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3 years ago

Describe how the potential energy of a river is changed into kinetic energy.

Physics

1 answer:

daser333 [38]3 years ago

8 0

Answer:

The energy possessed by a river when it is on top of a hill or a water fall is the potential energy . That is , as long as water is at some height above the ground level, it will have potential energy due to the influence of gravity . But once it starts flowing, the motion or movement is responsible for the kinetic energy.

Explanation:

Energy due to the position or state of an object is gravitational potential energy. Once the object starts moving, it has a certain velocity. Velocity possessed by a mass ( or an object ) will give rise to kinetic energy.

For a given object, It is conservation of energy which basically explains this. The total energy should always remain constant during a process.

For the river waters, the potential energy due to its state is converted to kinetic energy of its motion.

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A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

$F_k = F_{W,E}$

$kx_1 = mg$

The extension of the spring due to the weight of the object on Moon is a value of $x_2$ , then

$kx_2 = mg_m$

Recall that gravity on the moon is a sixth of Earth's gravity.

$kx_2 = m\frac{g}{6}$

$kx_2 = \frac{1}{6} mg$

$kx_2 = \frac{1}{6} kx_1$

$x_2 = \frac{1}{6} x_1$

We have that the displacement at the earth was $x_1 = 0.3m$ , then

$x_2 = \frac{1}{6} 0.3$

$x_2 = 0.05m$

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0

3 years ago

A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it

makkiz [27]

Answer:D

Explanation:

Given

mass of object $m=5 kg$

Distance traveled $h=10 m$

velocity acquired $v=12 m/s$

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy $=mgh=5\times 9.8\times 10=490 J$

Final Energy $=\frac{1}{2}mv^2+W_{f}$

$=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}$

where $W_{f}$ is friction work if any

$490=360+W_{f}$

$W_{f}=130 J$

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0

3 years ago

A toy rocket is launched with an initial velocity of 12.0 m/s in the horizontal direction from the roof of a 30.0-m-tall buildin

OlgaM077 [116]

Solution :

The motion in the y direction.

The time taken by the toy rocket to hit the ground,

$S=ut+\frac{1}{2}at^2$

S = distance travelled = 30 m

u = 0 m/s

a = $9.8 \ m/sec^2$

t= time in seconds

Therefore, $30 =\frac{1}{2}9.8 t^2$

t = 2.47 sec

Now motion in the x direction,

u = 12 m/sec

$a=1.6 t \ m/sec^3$

Upon integration 'v' with respect to 't'

$v=\frac{1.6t^2}{2}+12$

Once again integrating with respect to t,

$s=\frac{1.6t^3}{6}+12 t$

$s=\frac{1.6(2.47)^3}{6}+12 (2.47)$

= 0.0176+29.64

= 29.65 m

Therefore, the toy rocket will hit the ground at 29.65 m from the building.

7 0

3 years ago

a 6.7kg object moves with a velocity of 8 m/s. whats its kinetic energy?(A)26.8J(B)214.4J(C)167.5J(D)53.6J

Ne4ueva [31]

The answer in (B)214.4J

I hope this helped have a nice day!

6 0

3 years ago

Read 2 more answers

When one person shouts at a football game, the sound intensity level at the center of the field is 58.4 dB. When all the people

Tanzania [10]

Answer:

The number of people at game are approximately 22909

Explanation:

Given data

When one person shout $\beta _{1}=58.4dB$

When n number of person shout together $\beta _{n}=102dB$

The sound intensity level during one person shout is given by:

$\beta _{1}=10log(\frac{I_{1}}{I_{o}} )\\58.4=10log(\frac{I_{1}}{I_{o}} )\\5.84=log(\frac{I_{1}}{I_{o}} )\\\frac{I_{1}}{I_{o}} =10^{5.84}\\I_{1}=10^{5.84}*I_{o}$

The sound intensity level during n number of person shout is given by:

$\beta _{n}=10log(\frac{I_{n}}{I_{o}} )\\102=10log(\frac{I_{n}}{I_{o}} )\\10.2=log(\frac{I_{n}}{I_{o}} )\\\frac{I_{n}}{I_{o}}=10^{10.2}\\I_{n}=10^{10.2}*I_{o}$

Since each person generates same sound intensity and hence total number of persons can be determined as

$=\frac{I_{n}}{I_{1}}\\ =\frac{10^{10.2}I_{o}}{10^{5.84}I_{o}} \\=22909$

Hence

The number of people at game are approximately 22909

3 0

3 years ago

Read 2 more answers