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2 years ago

1. How much heat must be absorbed by 375 grams of water to raise its

Physics

1 answer:

antiseptic1488 [7]2 years ago

5 0

Answer:

39225J

Explanation:

Given parameters:

Mass of water = 375grams of water

Change in temperature = 25°C

Specific heat capacity of water = 4.184J/g°C

Unknown:

Amount of heat absorbed = ?

Solution:

To solve this problem, we use the expression below:

H = m c Ф

H is the heat absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

Insert the parameters and solve;

H = 375 x 4.184 x (25) = 39225J

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I believe the answer is Nuclear energy

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1 year ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

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3 years ago

An eagle carrying a trout flies above a lake along a horizontal path. The eagle drops the trout from a height of 6.1 m. The fish

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The answer is 7.1 I just took the test

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3 years ago

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A block is thrown with an initial velocity of 30.0 m/s at an angle of 25.0o above the horizontal. What is the highest elevation

Serga [27]

The highest elevation reached by the ball in its trajectory is 16.4 m.

To find the answer, we need to know about the maximum height reached in a projectile.

What's the mathematical expression of the maximum height reached in a projectile motion?

The maximum height= U²× sin²(θ)/g
U= initial velocity, θ= angle of projectile with horizontal and g= acceleration due to gravity

What's the maximum height reached by a block that is thrown with an initial velocity of 30.0 m/s at an angle of 25° above the horizontal?

Here, U = 30.0 m/s and θ= 25°
Maximum height= 30²× sin²(25)/9.8

= 16.4m

Thus, we can conclude that the highest elevation reached by the ball in its trajectory is 16.4 m.

Learn more about the projectile motion here:

brainly.com/question/24216590

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2 years ago

Two bowling balls each have a mass of 6.8 kg. They are located next to each other with their centers 21.8 cm apart. What gravita

12345 [234]

Answer:

6.49 x 10^-8 N

Explanation:

formula is

F= G * ((m1 * m2)/r^2)

F = 6.67x10^-11 * ((6.8*6.8/.218)

F = 6.49 x 10^-8 Newtons

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3 years ago