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2 years ago

How much energy does a 50 kg rock have if it is sitting on the edge of a 15 m cliff?

Physics

1 answer:

noname [10]2 years ago

3 0

Answer:

7350 J

Explanation:

The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

$U=mgh$

where

m is the mass of the rock

g is the gravitational acceleration

h is the height of the cliff

In this problem, we have

m = 50 kg

g = 9.8 m/s^2

h = 15 m

Substituting numbers into the formula, we find:

$U=(50 kg)(9.8 m/s^2)(15 m)=7350 J$

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Reginald slipped and broke his leg in his kitchen when he ran inside to grab a cookie. His mother had just mopped the floor. Wha

padilas [110]

Answer:

A. <u>The water decreases the friction between the floor and the feet </u>

Explanation: Think about it like this, when your in the shower and water is on your skin, you can scrub it fluidly, but when the water dries in your towel, there is more friction when your rub your skin, this is because the molecules in water aren't as compact as solids, so anything acting against it, is most likely to disperse from it.

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2 years ago

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor flows at a rate of 0.3 kg/s inside a p

grigory [225]

Answer: $h=160.84 W/m^2-K$

Explanation:

Given

mass flow rate=0.3 kg/s

diameter of pipe=5 cm

length of pipe=10 m

Inside temperature=22

Pipe surface =100

Temperature drop=30

specific heat of vapor(c)=2190 J/kg.k

heat supplied $Q=mc\Delta T=0.3\times 2190\times (30)$

Heat due to convection =hA(100-30)

$A=\pi d\cdot L$

$A=\pi 0.05\times 10=1.571 m^2$

$Q_{convection}=h\times 1.571\times (100-22)=122.538 h$

$Q=Q_{convection}$

19,710=122.538 h

$h=160.84 W/m^2-K$

5 0

2 years ago

Consider heat transfer between two identical hot solid bodies and their environments. The first solid is dropped in a large cont

sergeinik [125]

<h2>For Second Solid Lumped System is Applicabe</h2>

Explanation:

Considering heat transfer between two identical hot solid bodies and their environments -

If the first solid is dropped in a large container filled with water, while the second one is allowed to cool naturally in the air than for second solid, the lumped system analysis more likely to be applicable

The reason is that a lumped system analysis is more likely to be applicable in the air than in water as the convection heat transfer coefficient so that the Biot number is less than or equal to 0.1 that is much smaller in air

Biot number = the ratio of conduction resistance within the body to convection resistance at the surface of the body

∴ For a lumped system analysis Biot number should be less than 0.1

4 0

2 years ago

block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th

Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

8 0

2 years ago

A particle moves along a straight line with a velocity in meters per second given by v = 12 - 3t + 8t, where t is in seconds. Wh

elena-14-01-66 [18.8K]

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s

4 0

2 years ago