What happens when a force is applied to an object in stable equilibrium

Which of the following does not represent a significant negative environmental impact of using nuclear fuel? a. air pollution b.

sasho [114]

The answer is a. mining.

3 0

3 years ago

A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change

olga55 [171]

Answer:

The add mass = 5.465 kg

Explanation:

Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1 (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)

Therefore,

T ∝ √(m)

Therefore,

T₁/√m₁ = k

where k = Constant of proportionality.

T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.

Substituting into equation 4

m₂ = (2.07)²(0.5)/(1.18)²

m₂ = 4.285(1.392)

m₂ = 5.965 kg.

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

Thus the add mass = 5.465 kg

7 0

3 years ago

How does Newton's second law of motion can be used to calculate the acceleration of an object?

Alex777 [14]

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

4 0

4 years ago

0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t 0, the puck is moving to the right at 3.00 m/s.

balandron [24]

Answer: $(a)10.812\ m/s\ (b)\ 0.75\ m/s\ \text{left}$

Explanation:

Given

Mass of hockey puck $m=0.160\ kg$

Initial velocity of hockey puck is $u=3\ m/s$

First a horizontal force of $25\ N$ is applied to the right for $0.05\ s$

acceleration associated with it is

$\Rightarrow a=\dfrac{25}{0.160}\\\\\Rightarrow a=156.5\ m/s^2$

Using equation of motion i.e.

$\Rightarrow v=u+at\\\Rightarrow v=3+156.25\times 0.05\\\Rightarrow v=3+7.812\\\Rightarrow v=10.812\ m/s$

(b) When a force of $12\ N$ is applied for 0.05s

Using equation of motion i.e.

$\Rightarrow v=3-\dfrac{12}{0.160}\times 0.05\\\\\Rightarrow v=3-75\times 0.05\\\Rightarrow v=3-3.75\\\Rightarrow v=-0.75\ m/s\\\Rightarrow v=0.75\ \text{towards left}$

6 0

3 years ago

car travels 80 meters due north in 12 seconds then the car turns around and travels 30 Mi do South in 4 seconds calculate the av

Zinaida [17]

Answer:

1) 3.1 m/s

2) 7 m/s

Explanation:

Distance due north = 80 m

Distance due south = 30 m

Distance between north and south = (80 - 30) m = 50 m

Total time = (12 + 4) sec = 16 sec

1) Average speed = 50/16 = 3.1 m/s

2) Average velocity = Total distance/total time = (80 + 30) m/16 s = 110/16 = 7 m/s

8 0

4 years ago