0.6% of 36 = ? Please answer today This is due tomorrow!,

Find the minimum uncertainty in the speed of a uranium nucleus having mass 4e-25 kg if we know the position of the uranium nucle

Sholpan [36]

Answer:

13180 m/s

Explanation:

Mass is a measure of the amount of matter in an object. Mass is usually measured in grams (g) or kilograms (kg)

The nucleus is a membrane-bound organelle that contains genetic material (DNA) of eukaryotic organisms. As such, it serves to maintain the integrity of the cell by facilitating transcription and replication processes.

Having understood those terms, lets calculate the minimum uncertainty in the speed of a uranium nucleus having mass 4e-25 kg if we know the position of the uranium nucleus to within 1e-14 m, that is, to about its own size.

Please kindly check attachment for the step by step explaination.

5 0

3 years ago

When less saline water enters a marginal sea above a return flow of saltier water, the circulation pattern is called:?

Setler [38]

Mediterranean circulation

The Mediterranean sea is nearly disconnected from the main ocean. There is also less rain or freshwater from rivers supplementing it than evaporation. This causes the sea to have higher salinity of > 38 ppt than the open ocean's 34-36ppt.

4 0

3 years ago

A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are verti

NeX [460]

Answer:

a) T_A = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$ , b) T_B = g [m₂ ( $\frac{x}{d} -1$ ) + m₁ ( $\frac{L}{ 2d} -1$ ) ]

c) x = $d - \frac{m_1}{m_2} \ \frac{L}{2d}$ , d) m₂ = m₁ ( $\frac{ L}{2d} -1$ )

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

∑ τ = 0

T_A d - W₂ x -W₁ L/2 = 0

T_A = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$

b) the tension in string B

we write the expression of the translational equilibrium

∑ F = 0

T_A - W₂ - W₁ - T_B = 0

T_B = T_A -W₂ - W₁

T_ B = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$ - g m₂ - g m₁

T_B = g [m₂ ( $\frac{x}{d} -1$ ) + m₁ ( $\frac{L}{ 2d} -1$ ) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0

m₂ (d-x) = m₁ (0.5 L- d)

m₂ x = m₂ d - m₁ (0.5 L- d)

x = $d - \frac{m_1}{m_2} \ \frac{L}{2d}$

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}

$\frac{m_1}{m_2} \ (0.5L -d) = d$

$\frac{m_1}{m_2} = \frac{ d}{0.5L-d}$

m₂ = m₁ $\frac{0.5 L -d}{d}$

m₂ = m₁ ( $\frac{ L}{2d} -1$ )

5 0

3 years ago

In 1800, Herschel wondered if the different colors of the rainbow were equal in brightness and heat. He used a prism to separate

Papessa [141]

Answer:

D.

Explanation:

im guessing.

5 0

2 years ago

Help me please I need to turn it in before midnight...

ale4655 [162]

Answer:

The statement which explains how the total time spent in the air is affected as the projectile's angle of launch increases from 25 degrees to 50 degrees is;

C. Increasing the angle from 25° to 50° will increase the total time spent in the air

Explanation:

The equation that can be used to find the total time, T, spent in the air of a projectile is given as follows;

$T = \dfrac{2 \cdot u \cdot sin(\theta) }{g}$

Where;

T = The time of flight of the projectile = The time spent in the air

u = The initial velocity of the projectile

θ = The angle of launch of the projectile

g = The acceleration due to gravity ≈ 9.81 m/s²

Given that sin(50°) > sin(25°), when the angle of launch, θ, is increased from 25 degrees to 50 degrees, we have;

Let T₁ represent the time spent in the air when the angle of launch is 25°, and let T₂ represent the time spent in the air when the angle of launch is 50°, we have;

$T_1 = \dfrac{2 \cdot u \cdot sin(25^{\circ}) }{g}$

$T_2 = \dfrac{2 \cdot u \cdot sin(50^{\circ}) }{g}$

sin(50°) > sin(25°), therefore, we have;

$\dfrac{2 \cdot u \cdot sin(50^{\circ}) }{g} > \dfrac{2 \cdot u \cdot sin(25^{\circ}) }{g}$

Therefore;

T₂ > T₁

Therefore, increasing the angle at which the projectile is launched from 25° to 50° will increase the total time spent in the air.

7 0

2 years ago

0.6% of 36 = ? Please answer today This is due tomorrow!,​

0.6% of 36 = ? Please answer today This is due tomorrow!,