Which will increase the energy of motion of water molecules?

I need help plzzz!!!!!

Masteriza [31]

Answer:

I truly don't know

Explanation:

5 0

2 years ago

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height

butalik [34]

Answer:

a) I = -2257.6 Kg*m/s

b) F = -451,520N

Explanation:

part a.

we know that:

I = $P_f-P_i$

where I is the impulse, $P_f$ the final momentum and $P_i$ the initial momentum.

so:

I = $MV_f-MV_i$

where M is the mass, $V_f$ the final velocity and $V_i$ the initial velocity.

Therefore, we have to find the initial velocity or the velocity of the passenger just before the collition.

now, we will use the law of the conservation of energy:

$E_i=E_f$

so:

mgh = $\frac{1}{2}MV_i^2$

where g is the gravity and h the altitude. So, replacing values, we get:

(85kg)(9.8m/s^2)(36m)= $\frac{1}{2}(85kg)V_i^2$

solving for $V_i$ :

$V_i = 26.56m/s$

Then, replacing in the initial equation:

I = $MV_f-MV_i$

I = $(85kg)(0m/s)-(85kg)(26.56m/s)$

I = -2257.6 Kg*m/s

Then, the impulse is -2257.6 Kg*m/s, it is negative because it is upwards.

part b.

we know that:

Ft = I

where F is the average force, t is the time and I is the impulse. So, replacing values, we get:

F(0,005s) = -2257.6 Kg*m/s

solving for F:

F = -451520N

Finally, the force is -451,520N, it is negative because it is upwards.

3 0

2 years ago

A uniformly charged conducting sphere of 1.3 m diameter has a surface charge density of 8.1 µc/m2. (a) find the net charge on th

8_murik_8 [283]

<span>The surface charge density = q/A So q = surface charge density x Area The surface area of a sphere of radius R is 4*Pi*R^2. R = d/2 where d is diameter. This leaves us with 1.3/2 = 0.65. Area = 4 * pie * (0.65)^2 = 5.30998. So the net charge q = 8.1 * 10^(-6) * 5.30998 = 42.47998 * 10^(-6) The Total electric flux = Q/e_0 where , 8.854 Ă— 10â’12, e_0 is permitivity of free space. So Flux = 42.47998 * 10^(-6) / 8.854 * 10(â’12) = 4.833 * 10^(-6 - (-12)) = 4.833 * 10^(6)</span>

8 0

2 years ago

Elements that typically give up electrons

Naddika [18.5K]

Same answer as the first one above

8 0

2 years ago

The potential-energy function u(x) is zero in the interval 0≤x≤l and has the constant value u0 everywhere outside this interval.

VMariaS [17]

Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):
K = ½mv²
v = √(2K/m)
λ = h/(mv)
= h/(m√(2K/m))
= h/√(2Km)

So λ is proportional to 1/√K.
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:
K = 6U₀

Outer the potential well the potential energy is U₀, so
K = 5U₀
(because kinetic and potential energies add up to 6U₀)

Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is:
1/√(5U₀) : 1/√(6U₀)
= √6 : √5

5 0

2 years ago