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2 years ago

Is work done in this example: A student lifting a heavy backpack off of the floor.

Physics

2 answers:

vagabundo [1.1K]2 years ago

4 0

I don't know what u mean

Maru [420]2 years ago

3 0

You need to add more explaintion of what you are talking about

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State Pascal's principle of pressure . please help due tomorrow

Romashka [77]

Answer:

Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.

Explanation:

The pressure at any point in the fluid is equal in all directions.

3 0

3 years ago

What would a series circuit be used for?

igor_vitrenko [27]

Answer:

Explanation:

a series circuit would be an odd choice to power a battery or light a lamp when a direct would be much more efficient, and it's not converting types of energy, so C is the best possible answer

6 0

2 years ago

Which two stimuli did John B. Watson associate in his infamous “Little Albert” experiment? A. a white lab rat and the boy’s moth

MA_775_DIABLO [31]

I got answer c but im not 100% sure

7 0

3 years ago

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago

Object a travels in the +x-direction before hitting a stationary object

Leto [7]

The object’s resultant angle of motion with the +x-axis after the collision is 47°

<span>From object A:

1) x-momentum is 5.7 × 10^4 kilogram meters/second,
2) y-momentum is 6.2 × 10^4 kilogram meters/second.

Now, we know, tan</span>Ф = $\frac{y}{x}$

⇒tanФ = $\frac{6.2 × 10^4 }{5.7 × 10^4}$

⇒tanФ = 1.088

⇒ Ф = $tan^{-1}$ 1.088
= 47.4 ≈ 47

8 0

3 years ago

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