Answer:
34.3 m/s
Explanation:
Newton's Second Law states that the resultant of the forces acting on the car is equal to the product between the mass of the car, m, and the centripetal acceleration 
(because the car is moving of circular motion). So at the top of the hill the equation of the forces is:
where
(mg) is the weight of the car (downward), with m being the car's mass and g=9.8 m/s^2 is the acceleration due to gravity
R is the normal reaction exerted by the road on the car (upward, so with negative sign)
v is the speed of the car
r = 0.120 km = 120 m is the radius of the curve
The problem is asking for the speed that the car would have when it tires just barely lose contact with the road: this means requiring that the normal reaction is zero, R=0. Substituting into the equation and solving for v, we find:
Answer:
Pressure of liquid in container is given by, P= height × density × acceleration of gravity.
At the lower storey, the height of the liquid from the open end is great, since height is directly proportional to pressure, the pressure exerted by liquid is maximum hence increase in velocity of flow.
Unlike the upper storey where the height of water is less hence the pressure exerted by the liquid is minimum which decreases the velocity / speed of liquid flow
Momentum is conserved, so the sum of the separate momenta of the car and wagon is equal to the momentum of the combined system:
(1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s) = ((1250 + 448) kg) <em>v</em>
where <em>v</em> is the velocity of the system. Solve for <em>v</em> :
<em>v</em> = ((1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s)) / (1698 kg)
<em>v</em> ≈ (30.3 <em>i</em> + 12.0 <em>j</em> ) m/s
Answer:
what is that supposed to even mean
Explanation:
Answer:
a = - 3.75 m/s²
negative sign indicates deceleration here.
Explanation:
In order to find the constant deceleration of the car, as it stops, we will use the 3rd equation of motion. The 3rd equation of motion is as follows:
2as = Vf² - Vi²
a = (Vf² - Vi²)/2s
where,
a = deceleration of the car = ?
Vf = Final Velocity = 0 m/s (Since, the car finally stops)
Vi = Initial Velocity = 30 m/s
s = distance covered by the car = 120 m
Therefore,
a = [(0 m/s)² - (30 m/s)²]/(2)(120 m)
<u>a = - 3.75 m/s²</u>
<u>negative sign indicates deceleration here.</u>
