Who is wwe universal champion

How early do you think i should start saving and searching for scholarships and state aid

brilliants [131]

Never too early to start searching. Do some research about student savings versus parent savings though. If a student has savings, they will make you use it to pay for college, while the same amount of savings in the parents name may be exempt. Check it out.

8 0

3 years ago

Gas

Brut [27]

A: what is it called when a solid jumps straight to a gas?
B: what is it called when a liquid becomes a gas?
C: what is it called when a solid changes into a liquid?

Im sure you can at least figure out the last one! (hint: this happens when ice becomes water)

3 0

3 years ago

Read 2 more answers

[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r

blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity $V_{f}$ for an object traveling distance h taking the initial vertical component of velocity as $V_{i}$ the kinematics equation is written as

$V_{f}^{2}=V_{i}^{2}+2ah$ where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

$V_{f}^{2}=V_{i}^{2}+2gh$

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

$V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}$ = 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

$V_{f}^{2}=V_{i}^{2}+2gh$

Since we have final velocity of 25 m/s

$V_{i}^{2}=2gh-V_{f}^{2}$

$V_{i}=\sqrt{(V_{f}^{2}-2gh)}$

$V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}$ = 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0

3 years ago

The potential energy of two atoms in a diatomic molecule is approximated by U(r)=(a/r12)−(b/r6), where r is the spacing between

Dafna1 [17]

Answer:

A) F(r) = [12a/(r^(13))] - [6b/(r^(7))]

ii) Graphs are attached

B) Equilibrium Distance = (2a/b)^(1/6)

C) Minimum Energy = b²/4a

D) a = 6.67 x 10^(-138) Jm^(12)

b = 6.41 x 10^(-78) Jm^(6)

Explanation:

I've attached the explanation of A-C alongside the graphs

D) i) From the question, we are to make r equal to the derivative of "r" we got when F(r) = 0 which was r = (2a/b)^(1/6)

Thus, since we are given equilibrium distance as: 1.13 x 10^(-10), hence;

(2a/b)^(1/6) = 1.13 x 10^(-10)m

So, (2a/b)= [1.13 x 10^(-10)]^(6)

a = (b/2)[1.13 x 10^(-10)]^(6)

From earlier, we saw that b²/4a = U(r)

Thus since U(r) = 1.54 x 10^(-18) from the question, b²/4a = 1.54 x 10^(-18)

Putting a = (b/2)[1.13 x 10^(-10)]^(6);

We have;

(b²) / [(4b/2)[1.13 x 10^(-10)]^(6)]] = 1.54 x 10^(-18)

b/2 = [1.13 x 10^(-10)]^(6)] x 1.54 x 10^(-18)

So, b = 6.41 x 10^(-78) Jm^(6)

ii) Putting (6.41 x 10^(-78))² for b in;

a = (b/2)[1.13 x 10^(-10)]^(6)

We have, a = (6.41 x 10^(-78))²/ ( 4 x 1.54 x 10^(-18)

So a = 6.67 x 10^(-138) Jm^(12)

6 0

3 years ago

A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water 2.00s later.

Natali5045456 [20]

Answer:

19.62 ms

Explanation:

t = Time taken = 2 s

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive)

Equation of motion

$v=u+at\\\Rightarrow v=0+9.81\times 2\\\Rightarrow v=19.62\ m/s$

The speed of the pebble when it hit the water is 19.62 ms

3 0

3 years ago

Who is wwe universal champion​

Who is wwe universal champion