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Suatu larutan penyangga dibuat dengan mencampurkan 50mL larutan H3PO4 1 M dengan 50mL larutan NaOH 0,5 0 berapakah pH larutan It

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1 answer:

ValentinkaMS [3.4K]1 month ago

5 0

Larutan penyangga dibuat dengan menggabungkan 50mL larutan H3PO4 1 M dan 50mL larutan NaOH 0,5. Berapakah pH dari larutan tersebut?

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Bursitis can affect many joints and can be caused by a numerous sporting activities, such as cycling, tennis, and long-distance

Yuliya22 [3333]

Bursitis refers to the irritation or injury of the small fluid-filled sacs called bursae that are located around the joints. This condition can lead to pain during movement or when the affected area is pressed. Occasionally, a rash may also develop. Thank you for your inquiry; I trust this information will be beneficial.

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2 months ago

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The acceleration of segment D is m/s2. Rank segments A, B , C from least accelerations to greatest acceleration. Least

Ostrovityanka [3204]

Answer:

D, C, B, A

Explanation:

The derivative from a velocity-time graph provides the acceleration value.

Segment A

$\frac{dy}{dx} = \frac{15m/s}{1s} = 15m/s^2$

Segment B

$\frac{dy}{dx} = \frac{5m/s}{1s} = 5m/s^2$

Segment C

$\frac{dy}{dx} = \frac{0m/s}{2s} = 0m/s^2$

Segment D

$\frac{dy}{dx} = \frac{-20m/s}{1s} = -20m/s^2$

Sorted from the lowest to the highest acceleration:

D, C, B, A

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2 months ago

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Which formula is used to find fluctuation of the shape of body

Softa [3030]

Response:

varn=n1+1ehvkT–1

Explanation:

This formula comes from Einstein's theory.

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1 month ago

Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d

Maru [3345]

Answer:

4.05 m/s

Explanation:

We will express the varying velocities as vectors.

Newton moves northward at 3.90 m/s from Daniel's stationary position.

V_n = 3.9 j

Assuming Pauli runs relative to Daniel at velocity X.

The relative velocity of Newton as seen by Pauli will be

3.9 j - X

Given that

the relative velocity of Newton with respect to moving Pauli = 1.1 i (1.1 towards the east).

Thus,

3.9 j - X = 1.1 i

X = -1.1 i + 3.9 j.

Magnitude of X

X² = 1.1² + 3.9²

X = 4.05 m/s

Therefore, Pauli runs relative to Daniel at 4.05 m/s.

The direction will be west of north at an angle θ,

Tan θ = 1.1 / 3.9

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3 months ago

In an electricity experiment, a 1.0g plastic ball is suspended on a60-cm-long string and given an electric charge. A charged rod

Ostrovityanka [3204]

a) The magnitude of the electric force: $3.57\cdot 10^{-3} N$

b) String tension: 0.010 N

Clarification:

a)

When a charged rod approaches the ball, it remains suspended at an incline. This allows us to analyze the forces acting in two perpendicular orientations:

- Horizontally, the electric force $F_E$ acts in one direction, and the tension component in the string opposes it, $T sin \theta$ denoted by T, which is the angle relative to the vertical $\theta=20^{\circ}$

- Vertically, the weight of the ball $mg$ acts downward (with $m=1.0 g = 0.001 kg$ as the ball's mass and $g=9.8 m/s^2$ as gravitational acceleration), while the tension component acts upward $T cos \theta$

Consequently, since the ball is in equilibrium, the following two equations can be established:

$T sin \theta =F_E\\Tcos \theta = mg$

Dividing the two equations gives

$tan \theta=\frac{F_E}{mg}$

and solving for the electric force yields

$F_E=mg tan \theta=(0.001)(9.8)tan 20^{\circ}=3.57\cdot 10^{-3} N$

b)

To derive the tension in the string, one can use either of the above equations; for example, the equation for the horizontal component,

$T sin \theta =F_E$

Where

is the electric force

is the angle relative to vertical $F_E=3.57\cdot 10^{-3} N$ The tension in the string can be calculated as follows:

$\theta=20^{\circ}$

Discover more about electric force:

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1 month ago