Answer:
El avión recorrió 45 km en los 180 s.
Explanation:
La relación entre velocidad, distancia y tiempo se da de la siguiente manera;
Por lo cual los parámetros dados son los siguientes;
Velocidad = 900 km/h = 250 m / s
Tiempo = 180 s
Estamos obligados a calcular la distancia recorrida
De la ecuación para la velocidad dada arriba, tenemos;
Distancia recorrida = Velocidad pf viaje × Tiempo de viaje
Distancia recorrida = 900 km/h × 180 s = 900
Distancia recorrida = 900 km/h × 1 h/60 min × 1 min/60 s × 180 s = 45 km
Por lo tanto, el avión viajó 45 km en 180 s.
Answer:
u wanna do my edge bro the answer is b
Explanation:
Answer:
Speed of the speeder will be 28 m/sec
Explanation:
In first case police car is traveling with a speed of 90 km/hr
We can change 90 km/hr in m/sec
So 
Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m
After that car is accelerating with 
for 7 sec
So distance traveled by car in these 7 sec
So total distance traveled by police car = 224 m
This distance is also same for speeder
Now let speeder is moving with constant velocity v
so 
v = 28 m/sec
Answer:
- The magnitude of the vector
is 107.76 m
Explanation:
To find the components of the vectors we can use:
where 
is the magnitude of the vector, and θ is the angle over the positive x axis.
The negative x axis is displaced 180 ° over the positive x axis, so, we can take:
Now, we can perform vector addition. Taking two vectors, the vector addition is performed:
So, for our vectors:
To find the magnitude of this vector, we can use the Pythagorean Theorem
And this is the magnitude we are looking for.
4. The Coyote has an initial position vector of 
.
4a. The Coyote has an initial velocity vector of 
. His position at time 
is given by the vector
where 
is the Coyote's acceleration vector at time . He experiences acceleration only in the downward direction because of gravity, and in particular 
where 
. Splitting up the position vector into components, we have 
with
The Coyote hits the ground when 
:
4b. Here we evaluate 
at the time found in (4a).
5. The shell has initial position vector 
, and we're told that after some time the bullet (now separated from the shell) has a position of 
.
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where 
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for :
