Why can the brain be fossilized

A bottle of water with mass 0.9 kg is left out in the sun, the radiation from the sun warms up the water bottle. If the water bo

natita [175]

Answer:

Final temperature, T2 = 314.9 Kelvin

Explanation:

Given the following data:

Mass = 0.9kg

Initial temperature, T1 = 10°C to Kelvin = 10 + 273 = 283K

Quantity of heat = 120,000 J

Specific heat capacity = 4182 j/kgK

To find the final temperature;

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

Making dt the subject of formula, we have;

$dt = \frac {Q}{mc}$

Substituting into the equation, we have;

$dt = \frac {120000}{0.9*4182}$

$dt = \frac {120000}{3763.8}$

dt = 31.9K

Now, the final temperature T2 is;

But, dt = T2 - T1

T2 = dt + T1

T2 = 31.9 + 283

T2 = 314.9 Kelvin

8 0

3 years ago

A particle with a mass of 6.64 × 10–27 kg and a charge of +3.20 × 10–19 C is accelerated from rest through a potential differenc

blondinia [14]

Answer:

Explanation:

Given that,

Mass m = 6.64×10^-27kg

Charge q = 3.2×10^-19C

Potential difference V =2.45×10^6V

Magnetic field B =1.6T

The force in a magnetic field is given as Force = q•(V×B)

Since V and B are perpendicular i.e 90°

Force =q•V•BSin90

F=q•V•B

So we need to find the velocity

Then, K•E is equal to work done by charge I.e K•E=U

K•E =½mV²

K•E =½ ×6.64×10^-27 V²

K•E = 3.32×10^-27 V²

U = q•V

U = 3.2×10^-19 × 2.45×10^6

U =7.84×10^-13

Then, K•E = U

3.32×10^-27V² = 7.84×10^-13

V² = 7.84×10^-13 / 3.32×10^-27

V² = 2.36×10^14

V=√2.36×10^14

V = 1.537×10^7 m/s

So, applying this to force in magnetic field

F=q•V•B

F= 3.2×10^-19 × 1.537×10^7 ×1.6

F = 7.87×10^-12 N

6 0

3 years ago

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A 9.5 V battery supplies a 3.5 mA current to a circuit for 7.0 h. Part A How much charge has been transferred from the negative

pychu [463]

Answer:

88.2 C

Explanation:

The current can be defined as the rate of flow of charge in a conductor.

The relation between charge current and time is given as

I = Q/T

I = current, Q= charge and T = time

that is ampere = coulomb / second

The amount of charge passed is from the negative to the positive terminal

shall be given by:

Q = I * t = 3.5mA * 7h * 3600s/h = 88.2 C

Note: take care of the units.

7 0

3 years ago

Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th

vekshin1

Answer:

0.00417 kW/K or 4.17 W/K

Second law is satisfied.

Explanation:

Parameters given:

Rate of heat transfer, Q = 2kW

Temperature of hot reservoir, Th = 800K

Temperature of cold reservoir, Tc = 300K

The rate of entropy change is given as:

ΔS = Q * [(1/Tc) - (1/Th)]

ΔS = 2 * (1/300 - 1/800)

ΔS = 2 * 0.002085

ΔS = 0.00417 kW/K or 4.17 W/K

Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.

6 0

3 years ago

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.2m/s and the hatchback ca

love history [14]

Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. $s_{o}$ = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.

Truck:

$v_{i}$ = 33.2 m/s, a = 0 (since the velocity is constant), $s_{o}$ = 0

Using $s =s_{o}+v_{i}t+1/2at^{2}$

s = 33.2t .......... eq (1)

Hatchback:

$a=5m/s^{2}$ , $v_{i}$ = 0 m/s (since initial velocity is zero), $s_{o}$ = 0

Using $s =s_{o}+v_{i}t+1/2at^{2}$

putting in the data we will get

$s=(1/2)(5)t^{2}$

now putting 's' value from eq (1)

$2.5t^{2}-33.2t=0$

which will give,

t = 13.28 s

so both vehicles will meet up gain after 13.28 sec.

putting t = 13.28 in eq (1) will give

s = 440.896 m

So, both vehicles will meet up again at 440.896 m.

7 0

3 years ago

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