Answer:
Explanation:
mass attached m = .14 kg
force constant k = 5N / m
displacement
= amplitude of oscillation
A = .03 m
A ) period of motion =
= 2 x 3.14
T = 1.05 s
B ) maximum speed of block = angular velocity x amplitude
= (2π /T) x A
= (2 x 3.14 x .03) / 1.05
= .1794 m / s
17.94 cm /s
C )
maximum acceleration = angular velocity² x amplitude
= (2π /T)² x A
= (2π /1.05)² x .03
= 1.073 m / s²
D )
position
S = A cos ωt , ω is angular velocity
S = .03 cos(2πt /T)
= .03 cos 5.98 t
v =ω A sin(2πt /T)
= 5.98 x .03 sin5.98t
= .1794 sin5.98t
acceleration = ω²A sin5.98t
= 1.073 sin5.98t
Answer:
the jumper's speed just as he leaves the ground = 4.2 m/s
the force he must exert on the ground to perform the 0.90 m jump = 4.99 × 10³ N
Explanation:
Given that :
height h = 0.90 m
mass = 51 kg
distance s = 0.10 m
What is the jumper's speed just as he leaves the ground?
The velocity P.E = K.E
where;
g = 9.8
v = 4.2 m/s
the jumper's speed just as he leaves the ground = 4.2 m/s
What force must he exert on the ground to perform the 0.90 m jump?
Let's first determine the acceleration by using the equation of motion;
v² = u² + 2as
4.2² = 0 + 2(a)(0.10)
17.64 = 0.2 a
a = 17.64/0.2
a = 88.2 m/s²
The force F is now calculated by the relation:
F - mg = ma
F = mg+ ma
F = m(g+a)
F = 51(9.8 + 88.2)
F = 4988 N
F = 4.99 × 10³ N
In rutherford experiment he bombarded highly energetic alpha particles towards the gold foil
He observed that most of the alpha particles pass through the foil without any deviation while very few of the alpha particles are there which reflect back to its own path
while very less in number was there which were deflected from there path
so he concluded that the positive charge of atom is concentrated inside the nucleus and it is small space in tom and electrons are surrounded them
so correct answer would be
4. In an atom, most of the mass and the positive charge are located in a small core within the atom called the nucleus.
Answer:
The total frictional force is 358.0 newtons
Explanation:
Power is the amount of average work (W) an object does on a period of time (Δt):
Remember average work is average force (F) times displacement (Δs):
but displacement over time is average speed , then:
(1)
That is, the power of the car is the force the engine does times the speed of the car. As the question states, if the car is at constant velocity then the power developed is used to overcome the frictional forces exerted by the air and the road, that is by Newton's first law, the force the motor of the car does is equal the force of frictional forces. So, to find the frictional forces we only have to solve (1) for F:
Knowing that 1hp is 746W then 30hp=22380W and 1 mile = 1609m then 140 mph = 225308 = , then:
Answer: I'll help you when i end school
Explanation: ok