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tensa zangetsu [6.8K]
3 years ago
12

What is the value of the activation energy of the uncatalyzed reaction in reverse?

Physics
1 answer:
alexgriva [62]3 years ago
5 0
It would be: Activation Energy = 300 KJ

Hope this helps!
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. The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standin
pentagon [3]

Answer:

\theta = 67.22 degree

Explanation:

Let say the ball is projected at an angle with horizontal

So here two components of the velocity of the ball is given as

v_x = 8.15 cos\theta

v_y = 8.15 sin\theta

now the displacement in x direction is given as

x = v_x t

4.57 = (8.15 cos\theta)t

in y direction it is given as

y = y_o + v_y t - \frac{1}{2}gt^2

3.05 = 2.44 + (8.15 sin\theta) t - 4.9 t^2

now from above two equations

0.61 = 4.57 tan\theta - 4.9(\frac{4.57}{8.15 cos\theta})^2

0.61 = 4.57 tan\theta - 1.54(1 + tan^2\theta)

1.54 tan^2\theta - 4.57 tan\theta + 2.15 = 0

\theta = 67.22 degree

7 0
2 years ago
The famous black planet, haunch, has a radius of 106 m, a gravitational acceleration at the surface of 4 m/s2 , and the tangenti
levacccp [35]
Gravity on the surface = 4 m/s^2
Now, the acceleration due to centripetal motion, a = v^2/R

Where,
v= 10^3 m/s, R = 10^6 m
Then,
a = (10^3)^2/(10^6) = 1 m^2/s

The net gravitational acceleration = 4-1 = 3 m/s^2

The reading on the spring scale = ma = 40*3 = 120 N
6 0
3 years ago
What is the direction of transfer of energy in the waves produced?
NemiM [27]
Transverse waves occur when a disturbance causes oscillationsperpendicular (at right angles) to the propagation (the direction of energy transfer). Longitudinal waves occur when the oscillations are parallel to the direction of propagation.
7 0
3 years ago
Read 2 more answers
Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m
aev [14]

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

charge of second body q_2=-3.5\ mu C

Particle 1 is at origin and particle 2 is at x=0.6\ m

third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

0.6+x=1.1832x

x=3.27\ m

5 0
3 years ago
A 3.0 kg object moving 8.0 m/s in the positive x direction has a one-dimensional elastic collision with an object (mass = M) ini
finlep [7]
<h2>Option 2 is the correct answer.</h2>

Explanation:

Elastic collision means kinetic energy and momentum are conserved.

Let the mass of object be m and M.

Initial velocity object 1 be u₁,  object 2 be u₂

Final velocity object 1 be v₁,  object 2 be v₂

Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s

Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M

Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J

Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M

We have

            Initial momentum = Final momentum

            24 = 3v₁ + 6M

            v₁ + 2M = 8

             v₁ = 8 - 2M

            Initial kinetic energy = Final kinetic energy

            96 = 1.5 v₁² + 18 M

            v₁² + 12 M = 64

Substituting  v₁ = 8 - 2M

           (8 - 2M)² + 12 M = 64    

           64 - 32M + 4M² + 12 M = 64    

            4M² = 20 M

               M = 5 kg

Option 2 is the correct answer.  

6 0
2 years ago
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