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jeka57 [31]
3 years ago
15

A block of mass m begins at rest at the top of a ramp at elevation h with whatever pe is associated with that height.

Physics
2 answers:
natka813 [3]3 years ago
8 0
 <span>PE = mass * height * g = 87.60 J 

Frictional force = m * g * sin theta * uk = 4.7 kg * 9.81 m/sec^2 * sin 36.87 * 0.3 = 8.299 N 

Work done by friction = force * distance = 41.50 J 

Remaining energy to be converted to KE = 87.60 J - 41.50 J = 41.10 J</span>hope it helps
Darina [25.2K]3 years ago
8 0

Answer:

The PE associated with that height is given as, m×g×h

Explanation:

Potential energy (PE) is the energy possessed by an object when at rest. Its opposite is the kinetic energy (KE). When an object is at a certain height, it would possess potential energy because initially it is at rest. Also the value of the energy is determined by its mass, height and the force of gravity acting on it at that height.

So, the potential energy (PE) = mass× height× value of the gravitational force

                                                = m× g× h

Its unit is Joules. The potential energy of the mass would be converted to kinetic energy when it is in motion.

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A car and its passengers have a mass of 1200kg it is travelling at 12m/s.
Natali [406]

Answer:

<em>The increase of kinetic energy is 108,000 J</em>

Explanation:

<u>Kinetic Energy </u>

Is the energy an object has due to its state of motion. It's proportional to the square of the speed and the mass.

The equation for the kinetic energy is:

\displaystyle K=\frac{1}{2}mv^2

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

A car has a total mass of m=1,200 kg and travels at v1=12 m/s. Then it increases its speed at v2=18 m/s.

It's required to compute the increase of kinetic energy. We'll calculate both energies K1 and K2 and then subtract them.

\displaystyle K_1=\frac{1}{2}1,200*12^2=86,400\ J

\displaystyle K_2=\frac{1}{2}1,200*18^2=194,400\ J

The increase of kinetic energy is:

\Delta K=K_2-K_1 =194,400\ J-86,400\ J

\Delta K=108,000\ J

The increase of kinetic energy is 108,000 J

4 0
3 years ago
A girl is sitting in a sled sliding horizontally along some snow (there is friction present).
Over [174]

Answer:

159 N

Explanation:

The force of friction, Fr is a product of coefficient of feiction and the normal force. Therefore, Fr=uN where N is the normal force and u is coefficient of friction. Here, we have two coefficients of friction but since it is sliding, then we use coefficient of kinetic energy. Substituting 0.25 for u and 636 N for N then

Fr=0.25*636=159 N

Therefore, the force of friction is equivalent to 159 N

4 0
3 years ago
Are we even real? Do we exist or are we robots or a simulation or experiment?
ipn [44]

what kind of question is this....anyways...

Yes, we are real and yes we do exist. Without God there would be no creation or any type of life on earth. So, yes because of God Himself the creator of us we exist.

3 0
3 years ago
Read 2 more answers
It was a children versus grown-ups competition at school. One event required the adult to throw a basketball as far as he could.
aleksklad [387]
Based on Newton's Law, the event is not fair because:

<span>No, because a basketball is bigger than a baseball, and objects that are bigger accelerate slower.


I hope my answer has come to your help. God bless and have a nice day ahead!</span>

4 0
4 years ago
Read 2 more answers
A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does i
geniusboy [140]

Answers:

a) t=0.311 s

b) a=86.847 m/s^{2}

c) y=1.736 m

d) V=17.369 m/s

Explanation:

For this situation we will use the following equations:

y=y_{o}+V_{o}t+\frac{1}{2}at^{2} (1)  

V=V_{o} + at (2)  

Where:  

y is the <u>height of the model rocket at a given time</u>

y_{o}=0 is the i<u>nitial height </u>of the model rocket

V_{o}=0 is the<u> initial velocity</u> of the model rocket since it started from rest

V is the <u>velocity of the rocket at a given height and time</u>

t is the <u>time</u> it takes to the model rocket to reach a certain height

a is the <u>constant acceleration</u> due gravity and the rocket's thrust

<h2>a) Time it takes for the rocket to reach the height=4.2 m</h2>

The average velocity of a body moving at a constant acceleration is:

V=\frac{V_{1}+V_{2}}{2} (3)

For this rocket is:

V=\frac{27 m/s}{2}=13.5 m/s (4)

Time is determined by:

t=\frac{y}{V} (5)

t=\frac{4.2 m}{13.5 m/s} (6)

Hence:

t=0.311 s (7)

<h2>b) Magnitude of the rocket's acceleration</h2>

Using equation (1), with initial height and velocity equal to zero:

y=\frac{1}{2}at^{2} (8)  

We will use y=4.2 m :

4.2 m=\frac{1}{2}a(0.311)^{2} (9)  

Finding a:

a=86.847 m/s^{2} (10)  

<h2>c) Height of the rocket 0.20 s after launch</h2>

Using again y=\frac{1}{2}at^{2} but for t=0.2 s:

y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2} (11)

y=1.736 m (12)

<h2>d) Speed of the rocket 0.20 s after launch</h2>

We will use equation (2) remembering the rocket startted from rest:

V= at (13)  

V= (86.847 m/s^{2})(0.2 s) (14)  

Finally:

V=17.369 m/s (15)  

5 0
3 years ago
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