What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on
1 answer:
The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is .
The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.
The given data:
time, t = 1.1 s
initial speed, u = 1000 km/h =
final speed, v = 0 m/s
So we will be using the equation of motion, that is,
v = u + at
Hence , the deceleration of the rocket is .
To learn more about Attention here:
#SPJ4
You might be interested in
Answer:
Explanation:
This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".
In the y-stuff category:
v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)
Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)
a = -9.8 m/s/s
t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)
In the "x-stuff" category:
v₀ = 7.25 m/s (this is given)
Δx = ???
a = 0 (acceleration in this dimension is ALWAYS 0)
t = (we will solve for this in the y-dimension and plug it in here).
In the y dimension:
Δx = v₀t + and plugging in from the y-dimension info:
which simplifies to
so
which, to 2 significant digits is
t = 1.4 seconds
Now we will do the same in the x-dimension, using t = 1.4:
Δx = v₀t + and filling in the x-stuff:
Δx = Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:
Δx = 7.25(1.4) + 0 so
Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).
Answer:
80.4 N
Explanation:
As the block is at rest on the slope, it means that all the forces acting on it are balanced.
We are only interested in the forces that act on the block along the direction perpendicular to the slope. Along this direction, we have two forces acting on the block:
- The normal reaction N (contact force), upward
- The component of the weight of the block, , downward, where m is the mass of the block, g is the gravitational acceleration and
is the angle of the incline
Since the block is in equilibrium along this direction, the two forces must balance each other, so they must be equal in magnitude:
And by substituting the numbers into the equation, we find the size of the contact force normal to the slope: