Answer:
Option D
The air pressure inside the car is greater than the pressure outside.
Explanation:
When considering airflow over and around a surface, from Bernoulli's equation, air flow regions with higher velocity have a lower pressure, and regions with lower velocity have a higher pressure.
The air outside the convertible is moving faster than the air inside the convertible. This leads to a higher pressure zone just below the surface of the roof (inside the car) causing the roof of the convertible to bulge upwards
Answer:
Explanation:
Given that,
Basket ball is drop from height
H=10m
It is dropped on planet mass
And the acceleration due to gravity on Mars is given as
g= 3.7m/s²
Time taken for the ball to reach the ground
Initial velocity of the body is zero
u=0m/s
Using equation of motion: free fall
H = ut + ½gt²
10 = 0•t + ½ × 3.7 ×t²
10 = 0 + 1.85t²
10 = 1.85t²
Then, t² =10/1.85
t² = 5.405
t = √ 5.405
t = 2.325seconds
So the time the ball spend on the air before reaching the ground is 2.325 seconds
Answer:
6.18 m/s
Explanation:
Roller skate collision
The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;
x-axis component form (+x east);
+ 
+ 
= 
+
Ф
60 ·8 + 0 = (60 + 80)
Ф
480 = 140
Ф................. (I)
y-axis component form (+y north);
+ 
+ 
= 
+ 
Ф
0 + 80.9 = (60 + 80)
Ф
720=
140Ф
140Vf=
Ф......................................(2)
Substituting (2) into (1) to give the angle;
480 = 720tan Ф
Ф = arctan(0.67) =33.69°.......................(3)
Evaluating (1) with (3) gives the velocity magnitude
480 = 140Vfsin 33.69°
Vf=6.18 m/s
note 1:
This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.
Answer:
x = 1,185 m
, t = 4/3 s
, F = - 4 N
Explanation:
For this exercise we use Newton's second law
F = m a = m dv /dt
β - α t = m dv / dt
dv = (β – α t) dt
We integrate
v = β t - ½ α t²
We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t
v-v₀ = β t - ½ α t²
the farthest point of the body is when v = v₀ = 0
0 = β t - ½ α t²
t = 2 β / α
t = 2 4/6
t = 4/3 s
Let's find the distance at this time
v = dx / dt
dx / dt = v₀ + β t - ½ α t2
dx = (v₀ + β t - ½ α t2) dt
We integrate
x = v₀ t + ½ β t - ½ 1/3 α t³
x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³
The body comes out of rest
x = 3.5556 - 2.37
x = 1,185 m
The value of force is
F = β - α t
F = 4 - 6 4/3
F = - 4 N
