Answer:
Distance, d = 778.05 m
Explanation:
Given that,
Force acting on the car, F = 981 N
Mass of the car, m = 1550 kg
Initial speed of the car, v = 25 mi/h = 11.17 m/s
We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

Let d is the distance covered by car. Using second equation of motion as :

So, the car will cover a distance of 778.05 meters.
Answer:
the range or the ball is 48.81 m
Explanation:
given;
Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal.
find:
What is the range of the ball?
solution:
let Ф = 25°
Vo = 25 m/s
<u>consider x-motion using time of fight: x = Vox * t</u>
where x = R = range
t =<u> 2 Voy </u>
g
R =<u> Vo² sin (2Ф)</u>
g
plugin values into the formula:
R = <u>(25)² sin (2*25) </u>
9.81
R = 48.81 m
therefore, the range or the ball is 48.81 m
From an energy balance, we can use this formula to solve for the angular speed of the chimney
ω^2 = 3g / h sin θ
Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s
The formula for radial acceleration is:
a = rω^2
So,
a = 53.2 (0.987) = 52.494 /s^2
The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887
The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2
If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67°
Answer:
they help us allocate a particular place in case one needs to allocate or find a place or something
Answer:
a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)
b) The acceleration of point A is 3.25 m/s²
The acceleration of point E is 0.75 m/s²
Explanation:
a) The relative acceleration of B with respect to D is equal:

Where
aB = absolute acceleration of point B = 2.5 j (m/s²)
aD = absolute acceleration of point D = 1.5 j (m/s²)
(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam
(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)


We have that
(aB/D)t = BDα
Where α = acceleration of the beam
BDα = 1 m/s²
Where
BD = 2

b) The acceleration of point A is:

(aA/D)t = ADαj

The acceleration of point E is:
(aE/D)t = -EDαj
