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Vinil7 [7]
3 years ago
10

Opening a door is an example of force because...

Physics
1 answer:
Andrei [34K]3 years ago
8 0

Answer: The force moves the swing in a particular direction and the harder that you push the further the swing will go.

Explanation: example for when u go to open a door u must pull so it will swing open, plus u need greater strength to open that’s why it’s an example of force

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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
3 years ago
Will mark brainliest if its correct pls help
Nesterboy [21]

Answer:

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Explanation:

4 0
3 years ago
How does climate change?
Korvikt [17]

Answer:

Con el cielo del agua

Explanation:

espero que te ayude

5 0
2 years ago
Read 2 more answers
Consider a system to be two train cars traveling toward each other. What is the total momentum of the system before the train ca
Brut [27]

Let say the two train cars are of masses m_1 and m_2

now if the speed of two cars are v_1 and v_2

then we can say that the momentum of two cars before they collide is given by

P = m_1v_1 - m_2v_2

here two cars are moving in opposite direction so we can say that the net momentum is subtraction of two cars momentum.

Now since in these two car motion there is no external force on them while they collide

So the momentum of two cars are always conserved.

hence we can say that the final momentum of two cars will be same after collision as it is before collision

P = m_1v_1 - m_2v_2

5 0
3 years ago
Read 2 more answers
An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a
Liono4ka [1.6K]

Answer:

16.25^{\circ}

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Horizontal range is given by

R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}

The angle at which the arrow is to be released is 16.25^{\circ}.

4 0
3 years ago
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