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3 years ago

Answer quick!!! (Edge 2021)

Physics

2 answers:

mihalych1998 [28]3 years ago

7 0

Answer:

I believe the answer is B.

pantera1 [17]3 years ago

7 0

Answer:
B) Superficial External bean Radiation Therapy.

Explanation:

Surgery, radiation, chemotherapy and hormone therapy can all be used to relieve signs and symptoms. Medications may relieve symptoms such as pain and shortness of breath. Palliative treatment can be used at the same time as other treatments intended to cure your cancer.

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How does the end point differ from the equivalence point of a titration?

Gwar [14]

Answer:

Equivalence point and end point are terminologies in pH titrations and they are not the same. 

Explanation:

In a titration the substance added slowly to a solution usually through a pippette is called titrante and the solution to which it is added is called titrand. In acid-base titrations acid is added to base or base is added to acid.the strengths of the acid and base titrated determines the nature of the final solution.

At equivalence point the number of moles of the acid will be equal to the number of moles of the base as given in the equation. The nature of the final solution determines the pH at equivalence point. 

A pH less than 7 will be the result if the resultant is acidic and if it is basic the pH will be greater than 7. In a strong base-strong acid and weak base-weak acid titration the pH at the equivalence point will be 7 indicating neutral nature of the solution. 

3 0

3 years ago

The n = 8 to n = 4 transition in the bohr hydrogen atom occurs in the __________ region of the electromagnetic spectrum.

tiny-mole [99]

INFRARED REGION ..............

4 0

3 years ago

Read 2 more answers

Greeley [361]

Answer:

this is impossible for me

Explanation:

7 0

3 years ago

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$