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2 years ago

It takes 56.5 kilojoules of energy to raise the temperature of 150 milliliters of water from 5°C to 95°C. If you

Physics

1 answer:

Debora [2.8K]2 years ago

7 0

You'd get an extra 40/60 of the energy, or 2/3. Multiply 5/3 by the required energy to get the actual consumption.

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Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

stellarik [79]

Answer:

0.243 m/s

Explanation:

From law of conservation of motion,

mu+m'u' = V(m+m')................. Equation 1

Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.

make V the subject of the equation

V = (mu+m'u')/(m+m')................. Equation 2

Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s

Substitute into equation 2

V = (260000×0.32+52500×(-0.14))/(260000+52500)

V = (83200-7350)/312500

V = 75850/312500

V = 0.243 m/s

8 0

2 years ago

In transduction, the cochlea is part of this structure:

anyanavicka [17]

Answer:

wats up

Explanation:

8 0

2 years ago

HELP... 3.00 amu = _____ Mev. 3.22 x 10-3 2.79 x 103 3.10 x 102

DanielleElmas [232]

The correct answer is:
$2.79 \cdot 10^3 MeV$
Let's see why.

1 amu corresponds to the mass of the proton, which is:
$m_p = 1.66 \cdot 10^{-27} kg$
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
$E=mc^2 = (1.66 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2 = 1.49 \cdot 10^{-10} J$
Now we can convert it into electronvolts:
$E= \frac{1.49 \cdot 10^{-10}kg}{1.6 \cdot 10^{-19} J/eV} =9.34 \cdot 10^9 eV = 934 MeV$

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
$3 amu = 3 \cdot 934 MeV \sim 2790 MeV = 2.79 \cdot 10^3 MeV$

5 0

2 years ago

Read 2 more answers

When something is _______ it is not moving

Solnce55 [7]

Stationary is the answer

3 0

2 years ago

Read 2 more answers

A 28.0 kg child plays on a swing having support ropes that are 2.30 m long. A friend pulls her back until the ropes are 45.0 ∘ f

Dimas [21]

Answer

A)184.9J

B)=3.63m/s

C) Zero

Explanation:

A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

U=Mgh

Where m=28kg

g= 9.8m/s

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Her Potential Energy will now

= 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

where

m = 28.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

V=√2k/m

V=√(2×184.9/28

=3.63m/s

C)we can find work done by the tension in the rope is given using expresion below

W= Tdcosx

where W= work done

T is the tension

d = displacement of the child

x= angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.

6 0

2 years ago