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Alja [10]
3 years ago
12

A sled slides down a hill with friction between the sled and hill but negligible alr resistance. Which of the following must be

correct about the resulting change in energy of the sled Earth system?
A) The sum of the kinetic energy and the gravitational potential energy changes by an amount equal to the energy dissipated by friction,
B)The gravitational potential energy decreases and the kinetic energy is constant.
C)The decrease in the gravitational potential energy is equal to the increase in kinetic energy
D)The gravitational potential energy and the kinetic energy must both decrease.
Physics
1 answer:
Andreyy893 years ago
8 0

Answer:

A) The sum of the kinetic energy and the gravitational potential energy changes by an amount equal to the energy dissipated by friction,

Explanation:

  • The kinetic energy is the energy that the object has and is defied by the work that is needed to accelerate the body.
  • The gravitational potential is a mechanism by which an equal amount of energy is being transferred per unit mass that is needed for the object to move from the specific location.
  • Hence when the sled moves down the hill with the force of gravity it has negligible resistance as an equal amount of energy is dissipated.
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Find a numerical value for ρearth, the average density of the earth in kilograms per cubic meter. use 6378km for the radius of t
Andre45 [30]

Answer:

5501 kg/m^3

Explanation:

The value of g at the Earth's surface is

g=\frac{GM}{R^2}=9.70 m/s^2

where G is the gravitational constant

M is the Earth's mass

R=6378km = 6.378 \cdot 10^6 m is the Earth's radius

Solving the formula for M, we find the value of the Earth's mass:

M=\frac{gR^2}{G}=\frac{(9.81 m/s^2)(6.378\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=5.98\cdot 10^{24}kg

The Earth's volume is (approximating the Earth to a perfect sphere)

V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6.378\cdot 10^6 m)^3=1.087\cdot 10^{21} m^3

So, the average density of the Earth is

\rho = \frac{M}{V}=\frac{5.98\cdot 10^{24} kg}{1.087\cdot 10^{21} m^3}=5501 kg/m^3

4 0
4 years ago
Read 2 more answers
Water flows over a section of Niagara Falls at a rate of 2.2 × 106 kg/s and falls 65 m. What is the power wasted by the waterfal
Irina-Kira [14]

Answer:

P = 1401.4 x 10⁶ W

Explanation:

Given that

Water flow rate m = 2.2 x 10⁶ kg/s

Height ,h= 65 m

Acceleration due to gravity ,g= 9.8 m/s²

The power P is given as

P= m g h

P=Power

m=Water flow rate

h=height

Now by putting the values in the above equation

P = 2.2 x 10⁶ x 9.8 x 65 W

P = 1401.4 x 10⁶ W

Therefore the power wasted will be 1401.4 x 10⁶ W.

7 0
3 years ago
How long would it take for someone to bike to school if the school is 10 km away and the
damaskus [11]

Answer:

It would take the cyclist 2 hours to cycle to school.

Explanation:

5km/hr for 2 hours would be 10km.

8 0
3 years ago
How does the circuit change when the wire is added? a closed circuit occurs and makes all bulbs turn off. an open circuit occurs
Vitek1552 [10]

The circuit change when a wire is added is, an open circuit occurs and makes all bulbs turn off.

<h3>What is a closed circuit?</h3>

A closed circuit is a type of circuit connection in which the wire connection is complete and current flow occurs, turning the light bulbs on in the process.

<h3>What is an open circuit?</h3>

An open circuit is a type of circuit connection in which the wire connection is incomplete and current cannot flow, turning off the light bulbs.

Thus, the circuit change when the wire is added is, an open circuit occurs and makes all bulbs turn off.

Learn more about open circuit here: brainly.com/question/20351910

#SPJ4

3 0
2 years ago
The speed of a 2 kg mass on a spring is 5 m/s as it passes through its equilibrium position. What is its frequency if the amplit
Lena [83]

Solution :

Given :

Mass of the object attached to the spring, m = 2 kg

Velocity of the object as it moves, V = 5 m/s

Amplitude of the object as it swings, A = 2.5 m

We have to find the frequency of the object.

We known,

$V_{max} = \omega A = 2 \pi f A$

Therefore, $f= \frac{V_{max}}{2 \pi A}$

$f= \frac{5}{2 \times 3.14 \times 2.5}$

f   = 0.32 Hz

Therefore the frequency of the object is 0.32 hertz when the amplitude of the 2 kg mass is 2.5 m moving a speed of 5 m/s.

3 0
3 years ago
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