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:
brainly.com/question/28500124
#SPJ4
Answer:
1470kgm²
Explanation:
The formula for expressing the moment of inertial is expressed as;
I = 1/3mr²
m is the mass of the body
r is the radius
Since there are three rotor blades, the moment of inertia will be;
I = 3(1/3mr²)
I = mr²
Given
m = 120kg
r = 3.50m
Required
Moment of inertia
Substitute the given values and get I
I = 120(3.50)²
I = 120(12.25)
I = 1470kgm²
Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²
<span>A: put an atom on a poster in the exhibit
Good luck. The poster itself is made of trillions of trillions of trillions
of atoms. You could not see the extra one any easier than you could
see the ones that are already there, and even if you could, it would be
lost in the crowd.
B: use a life size drawing of an atom
Good luck. Nobody has ever seen an atom. Atoms are too small
to see. That's a big part of the reason that nobody knew they exist
until less than 200 years ago.
D: set up a microscope so that visitors can view atoms
Good luck. Atoms are way too small to see with a microscope.
</span><span><span>C: Display a large three dimensional model of an atom.
</span> </span>Finally ! A suggestion that makes sense.
If something is too big or too small to see, show a model of it
that's just the right size to see.
Answer:
2697.75N/m
Explanation:
Step one
This problem bothers on energy stored in a spring.
Step two
Given data
Compression x= 2cm
To meter = 2/100= 0.02m
Mass m= 0.01kg
Height h= 5.5m
K=?
Let us assume g= 9.81m/s²
Step three
According to the principle of conservation of energy
We know that the the energy stored in a spring is
E= 1/2kx²
1/2kx²= mgh
Making k subject of formula we have
kx²= 2mgh
k= 2mgh/x²
k= (2*0.01*9.81*5.5)/0.02²
k= 1.0791/0.0004
k= 2697.75N/m
Hence the spring constant k is 2697.75N/m
Answer:
3.4 x 10⁴ m/s
Explanation:
Consider the circular motion of the electron
B = magnetic field = 80 x 10⁻⁶ T
m = mass of electron = 9.1 x 10⁻³¹ kg
v = radial speed
r = radius of circular path = 2 mm = 0.002 m
q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
For the circular motion of electron
qBr = mv
(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v
v = 2.8 x 10⁴ m/s
Consider the linear motion of the electron :
v' = linear speed
x = horizontal distance traveled = 9 mm = 0.009 m
t = time taken = = = 4.5 x 10⁻⁷ sec
using the equation
x = v' t
0.009 = v' (4.5 x 10⁻⁷)
v' = 20000 m/s
v' = 2 x 10⁴ m/s
Speed is given as
V = sqrt(v² + v'²)
V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)
v = 3.4 x 10⁴ m/s