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3 years ago

*Please Help!

1 answer:

8 0

We calculate by using the equation as follows:

Free Energy = DeltaG

The formula is DeltaG = Enthalpy - (Temperature)*(Entropy) 

DeltaG = 1.25 * 10^5 J - (290)(300) 

DeltaG or Free Energy = 38000 J/mol or 38kJ/mol. 

This means that this system is not spontaneous at 290 K

Hope this answers the question. Have a nice day.

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A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_{max}= \dfrac{v^2 sin^2(\theta)}{g}$

now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

$\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0

3 years ago

2. In the pendulum lab, the period (swing time) was O a. the independent variable O b.graphed on the vertical-axis O c.graphed o

Nadya [2.5K]

B is appropriate answer

7 0

3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

4 years ago

After a laser bean passes through two thin parallel slits, thefirst completely dark fringes occur at ± 15.00with the original di

PSYCHO15rus [73]

Answer:

143 °

Explanation:

a ) If d be the distance between slits , λ be wavelength of light used and at angle θ nth dark fringe is formed then

d sinθ = ( 2n+1) λ/2

for first dark fringe

d sinθ = λ/2

d /λ = 1/ 2 sinθ

1 / 2 sin15

= 1.93

b )

For intensity of fringe at angle θ, the relation is

I = I₀ cos²θ

I / I₀ = cos²θ/2

Given I / I₀ =0. 1

0.1 = cos²θ/2

θ/2 = 71.5

θ = 143 °

4 0

3 years ago

What is the value of work done on an object when a 70-newton force moves it 9.0 meters in the same direction as the force?

marta [7]

630 watts - 9 x 70 = 630

4 0

4 years ago

Read 2 more answers