A decrease in the magnitude of velocity is called

When a light wave enters into a medium of different optical density,

olga nikolaevna [1]

I think it should be option (b)

6 0

3 years ago

A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

b)

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

c)

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$

Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

$\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2} = 4J (6)$

⇒ $\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} = 4J - 3.6 J = 0.4 J (7)$

⇒ $\Delta x_{1} = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)$

6 0

2 years ago

A speeding car traveling at 24.8 m/s passes a police car that is at rest. The police car begins its pursuit at the instant the s

marusya05 [52]

Answer:

the acceleration required is 1.37m/s^2

Explanation:

The car is having a constant velocity movement, so if we calculate the time to reach 897m, we can use it to find the acceleration the policeman need to apply to reach the car.

$x=v*t\\t=\frac{x}{v}\\t=\frac{897m}{24.8m/s}\\t=36.17s$

the policeman is traveling with a constant acceleration starting from rest so:

$x=\frac{1}{2}*a*t^2\\\\a=\frac{2*x}{t^2}\\\\a=1.37 m/s2$

7 0

2 years ago

A bag of cement has a mass of 62 g. What is the mass of the bag of cement in S.I. units (kg)?

NNADVOKAT [17]

The mass of this bag of cement in S.I. units (kg) is equal to 0.062 kilograms.

<u>Given the following data:</u>

Mass of cement = 62 grams.

To calculate the mass of this bag of cement in S.I. units (kg):

<h3>How to convert to S.I. units.</h3>

In Science, kilograms (kg) is the standard unit of measurement or S.I. units of the mass of a physical object. Thus, we would convert the value of the mass of this bag of cement in grams to kilograms (kg) as follows:

<u>Conversion:</u>

1000 grams = 1 kilograms.

62 grams = X kilograms.

Cross-multiplying, we have:

X = $\frac{62}{1000}$