The standard wave format for any wave is

What type of energy is the sum of an object’s potential and kinetic energy?

nikklg [1K]

Answer: its mechanical energy

Explanation:

8 0

3 years ago

A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F

ivanzaharov [21]

Answer: $32.65\ m$

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is $\mu_s=0.4$

The coefficient of kinetic friction is $\mu_k=0.2$

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction $\mu_kg=0.2\times 9.8\ m/s^2$

Using the equation of motion $v^2-u^2=2as\\$

insert the values

$0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m$

4 0

3 years ago

Object A is positively charged. Object A and Object B

ozzi

Answer: object B is negatively charged, object C is positively charged and object D is also positively charged

Explanation: since unlike charges attract and like charges repel, for object A which is positively charged and B to attract B must be negatively charged and then for B which is negatively charged and C to attract C must be positively charged and for C and D to repel they have to be of thesame charge which means D is positive as well.

4 0

2 years ago

A net force of 60 N north acts on an object with a mass of 30 kg. Use Newton's second law of

earnstyle [38]

Answer:

Explanation:

F = ma. For us, this looks like

60 = 30a and

a = 2 m/s/s

If the force goes up to, say, 90, then

90 = 30a and

a = 3...if the force goes up, the acceleration also goes up.

If the mass goes up to say, 60, and the force stays the same, then

60 = 60a and

a = 1...if the mass goes up, the acceleration goes down.

7 0

3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

2 years ago