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zubka84 [21]
3 years ago
6

Can I put an “I” for an argumentative essay?

Physics
1 answer:
masha68 [24]3 years ago
4 0

Answer:

There is no place for first person in formal academic argumentative essay. However, you can use first person only if it has been asked to provide your personal opinion or reflection about the topic.

Explanation:

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The ramp on the back of a moving van is in example of what type <br> simple machine
abruzzese [7]
Inclined Plane.Because an inclined plane is a flat,sloped surface. And a ramp is a perfect example of an inclined plane.
4 0
3 years ago
Read 2 more answers
A positive test charge q is released from rest at distance r away from a charge of Q and a distance 2r away from a charge of 2Q.
Luba_88 [7]

Answer: Option (b) is the correct answer.

Explanation:

It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.

Then test charge to the right immediately after being released.

Therefore, the net force will be as follows.

            F = \frac{kqQ}{r^{2}} - kq\frac{(2Q)}{(2r)^{2}}

               = \frac{4KqQ - 2KqQ}{4r^{2}}

               = \frac{KqQ}{2r^{2}}

           F = \frac{KqQ}{2r^{2}} > 0

Thus, we can conclude that the test charge move to the right immediately after being released.

7 0
3 years ago
A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed
Liono4ka [1.6K]

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

F = \dfrac{mv^2}{r}

m is the mass, v is the velocity and r is the radius.

It follows that F \propto v^2, provided m and r are constant.

When v is doubled, the new force, F_1, is

F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F

Hence, the tension in the string is quadrupled.

8 0
3 years ago
Tridil is infusing at 15 ml/hr on an infusion pump. The drug is mixed 50 mg in 500 ml D5W. How many MCG/minute is the patient re
olga nikolaevna [1]

Answer:

patient receiving drug 25 MCG/minute

Explanation:

given data

infusing = 15 ml/hr

drug = 50 mg

D5W = 500 ml

to find out

How many MCG/minute

solution

we know infusing rate is 15 ml/hr = 0.25 ml/min

so 0.25 ml drug content = 50 /500 × 0.25

0.25 ml drug content = 0.025 mg

so here

rate of drug will be 0.025 mg

rate of drug = 0.025 mg = 25 ×10^{-6} gm/min

rate of drug = 25 MCG/minute

so patient receiving drug 25 MCG/minute

8 0
2 years ago
Meher is riding a bicycle on a slope. explain the different motions taking place during this time​
gtnhenbr [62]

Answer:

Mehar cant ride down the slope

Explanation:

She does not has a bicycle

7 0
2 years ago
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