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adelina 88 [10]
3 years ago
10

Which of the following remains constant during the motion of a projectile fired from a planet?

Physics
2 answers:
trapecia [35]3 years ago
8 0

Answer:

C. Horizontal component of velocity

Explanation:

Object in motion stays in motion,

nothing works against its motion in the horizontal direction, unlike in the vertical direction, gravity pulls object down.

romanna [79]3 years ago
6 0
The real answer is b kinetic energy .
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Circuit A in a house has a voltage of 208 V and is limited by a 40.0-A circuit breaker. Circuit B is at 120.0 V and has a 20.0-A
Setler79 [48]

Given Information:  

Voltage of circuit A = Va = 208 Volts

Current of circuit A = Ia = 40 Amps

Voltage of circuit B = Vb = 120 Volts

Current of circuit B = Ib = 20 Amps

Required Information:  

Ratio of power = Pa/Pb = ?

Answer:  

Ratio of power = Pa/Pb = 52/15

Explanation:  

Power can be calculated using Ohm's law

P = VI

Where V is the voltage and I is the current flowing in the circuit.

The power delivered by circuit A is

Pa = Va*Ia

Pa = 208*40

Pa = 8320 Watts

The power delivered by circuit B is

Pb = Vb*Ib

Pb = 120*20

Pb = 2400 Watts

Therefore, the ratio of the maximum power delivered by circuit A to that delivered by circuit B is

Pa/Pb = 8320/2400

Pa/Pb = 52/15

4 0
3 years ago
At what speed does a falling hailstone travel? Does the speed depend on the distance that the hailstone falls?
Anton [14]

Answer:The speed if hailstone dependly largely on its size. A hailstone with a diameter of 0.39 inches,falls wit a speed of 20mph while a hailstone with 3.1 inches in diameter falls at a speed of 110mph.

No speed does not depend on the distance that the hailstone falls.

Explanation: There are other factors that affect the speed of the falling hailstone apart from its size.They are:

1. Friction between the air and the hailstone

2. Wind condition( windy or moist air)

3. The rate at which it melts falling.

7 0
3 years ago
Which of the following creates an adhesive force that prevents separation of the parietal and visceral pleurae during ventilatio
Diano4ka-milaya [45]

Answer:

Negative intrapleural pressure is the correct answer

Explanation:

Intrapleural pressure is more subatmospheric in the uppermost part of the thorax than in the lowermost parts in the standing horse.

Air moves from a region of higher pressure to one of lower pressure.  Therefore, for air to be moved into or out of the lungs, a pressure difference between the atmosphere and the alveoli must be established. If there is no pressure difference, no airflow will occur.

Under normal circumstances, inspiration is accomplished by causing alveolar pressure to fall below atmospheric pressure. When the mechanics of breathing are being discussed, atmospheric pressure is conventionally referred to as 0 cm H2O, so lowering alveolar pressure below atmospheric pressure is known as negative-pressure breathing.

3 0
3 years ago
Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to
Debora [2.8K]

Here we want to study how the linear charge density changes as we change the measures of our body.

We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.

<em>I will take two assumptions:</em>

The charge is homogeneous, so the density is constant all along the wire.

As we work with a linear charge density we work in one dimension, so the wire "has no radius"

Originally, the wire has a charge Q and a length L.

The linear charge density will be given by:

λ = Q/L

Now the length of the wire is stretched to 10 times the original length, so we have:

L' = 10*L

We want to find the value of Q' such that λ' (the <u>linear density of the stretched wire</u>) is still equal to λ.

Then we will have:

λ' = Q'/L' = Q'/(10*L) = λ = Q/L

Q'/(10*L) = Q/L

Q'/10 = Q

Q' = 10*Q

So the new <u>charge must be 10 times the original charge</u>, this means that we need to add 9*Q of charge to keep the linear charge density unchanged.

If you want to learn more, you can read:

brainly.com/question/14514975

6 0
3 years ago
If a scuba diver fills his lungs to full capacity of 5.5 L when 10 m below the surface, to what volume would his lungs expand if
umka21 [38]

Answer:

The volume at the surface is 10.97 L.

Explanation:

Given that,

Volume = 5.5 L

Height = 10 m

Density of sea water= 1025 kg/m³

We need to calculate the pressure at that point

Using formula of pressure

P'=P+\rho gh

Put the value into the formula

P'=1.01\times10^{5}+1025\times9.8\times10

P'=201450\ Pa

We need to calculate the volume at the surface

Using equation of ideal gas

PV= RT

So, for both condition

PV=P'V'

Put the value into the formula

V=\dfrac{201450\times5.5}{1.01\times10^{5}}

V=10.97\ L

Hence, The volume at the surface is 10.97 L.

3 0
3 years ago
Read 2 more answers
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