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pshichka [43]
2 years ago
6

How does the period of a wave change if the frequency increases by a factor of three?

Physics
1 answer:
ale4655 [162]2 years ago
3 0

Answer:

Period becomes 1/3 of the original

Explanation:

Period = 1/f      if f becomes 3f   then period becomes 1/3 of the original

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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
What is the kinetic energy of a 0.135 kg baseball thrown at 40 m/s
Ronch [10]
KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J
4 0
3 years ago
Given the diagram showing gas molecules in different containers, a reasonable inference could be made that there is a relationsh
Anvisha [2.4K]

Answer:

The correct answer should be A

Explanation:

4 0
3 years ago
Read 3 more answers
Difference between electrolytes and nonelectrolytes.
Trava [24]

Answer:

Electrolytes are salts or molecules that ionize completely in solution. As a result, electrolyte solutions readily conduct electricity. Nonelectrolytes do not dissociate into ions in solution; nonelectrolyte solutions do not, therefore, conduct electricity

Explanation:

8 0
1 year ago
Read 2 more answers
Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.
rjkz [21]

Answer:

Momentum of block B after collision =-50\ kg\ ms^{-1}

Explanation:

Given

Before collision:

Momentum of block A = p_{A1}= -100\ kg\ ms^{-1}

Momentum of block B = p_{B1}= -150\ kg\ ms^{-1}

After collision:

Momentum of block A = p_{A2}= -200\ kg\ ms^{-1}

Applying law of conservation of momentum to find momentum of block B after collision p_{B2}.

p_{A1}+p_{B1}=p_{A2}+p_{B2}

Plugging in the given values and simplifying.

-100-150=-200+p_{B2}

-250=-200+p_{B2}

Adding 200 to both sides.

200-250=-200+p_{B2}+200

-50=p_{B2}

∴ p_{B2}=-50\ kg\ ms^{-1}

Momentum of block B after collision =-50\ kg\ ms^{-1}

6 0
2 years ago
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