How does the density of most metals compare to most non metals?

A force of 45 newtons is applied on an object, moving it 12 meters away in the same direction as the force. What is the magnitud

NARA [144]

<h2>Answer: 540 J</h2>

Explanation:

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:

$W=(F)(d)$ (1)

In this case both (the force and the distance in the path) are parallel (this means they are in the same direction), so the work $W$ performed is the product of the force exerted to push the box $F=45N$ by the distance traveled $d=12m$ .

Hence:

$W=(45N)(12m)$ (2)

$W=540Nm=540J$

6 0

3 years ago

According to ____, the ability to detect a stimulus depends not only on the intensity of the stimulus but also on other variable

irakobra [83]

Answer:

Signal detection theory

Explanation:

Signal detection theory states that stimulus ca be detected according to its intensity and a person's psychological and/or physical state. This means that we can notice things according to how strong they are but also, a person's characteristics like experience and physiological state like fatigue can affect the ability to detect them.

Because of this, the answer is that according to signal detection theory, the ability to detect a stimulus depends not only on the intensity of the stimulus but also on other variables such as the level of noise in the system and your expectations.

8 0

3 years ago

You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C

jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0

3 years ago

Elements that easily transmit electricity and heat display the property know as ______

andre [41]

It is known as conductivity

7 0

3 years ago

Read 2 more answers

An astronaut is moving in space when a big explosion occurs about 50 meters behind him. How will the astronaut come to know abou

LUCKY_DIMON [66]

Answer:

The correct answer is B.

The astronaut will know due to the light from the explosion.

Explanation:

Sound and vibrations require a medium such as air to travel through. Space, there is no air. Only a vacuum. So sound and vibrations are unable to travel. Light requires no medium to travel. It can go through a vacuum.

Therefore the Astronaut will see a bright flash of light as it travels from the explosion to outer space. It is also important to note that light can travel very far because nothing else interacts with its wave particles and as such, it cannot be impeded.

Cheers!

7 0

3 years ago