= 6 s
Problem Development
We calculate the acceleration replacing the data in the formula (2) :
0 = 25 + a*6
a= -25/6 = -4.16 m/s² ( The negative sign indicates that the car is braking)
We calculate the force is required to stop the car replacing the data in the formula (1)
-F = 1400 kg*(-4.16 m/s²)
F=5833.3 N
