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Bumek [7]
3 years ago
11

Which therapy is associated with light, but not sound?

Physics
2 answers:
o-na [289]3 years ago
8 0

Answer:

Option D....ablating tumor

dimulka [17.4K]3 years ago
8 0

Answer:

Its's not A

Explanation:

Because i got it wrong :'(

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A particle moves along a straight path through displacement while force acts on it. (Other forces also act on the particle.) Wha
allsm [11]

a) c = 1.85

b) c = 0.8

c) c = 2.33

Explanation:

a)

The displacement of the particle is given by

d=2.2i+cj

While the force applied on the particle is

F=3.2i-3.8 j

So we have a problem in 2-dimensions.

The work done on the particle is given by the scalar product between force and displacement:

W=F\cdot d (1)

Here the work done on the particle is zero, so

W = 0

Therefore from eq(1) we find:

0=(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=7.04\\c=\frac{7.04}{3.8}=1.85

b)

In this problem, the work done on the particle is

W=4.0 J

The force and displacement are still

d=2.2i+cj (displacement)

F=3.2i-3.8 j (force)

Therefore, by calculting the scalar product between force and displacement and equating it to the work done (4.0 J), we find:

W=F\cdot d

4.0 =(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=3.04\\c=\frac{3.04}{3.8}=0.8

c)

In this problem instead, the work done on the particle is negative:

W=-1.8 J

As before, the force and displacement are

d=2.2i+cj (displacement)

F=3.2i-3.8 j (force)

And so again, we calculate the scalar product between  force and displacement and we equate it to the work done on the particle, -1.8 J.

Doing so, we find:

W=F\cdot d

-1.8=(3.2i-3.8j)\cdot (2.2i+c)=7.04-3.8c\\3.8c=8.84\\c=\frac{8.84}{3.8}=2.33

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4 years ago
What is the volume of an object that has Mass=100g and Density= 745g/mL.
lana66690 [7]

Answer:0.13ml

Explanation:

6 0
3 years ago
Plz i need help for the 5 problems. plz show the work!!!
Artemon [7]

Answer:

1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

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_______________have a negative charge and are located on the outside of the nucleus.
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May 5th is Adachi day change your profile picture

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When light does not pass through or bounce off an object it is what
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 it it called reflection
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