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3 years ago

How does energy change (transforms) as the mass is dropping?

Physics

1 answer:

olganol [36]3 years ago

4 0

Answer:

Mass has total mechanical energy, which is the sum of kinetic and potential energy. as the mass is dropping, potential energy is converted into kinetic energy so mechanical energy is preserved If there is no friction. If there is friction, some of the mechanical energy is lost as heat energy so it changes.

Explanation:

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Hello i need help with this!!

mr Goodwill [35]

Answer:

Yes

Explanation:

If lamp A burnt out there would still be a wire above it that connects lamp B and C to the power source

3 0

1 year ago

a person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500m/s^2. what is the velocity of the stroll

grandymaker [24]

You're going to use the constant acceleration motion equation for velocity and displacement:
(V)final²=(V)initial²+2a(Δx)

Given:
a=0.500m/s²
Δx=4.75m
(V)intial=0m
(V)final= UNKNOWN

(V)final= 2.179m/s

5 0

3 years ago

Read 2 more answers

After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg

kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s$

v₂ = 4.2 m/s

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)$

v₂ = 5 m/s

5 0

3 years ago

A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.

tankabanditka [31]

Answer:

$v_o=40.14\ m/s$

Explanation:

Horizontal Launch

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

The horizontal speed is constant and equal to the initial speed $v_o$
The vertical speed is zero at launch time, but increases as the object starts to fall
The height of the object gradually decreases until it hits the ground
The horizontal distance where the object lands is called the range

We have the following formulas

$\displaystyle v_x=v_o$

$\displaystyle x=v_o.t$

$\displaystyle v_y=g.t$

$\displaystyle y=\frac{gt^2}{2}$

Where $v_o$ is the initial horizontal speed, $v_y$ is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

$\displaystyle y=\frac{gt^2}{2}$

Rearranging and solving for t

$\displaystyle 2y=gt^2$

$\displaystyle t^2=\frac{2\ y}{g}$

$\displaystyle t=\sqrt{\frac{2\ y}{g}}$

We then replace this value in

$\displaystyle x=v_o.t$

To get

$\displaystyle v_o=\frac{x}{t}$

$\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}$

$\displaystyle v_o=\sqrt{\frac{g}{2y}}.x$

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing $v_o$

$\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6$

The launch velocity is

$\boxed{v_o=40.14\ m/s}$

7 0

3 years ago

Three persons wants to push a wheel cart in the direction marked x in Fig. The two person push with horizontal forces F1 and F2

Svetllana [295]

Answer:

I had to search the Figure on Google to solve this question.

a) The magnitude of the force F₃ is:

$F_{3} = 87.47 N$

And the direction of F₃:

$\alpha = 79.04 ^{\circ}$ (with respect to the y-direction, in the third quadrant)

b) P = 4.22 N

Explanation:

I had to search the Figure on Google to solve this question.

a) We can find the force of the third person as follows:

$\Sigma F_{x} = F_{1x} + F_{2x} + F_{3x} = 0$

$\Sigma F_{y} = F_{1y} - F_{2y} + F_{3y} = 0$

So, in x-direction we have:

$\Sigma F_{x} = 45 N*cos(70) + 75 N*cos(20) + F_{3x} = 0$

$F_{3x} = -85.87 N$

In y-direction we have:

$\Sigma F_{y} = 45 N*sin(70) - 75 N*sin(20) + F_{3y} = 0$

$F_{3y} = -16.63 N$

The magnitude of the force F₃ is:

$F_{3} = \sqrt{F_{3x}^{2} + F_{3y}^{2}} = \sqrt{(-85.87 N)^{2} + (-16.63 N)^{2}} = 87.47 N$

To find the direction of F₃ we need to calculate its angle with respect to the y-direction (in the third quadrant):

$tan(\alpha) = \frac{|F_{3x}|}{|F_{3y}|} = \frac{85.87 N}{16.63 N}$

$\alpha = 79.04 ^{\circ}$

b) If the third person exerts the force found in part (a) the car will stop, so the only way for the cart to accelerate at 200 m/s² is that the third person does not exert the force found in a.

To find the weight of the cart when it accelerates at 200 m/s², we need to consider: F₃ = 0.

First, we need to find the cart's mass. Since the car is moving in the x-direction we have:

$\Sigma F_{x} = F_{1x} + F_{2x} = ma$