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mart [117]
3 years ago
10

Why did the Framers establish the Legislative Branch in Article I of the Constitution?

Physics
1 answer:
shepuryov [24]3 years ago
7 0

Answer:

I think it will be A.

Explanation:

A.

they felt people would more easily accept a strong military if it was controlled by the legislative branch.

You might be interested in
Which particle is a negatively charged ion?
wel

Answer:

option D is correct

Explanation:

cl has 17 electrons if it has 18 it becomes cl-

6 0
3 years ago
wo lacrosse players collide in midair. Jeremy has a mass of 120 kg and is moving at a speed of 3 m/s. Hans has a mass of 140 kg
Julli [10]

2.71 m/s fast Hans is moving after the collision.

<u>Explanation</u>:

Given that,

Mass of Jeremy is 120 kg (M_J)

Speed of Jeremy is 3 m/s (V_J)

Speed of Jeremy after collision is (V_{JA}) -2.5 m/s

Mass of Hans is 140 kg (M_H)

Speed of Hans is -2 m/s (V_H)

Speed of Hans after collision is (V_{HA})

Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is  

= =\mathrm{M}_{1} \times \mathrm{V}_{\mathrm{J}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{H}}

Substitute the given values,

= 120 × 3 + 140 × (-2)

= 360 + (-280)

= 80 kg m/s

Linear momentum after the collision of Jeremy and Hans is  

= =\mathrm{M}_{\mathrm{J}} \times \mathrm{V}_{\mathrm{JA}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{HA}}

= 120 × (-2.5) + 140 × V_{HA}

= -300 + 140 × V_{HA}

We know that conservation of liner momentum,

Linear momentum before the collision = Linear momentum after the collision

80 = -300 + 140 × V_{HA}

80 + 300 = 140 × V_{HA}

380 = 140 × V_{HA}

380/140= V_{HA}

V_{HA} = 2.71 m/s

2.71 m/s fast Hans is moving after the collision.

4 0
3 years ago
2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th
Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = 5.33 × 10^{-5} T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ B = \frac{\mu_{0}i }{2\pi R}

Where B = magnetic field due to long wires, \mu_{0} = 4\pi \times10^{-7}, R = perpendicular distance from wire to given point

From any one wire R_{1}  = 5 cm, R_{2}  = 3 cm

so we write,

∴ B = B_{1} + B_{2}

 B = \frac{\mu_{0} i}{2\pi R_{1} } +  \frac{\mu_{0} i}{2\pi R_{2} }

 B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]

 B = 5.33\times10^{-5}  T

Therefore, the magnitude of magnetic field at given point = 5.33\times10^{-5} T

3 0
3 years ago
What is power, and what is its relationship to voltage and amperage? (4 points)
tigry1 [53]

Answer:

The relationship between amperage, voltage, and power is that power equals the amperage quantity times the amount of voltage.

Explanation: brainliest pls

5 0
3 years ago
Read 2 more answers
How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

Read more about the flows rate

brainly.com/question/27880305

#SPJ1

5 0
2 years ago
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